Toyah Battersby

Toyah Habeeb (also Battersby) is a fictional character from the British soap opera Coronation Street, played by Georgia Taylor. Toyah and her family were introduced to the serial by executive producer Brian Park in a bid to increase falling ratings. Actress and singer Toyah Willcox gave the producers permission to name the youngest member of the Battersby family after herself, since the character was said to be born during the height of her career. Taylor originally auditioned for the part of Zoe Tattersall, but was cast as Toyah instead. The role marked her acting debut. She made her first screen appearance during the episode broadcast on 4 July 1997.

Toyah was portrayed as cocky, naive, too trusting and occasionally needing to be "a bit more clued up" about things. Taylor called Toyah a decent person and someone who was loyal to her friends and family. Taylor initially disliked Toyah's image, particularly her make-up and clothes, which were too revealing. Toyah's first major storyline saw her being abducted by a stranger. At the time, Taylor said it was one of the hardest storylines she had to film and the Independent Television Commission believed it should have carried a warning before it was broadcast. A few months later, Toyah became the show's first female character to be involved in a storyline surrounding the subject of under-age sex. The storyline was criticised by watchdogs, but Taylor defended it, calling it "educational".

Toyah began a relationship with Spider Nugent (Martin Hancock) in 1999, which lasted until Hancock's departure in July 2000. The following year, the character was the focus of one of Coronation Street's most controversial storylines, which saw her beaten and raped by Phil Simmonds (Jack Deam). The storyline marked the first time in 40 years that the show had included a rape. The storyline was criticised by some former Coronation Street cast members, who thought the producers were using the subject of rape to boost ratings. Taylor decided to leave the show in July 2002 and her character departed on-screen the following year on 5 February 2003. Toyah was well received by critics and Taylor received the Best Young Actor accolade at the Inside Soap Awards and Best Dramatic Performance at The British Soap Awards.

Following a thirteen-year absence from Coronation Street, Taylor announced that she had reprised the role in September 2016. She returned to filming the following month and Toyah's return scenes aired on 25 December 2016. Toyah's return storyline revolved around her decision to end her marriage to Toby Chapman (Andrew Dowbiggin) as she is having an affair with Peter Barlow (Chris Gascoyne) whom she later has a relationship with and the pair becoming landlords of The Rovers Return Inn. Storylines since have included her desire to have children resulting in trying to pass Eva Price's (Catherine Tyldesley) baby off as a result of IVF treatment and her relationship and later marriage to Imran Habeeb (Charlie De Melo) which ends in Toyah causing his death. She was later arrested for his murder and found not guilty.

## Storylines

### 1997–2003

Toyah and her family move into 5 Coronation Street and soon upset the neighbours with their bad behaviour, most notably Curly Watts (Kevin Kennedy). Toyah and her stepsister, Leanne (Jane Danson), steal from the local newsagent and indulge in under-age drinking. Toyah develops a crush on Spider Nugent (Martin Hancock) and turns vegetarian to impress him. She also asks him to help save a turkey from being slaughtered for Christmas. After arguing with her mother, Janice (Vicky Entwistle), and stepfather, Les (Bruce Jones), Toyah runs away to London to find her biological father, Ronnie Clegg (Dean Williamson). However, Toyah comes across Neil Flynn (Tim Dantay), who pretends to be her father and then abducts her. Neil takes Toyah to some woods, but while he is distracted, she escapes and is found by Janice and Les.

Toyah gets a job at Roy's Rolls, the local café, and begins dating Philip "Dobber" Dobson (John Donnelly). Dobber is a bad influence on Toyah, persuading her to steal from the café and threatening to dump her if she did not have sex with him. Toyah loses her virginity to Dobber but eventually sees through him and they break up. With the help of a former teacher, Ken Barlow (William Roache), Toyah's grades at school soon improve and she even wins a writing competition. While spending the night watching a lunar eclipse together, Toyah and Spider have sex. They soon begin a relationship and stay at Emily Bishop's (Eileen Derbyshire) house for a while until they find a place of their own. Toyah gets a second job as a barmaid at The Rovers Return and start to drift apart so when Spider asks her to go to India with him, she turns him down. Spider leaves Weatherfield alone. Toyah later has a one-night stand with Andy McDonald (Nicholas Cochrane).

After a night out with Sam Kingston (Scott Wright), Toyah decides to walk home alone, but is attacked and raped. Toyah is found by Jason Grimshaw (Ryan Thomas), who takes her home. Janice initially thinks Toyah has been robbed and is shocked when Toyah tells her she was raped. Toyah gives a statement to the police and is examined at the hospital. Several male residents are suspected, particularly Peter Barlow (Chris Gascoyne), but they are ruled out by DNA tests. Toyah struggles to deal with what has happened to her and she develops agoraphobia. While being comforted at home by her friend, Phil Simmonds (Jack Deam), Toyah recognises his voice as he calls out her name and realises that he had raped her. Phil realises that she knows and stops her from escaping. Peter hears Toyah's screams and rescues her after knocking Phil unconscious. Phil is found guilty of rape and sent to prison. Toyah has a brief romance with Sam and moves into a flat with Fiz Brown (Jennie McAlpine) and Maria Sutherland (Samia Ghadie).

While on holiday in Blackpool, Toyah meets Goran Milanovic (Matt Zarb), an illegal immigrant. After a few dates, he proposes marriage, but she refuses before starting to date her college tutor, John Arnley (Paul Warriner). She is initially reluctant to introduce him to her flatmates and when she does, they are surprised to see that he is a lot older. John asks Toyah to move in with him and she agrees. When Toyah and Fiz argue about John, Fiz tells her that he and Maria had sex and Maria got pregnant. Toyah confronts Maria, who gives her the impression that John forced himself on her. Remembering her rape, Toyah attacks John in front of his class before dumping him. Toyah decides to leave Weatherfield after reuniting with Spider and they move to London.

### 2016–present

Toyah returns on Christmas Day, telling Leanne that she has left her husband Toby Chapman (Andrew Dowbiggin), as she has been having an affair. Toyah later meets Peter Barlow, her new partner, and they agree to tell Leanne that they are together. Toby visits Toyah to try and save their marriage but she refuses to listen or reconcile. Toyah then goes with Leanne to the hospital after she has a migraine. Toyah and Peter decide to keep their relationship a secret, but Peter's son, Simon Barlow (Alex Bain), sees them kiss using his drone. Peter tells Toyah that Simon knows and they ask him to keep it a secret for Leanne's sake.

Toyah argues with Leanne's half-sister, Eva Price (Catherine Tyldesley), for not passing on a message about starting a shift at the pub, but they agree to call a truce. Peter leaves cigarettes at Leanne's flat and Toyah tries to take the blame, but Simon is forced to cover for her. He later tells Eva who blackmails Toyah: stop Peter from shutting down Aidan Connor's (Shayne Ward) factory or Eva will tell Leanne about Peter and Toyah. Peter agrees to Eva's demands and they continue their affair. Toby returns in January 2017 and demands Toyah sign a paper to destroy their frozen embryos. Toyah is reluctant and after a heart to heart, they kiss. However, Toyah makes it clear that she does not love Toby and is staying with Peter, causing Toby to lash out and leave. Peter learns what has happened and decides to drop off Chloe Tipton (Jo-Anne Knowles), making Toyah jealous. They later make up and Leanne catches Peter and Toyah kissing. During a heated argument, Leanne disowns Toyah and refuses to talk to her again, despite her efforts. Leanne forgives Toyah after she helps deliver baby Oliver in a lift after it breaks down.

Peter buys The Rovers Return from Steve (Simon Gregson) and Liz McDonald (Beverley Callard). After another round of IVF fails, Toyah is told that future treatment is not recommended as problems with her uterus mean that she is very unlikely to ever get pregnant. She and Peter look into surrogacy after talking to Izzy Armstrong (Cherylee Houston). The couple are accepted as clients by a surrogacy agency and Jackie, a surrogate mother, agrees to carry their baby after meeting them. She strikes up a feud with Peter's ex-wife, Carla Connor (Alison King), upon her return to Weatherfield. Toyah is later devastated when Jackie miscarries the baby and she is shocked to discover that Eva is pregnant with her ex-fiancé Aidan's baby. Toyah supports Eva's decision to have an abortion. However, Toyah concocts a plan shortly before the termination where Eva would give birth to her baby and pretend the baby is Toyah and Peter's through Jackie. Eva gives birth to her baby girl who she Toyah and Peter name Susie as they go through with the plan of pretending Eva's daughter is theirs, with Peter none the wiser. Peter later finds out and breaks up with Toyah selling the pub. Eva leaves the street with Susie meaning that Toyah no longer has a child. She starts to move on with her life and gets a job at the medical centre and has a fling with Imran Habeeb (Charlie de Melo) unaware Leanne is also having a fling with him. When they find out, they hatch a plan and pretend they want a threesome with him but the plan backfires when he realises what they are up to. Toyah begins a relationship with Imran and they later foster children together including Kelly Neelan (Millie Gibson) and become embroiled in her murder charge for her role in Seb Franklin’s (Harry Visioni) death.

In October 2021, unbeknownst to Toyah, Imran has a one-night stand with Seb's mother Abi (Sally Carman), resulting in Abi getting pregnant and, in 2022, giving birth to their son, Alfie Franklin. Toyah finds out about Imran’s paternity to Alfie and after marrying him to gain custody they get custody of baby Alfie, which included Imran lying and stitching Abi up by hiring colleague Ben Chancellor (Jon-Paul Bell) to take pictures of her supposedly doing a drug deal for evidence in the custody battle, however this turns out to be Abi paying off an old friend Dean (Anthony Crank) to stay away from her. (Dean had caused trouble for Abi earlier in the year). In June 2022, Imran and Toyah caught on to the fact that Abi is planning to flee abroad with Baby Alfie. Toyah considers going to the police. Imran's guilty conscience took hold as he eventually persuaded Toyah against getting the police involved and he tells her the truth about him framing Abi as a drug addict. The pair were on their way home when a crash occurred with them in the car. Imran later dies of a cardiac arrest caused by the crash and Toyah survives, but ends up in hospital. It later emerges Toyah caused the crash on purpose and she is arrested and charged with careless driving.

In July 2023, Toyah is kidnapped.

## Creation and casting

The four-strong Battersby family was created and introduced to Coronation Street in a bid to increase falling ratings. News of the family's arrival came days after the series producer, Brian Park, had axed three regular characters. He stated "The Battersbys are an exciting injection of fresh blood. They follow in the great tradition of the Ogdens and Duckworths, balancing the mix of comedy and drama for which Coronation Street is justly renowned." Actress and singer Toyah Willcox gave the producers permission to name the youngest member of the family after herself, since the character was born in 1982 at the height of her career. Willcox later regretted her decision.

Actress Georgia Taylor was cast as Toyah and the role marked her acting debut. Taylor was studying for her A-levels when she originally auditioned for the part of Zoe Tattersall. She was unsuccessful, so she then auditioned for Toyah and won the role. Of her casting, she explained "When I got the call to say that I'd got the part I just sat there open-mouthed, hugging my best friend. I could barely believe it. It is all very different than being on stage – it's not quite so intense because your audience isn't right there in front of you. But it's a very different type of satisfaction, working much faster with storylines and characters developing all the time." At the time of her casting, Taylor was two years older than the character she portrayed.

## Development

### Characterisation

A month before she made her debut on-screen, Toyah was described as being Janice's (Vicky Entwistle) 14-year-old illegitimate daughter, who would become "the scourge of every Weatherfield Comprehensive teacher". During an interview with the Daily Record's Polly Graham, Taylor stated "Toyah's gobby and cocky and makes her parents' life a misery. But it's brilliant playing her, it's more fun than playing somebody nice." Graham explained that both Toyah and her stepsister, Leanne (Jane Danson) were well known to the police for under-age drinking and burglary. Taylor said that she hated her character's clothes, as they were too revealing for her. She explained "In the beginning, the clothes that Toyah wore were terrible. I was very conscious of my bum hanging out from these tiny mini-skirts. There are a lot of girls who wear those tarty skirts, but I never felt comfortable in that sort of thing." Taylor also disliked Toyah's make-up, which was initially bright pink lipstick, a lot of black eyeliner and no foundation.

Toyah's image eventually began to change and she became more aware of the way she looked when she got a boyfriend and started college. Toyah also took extra English lessons from Ken Barlow (William Roache) in a bid to better herself. Taylor later commented that her character got on her nerves due to her naivety. She also thought that Toyah had a romantic view of love, was too trusting and occasionally needed to be "a bit more clued up". The actress added that Toyah was "a very decent person", who was loyal to her friends. Digital Spy's Kris Green described Toyah as "the more mellow of the Battersbys" and thought that it was due to Les (Bruce Jones) not being her biological father. The Daily Mirror's Charlie Catchpole commented that Toyah went "from lumpy, sulky schoolgirl to confident, radiant young woman".

### Abduction

Following an argument with Janice and Les, Toyah runs away to London to find her real father, Ronnie. However, she was "led on" by Neil Flynn (Tim Dantay), a stranger pretending to be Ronnie, and was later abducted by him. Taylor said that the storyline was one of the hardest she had to film: "In the story, I come down to London to track down my real father and things go badly wrong. Toyah hitchhikes all the way and just leaves herself so wide open. She is so naive and trusting and open to being used and abused." The storyline was filmed on location in the city of London in early August. An hour long episode focusing on Janice and Les' search for Toyah was broadcast a month later.

Jones, who played Les, revealed that the runaway storyline was tough on him and he began crying when he read the scripts. He continued "The four of us in the Battersby family read our scripts together to work out how we are going to play our parts. On this occasion, all four of us were sobbing. It is not like a normal Coronation Street story, it's more like a Taggart or a Cracker." Jones thought the "scrapes" that Toyah found herself in were frightening and hoped that if the storyline could stop one teenager from running away from home, then it would be worthwhile. The storyline attracted the attention of the Independent Television Commission, who believed that the episode featuring Toyah's abduction should have carried a warning before it was broadcast.

### Under-age sex

In October 1998, Toyah became the first female Coronation Street character to be involved in a storyline surrounding the subject of under-age sex. Toyah, who was sixteen at the time, had a holiday romance with Philip "Dobber" Dobson (John Donnelly). She quickly introduced him to her parents and was prepared to do anything for him, like stealing from the local café. When Dobber asked Toyah to have sex with him, she agreed, despite knowing that she should wait. Toyah was scared that if she said no, Dobber would break up with her. Toyah was persuaded to lose her virginity to Dobber in the back of his car. The storyline was branded "controversial" and "shocking", while watchdogs stated that it sent out the wrong ideas to young schoolgirls. Taylor defended the storyline and her character, saying "We have taken a strong moral line and are not promoting teenage sex. It's quite the opposite. Toyah's actions should be a lesson for other girls not to fall into the same trap. It was an educational storyline, and I feel it was done in a responsible way."

The actress went on to explain that the storyline showed how Toyah regretted her actions and that young female viewers could learn from her mistakes. She added "Anyone watching can see Dobber is just using Toyah and I hope the message gets across to other girls not to fall into the same trap." Toyah did not use protection with Dobber, so her sister, Leanne, persuaded her to go to the doctor for the morning-after pill. Taylor praised Leanne for being Toyah's "conscience and the voice of common sense". She thought that Leanne had given her good and responsible advice on how to deal with what happened. The Daily Record's John Millar agreed with Taylor and wrote that the storyline was "building to a stark warning for the millions of teenagers tuning into the Street." Millar believed at no time did the viewers get the impression that Toyah was doing something "admirable" and proclaimed that Toyah was "the sort of girl who needs care and affection, which explains why the poor soul was coaxed into losing her virginity after a holiday romance convinced her that she was in love."

### Relationship with Spider Nugent

Five months after moving onto Coronation Street, Toyah developed a crush on her neighbour, Spider Nugent (Martin Hancock) and even turned vegetarian to try and impress him. Hancock told All About Soap's Hilary Kingsley that Spider initially saw Toyah as "a chubby schoolgirl, a pesky brat who had a crush on him." However, when she listened to his issues and helped him with his battles against the local supermarket and the council, Spider began to admire her instead. Hancock explained "She was so efficient and hard-working that she became a friend. And then, over the summer, it hit him, "Blimey! She's turned out nice!" While Spider and Toyah watched the 1999 solar eclipse together, they ended up having sex for the first time. Hancock observed that both Spider and Toyah were overcome by the excitement to the point that there was "an explosion of lust." However, in the morning, Spider was torn about what had happened. He admired Toyah, but knew that if they became a couple, it would cause a lot of problems, particularly as there was a ten-year age gap between them. Taylor explained that she was happy that her character was getting together with Spider, although their scenes together often took place in the "grottiest" of locations.

Spider decided to "dodge his responsibilities" and began staying away from Toyah, who was not happy about it as she wanted them to make a commitment. Spider eventually changed his mind and Hancock commented "The fact is that he really fancies Toyah and he figures that he may as well be hung for a sheep as a goat." The actor reckoned that Spider and Toyah brought out the best in each other. Spider and Toyah soon began a proper relationship and briefly lived together. However, when Spider announced that he wanted to settle down and buy a house with Toyah, the couple realised that they had started to drift apart. Spider thought that if they got a mortgage together, it would help save the relationship, but Toyah did not want that at all. When Spider was offered a promotion, he realised that he did not want to work for the DSS anymore and quit. He then told Toyah that he had bought them tickets to India so that they could go and "find themselves." However, Toyah insisted that she did not want to leave Weatherfield. Nicole Carmichael from Inside Soap wrote that while Spider did not want to lose Toyah, he felt that he could not stay and soon departed.

### Rape

In April 2001, Toyah was the focus of one of Coronation Street'''s most controversial storylines. After enjoying a night out with her friend Sam Kingston (Scott Wright), Toyah is raped, beaten and left for dead in an alleyway by an unknown assailant as she walked home alone. The storyline began during a special 45-minute episode that was broadcast during the Easter weekend. Viewers saw Toyah being grabbed by her attacker, but not the actual act of rape. Instead Toyah is found by Jason Grimshaw (Ryan Thomas) just as he set out for an early morning run. Jason takes a "traumatised" Toyah back to her house, where she tells her mother what had happened. Toyah then has to repeat the story to the police and reveals that she does not know who raped her because she was semi-conscious during the attack. Toyah undergoes an examination and counselling at the local hospital. The scenes were shot at the St Mary's Hospital Sexual Assault Centre in Manchester, where the writers spent four months researching victims' experiences and working with the clinical director and the staff to get the scenario right.

The storyline was criticised by some former Coronation Street cast members who believed the producers were using the subject of rape to gain higher ratings than rival soap opera EastEnders. Taylor defended the storyline and hoped that it would help other women who had been attacked to seek help. She stated "The main focus of the storyline is the psychological effect that the attack has on Toyah. There are many women out there who have been raped or sexually assaulted and we owe it to them to make sure that the story-line is handled sensitively and accurately. There is no denying that rape is a sensitive and controversial storyline. But I have put my heart and soul into it trying to portray it accurately and I only hope that women who have experienced it will think I have given it justice." A show spokesperson commented that the storyline would be "handled sensitively" and that it would not become "a whodunnit". Some of the male residents came under suspicion and were forced to give DNA samples. Peter Barlow (Chris Gascoyne) briefly became the prime suspect, but DNA soon eliminated him from enquires.

Toyah was aware that her attacker knew her as she remembered him calling her name, but could not put a name to the voice. She also struggled to deal with her "physical and emotional scars" and soon developed agoraphobia. Toyah found support from her mother and neighbour Charlie Ramsden (Clare McGlinn), who revealed that she had once been raped. Taylor told Inside Soap's Allison Maund that Charlie's advice really helped Toyah to start getting back on her feet. However, the process was hampered when Toyah received an anonymous phone call, which left her feeling "extremely frightened". The phone call sparked "a terrifying event" in which Toyah's rapist was finally revealed to be her friend Phil Simmonds (Jack Deam). When Phil came round to comfort Toyah, she recognised his voice as he called out her name and realised that he was her rapist. While Toyah tried to escape, Phil grabbed her and held her hostage. Toyah's screams were heard by Peter, who broke the door down and knocked Phil unconscious. A show spokesperson commented that Toyah would be left to worry about whether Phil would deny the rape and be convicted.

A couple of months later, Toyah received a letter from Phil asking her to visit him in prison and ,in return, he would plead guilty to raping her. Until the letter arrived, Toyah had been piecing her life back together by getting her job back at the Rovers and starting a romance with Sam. Of how Toyah felt about the letter, Taylor told Adrian Lobb from Inside Soap, "By the time Toyah gets the letter, she has grown a little stronger. But it comes as a complete surprise. Toyah doesn't trust Phil, obviously, and she thinks he's playing games. While part of her wants to confront him about the attack, there's another side of her that wants to let it go, and just leave it to the courts to deal with him." Toyah agreed to Phil's request to visit him and Taylor said that Toyah believed that seeing Phil in person would help her to come to terms with what happened. When Toyah came face to face with Phil, her first thought was how pathetic he looked. When Phil tried to make excuses for what he had done, Toyah did not feel any pity and told him to plead guilty, which he agreed to do. Toyah left the prison feeling "tearful, but triumphant" and Taylor commented that while she would never fully recover from her ordeal, Toyah was determined to look towards the future.

### Departure

On 30 July 2002, Neil Wilkes from Digital Spy reported that Taylor had quit Coronation Street after five years. Wilkes revealed that her contract would expire in October that year, but the actress had chosen to stay with the serial beyond that to give the writers time to come up with a storyline for her character's departure. Series producer Kieran Roberts added that he was sorry to see Taylor go, but understood her decision to move on. Taylor later explained that she chose to leave because she and the writers struggled to come up with storylines for Toyah. She told the Liverpool Echo's Janet Tansley "I was asked where she would go from there and couldn't think what she could do. If I couldn't see a future for her, I couldn't play her." Taylor filmed her final scenes on 3 January 2003.

Despite a rumour that Toyah would become a victim of serial killer Richard Hillman (Brian Capron), the producers decided that she needed "a fiery fling" before she left. Therefore, Toyah's last storylines saw her embark on "a passionate affair" with her university tutor John Arnley (Paul Warriner). However, the relationship quickly turned into a love triangle when John had an affair with Toyah's flatmate Maria Sutherland (Samia Ghadie). When Toyah discovered the affair, she decided to leave the Street and made her on-screen exit in February 2003.

### Reintroduction

On 20 September 2016, it was announced Taylor had agreed to reprise her role after an absence of 13 years. Taylor began filming her scenes in October 2016 and made her on-screen return in the episode broadcast on 25 December 2016. Toyah has often been mentioned and referred to by her family members since her departure. Taylor expressed her happiness at her return, saying "I am thrilled to be re-joining my old friends on the cobbles and hugely excited to be given the opportunity to explore the character of Toyah 13 years on. Coronation Street has always had a place in my heart and I'm delighted to be working again with my good friend Jane Danson and bringing the Battersby sisters back together." The actress also said that the upcoming storylines played a part in her return to the show. She described them as "fantastic, clever, and detailed". The show's series producer, Kate Oates, added that Toyah would have some secrets upon her return that she has kept from her sister, Leanne. The character's return storyline saw her showing up in Weatherfield after deciding to end her marriage to husband Toby (Andrew Dowbiggin), because she has been having an affair. Toby later follows Toyah to the Street, where he asks her to reconsider her decision and try for a baby. However, Toyah makes it clear that their marriage is finished and asks him to leave.

## Reception

In 1998, Taylor earned the "Best Young Actor" accolade at the Inside Soap Awards. In 2001, Taylor won "Best Dramatic Performance" at The British Soap Awards. She was also nominated for "Most Popular Actress" at the 7th National Television Awards. At the Inside Soap Awards, Taylor earned a ""Best Actress" nomination, while Toyah's rape was nominated in the "Most Dramatic Storyline" category. In July 2018, Taylor was longlisted for the "Best Actress" award at the Inside Soap Awards. She did not progress to the viewer-voted shortlist. For her portrayal of Toyah, Taylor was nominated for Best Soap Actor (Female) at the 2018 Digital Spy Reader Awards; she came in eleventh place with 1.5% of the total votes.

The People's Gavin Blyth and Jon Clarke stated that Toyah and Leanne were "rough, tough and make other streetwise teenagers look like they are fresh out of nursery." The characters were credited with increasing the show's ratings from 13 million to 19 million viewers. When Toyah met Dobber, the Daily Mirror's Tony Purnell commented "If anyone was going to find Mr Wrong on holiday it was Toyah Battersby." John Millar from the Daily Record wrote "The Street should be congratulated if the plight of Toyah Battersby – a brilliant performance by young Georgia Taylor, by the way – means just one schoolgirl says no to her lusting boyfriend."

An Inside Soap columnist observed that during the controversy over whether the show was right to run a storyline about rape, Taylor's "magnificent" portrayal of a rape victim seemed to have been forgotten. The Daily Mirror's Tony Stewart also praised Taylor's performance throughout the storyline, saying it was "so heart-achingly convincing". Stewart also wrote that Toyah's attack was "a chilling reminder of how brutal life can be". Stewart's colleague, Sue Crawford, branded Toyah a "disaster magnet". In June 2011, Rachel Tarley from the Metro included Toyah's rape in her list of the top 10 most controversial soap storylines.

Lorna Cooper, editor for MSN TV, included Toyah in a feature about soap's forgotten characters. Cooper said Toyah "didn't have the best of luck during her 1997–2003 stint on Coronation Street." A writer for Virgin Media quipped "Wayward teen Toyah rampaged through The Street in the 1990s. She fell for eco-warrior Spider but her life fell apart after she was raped by evil Phil Simmonds. She then followed Spider to London." In November 2012, Paul Millar from Digital Spy stated "Toyah Battersby was one of Coronation Street's best-loved characters – certainly the best-loved Battersby – during the time in which she graced the cobbles."

Laura-Jayne Tyler of Inside Soap praised Toyah's return, stating "We're loving every single second of Toyah being back, and actress Georgia Taylor is rocking her return like she's never been away. Having said that, we're not sure that she's fully embracing Toyah's steamy snogging sessions with Peter Barlow. If it were us, we'd be getting properly stuck in!" In September 2017, Katie Fitzpatrick of the Manchester Evening News'' noted how viewers found Toyah's middle name, Laverne, "hilarious" when it was revealed on-screen.

Fictional character from Coronation Street

Fictional vegan and vegetarian characters

Television characters introduced in 1997


Newton's theorem of revolving orbits

In classical mechanics, Newton's theorem of revolving orbits identifies the type of central force needed to multiply the angular speed of a particle by a factor k without affecting its radial motion (Figures 1 and 2). Newton applied his theorem to understanding the overall rotation of orbits (apsidal precession, Figure 3) that is observed for the Moon and planets. The term "radial motion" signifies the motion towards or away from the center of force, whereas the angular motion is perpendicular to the radial motion.

Isaac Newton derived this theorem in Propositions 43–45 of Book I of his Philosophiæ Naturalis Principia Mathematica, first published in 1687. In Proposition 43, he showed that the added force must be a central force, one whose magnitude depends only upon the distance r between the particle and a point fixed in space (the center). In Proposition 44, he derived a formula for the force, showing that it was an inverse-cube force, one that varies as the inverse cube of r. In Proposition 45 Newton extended his theorem to arbitrary central forces by assuming that the particle moved in nearly circular orbit.

As noted by astrophysicist Subrahmanyan Chandrasekhar in his 1995 commentary on Newton's Principia, this theorem remained largely unknown and undeveloped for over three centuries. Since 1997, the theorem has been studied by Donald Lynden-Bell and collaborators. Its first exact extension came in 2000 with the work of Mahomed and Vawda.

## Historical context

The motion of astronomical bodies has been studied systematically for thousands of years. The stars were observed to rotate uniformly, always maintaining the same relative positions to one another. However, other bodies were observed to wander against the background of the fixed stars; most such bodies were called planets after the Greek word "πλανήτοι" (planētoi) for "wanderers". Although they generally move in the same direction along a path across the sky (the ecliptic), individual planets sometimes reverse their direction briefly, exhibiting retrograde motion.

To describe this forward-and-backward motion, Apollonius of Perga (c. 262 – c. 190 BC) developed the concept of deferents and epicycles, according to which the planets are carried on rotating circles that are themselves carried on other rotating circles, and so on. Any orbit can be described with a sufficient number of judiciously chosen epicycles, since this approach corresponds to a modern Fourier transform. Roughly 350 years later, Claudius Ptolemaeus published his Almagest, in which he developed this system to match the best astronomical observations of his era. To explain the epicycles, Ptolemy adopted the geocentric cosmology of Aristotle, according to which planets were confined to concentric rotating spheres. This model of the universe was authoritative for nearly 1500 years.

The modern understanding of planetary motion arose from the combined efforts of astronomer Tycho Brahe and physicist Johannes Kepler in the 16th century. Tycho is credited with extremely accurate measurements of planetary motions, from which Kepler was able to derive his laws of planetary motion. According to these laws, planets move on ellipses (not epicycles) about the Sun (not the Earth). Kepler's second and third laws make specific quantitative predictions: planets sweep out equal areas in equal time, and the square of their orbital periods equals a fixed constant times the cube of their semi-major axis. Subsequent observations of the planetary orbits showed that the long axis of the ellipse (the so-called line of apsides) rotates gradually with time; this rotation is known as apsidal precession. The apses of an orbit are the points at which the orbiting body is closest or furthest away from the attracting center; for planets orbiting the Sun, the apses correspond to the perihelion (closest) and aphelion (furthest).

With the publication of his Principia roughly eighty years later (1687), Isaac Newton provided a physical theory that accounted for all three of Kepler's laws, a theory based on Newton's laws of motion and his law of universal gravitation. In particular, Newton proposed that the gravitational force between any two bodies was a central force F(r) that varied as the inverse square of the distance r between them. Arguing from his laws of motion, Newton showed that the orbit of any particle acted upon by one such force is always a conic section, specifically an ellipse if it does not go to infinity. However, this conclusion holds only when two bodies are present (the two-body problem); the motion of three bodies or more acting under their mutual gravitation (the n-body problem) remained unsolved for centuries after Newton, although solutions to a few special cases were discovered. Newton proposed that the orbits of planets about the Sun are largely elliptical because the Sun's gravitation is dominant; to first approximation, the presence of the other planets can be ignored. By analogy, the elliptical orbit of the Moon about the Earth was dominated by the Earth's gravity; to first approximation, the Sun's gravity and those of other bodies of the Solar System can be neglected. However, Newton stated that the gradual apsidal precession of the planetary and lunar orbits was due to the effects of these neglected interactions; in particular, he stated that the precession of the Moon's orbit was due to the perturbing effects of gravitational interactions with the Sun.

Newton's theorem of revolving orbits was his first attempt to understand apsidal precession quantitatively. According to this theorem, the addition of a particular type of central force—the inverse-cube force—can produce a rotating orbit; the angular speed is multiplied by a factor k, whereas the radial motion is left unchanged. However, this theorem is restricted to a specific type of force that may not be relevant; several perturbing inverse-square interactions (such as those of other planets) seem unlikely to sum exactly to an inverse-cube force. To make his theorem applicable to other types of forces, Newton found the best approximation of an arbitrary central force F(r) to an inverse-cube potential in the limit of nearly circular orbits, that is, elliptical orbits of low eccentricity, as is indeed true for most orbits in the Solar System. To find this approximation, Newton developed an infinite series that can be viewed as the forerunner of the Taylor expansion. This approximation allowed Newton to estimate the rate of precession for arbitrary central forces. Newton applied this approximation to test models of the force causing the apsidal precession of the Moon's orbit. However, the problem of the Moon's motion is dauntingly complex, and Newton never published an accurate gravitational model of the Moon's apsidal precession. After a more accurate model by Clairaut in 1747, analytical models of the Moon's motion were developed in the late 19th century by Hill, Brown, and Delaunay.

However, Newton's theorem is more general than merely explaining apsidal precession. It describes the effects of adding an inverse-cube force to any central force F(r), not only to inverse-square forces such as Newton's law of universal gravitation and Coulomb's law. Newton's theorem simplifies orbital problems in classical mechanics by eliminating inverse-cube forces from consideration. The radial and angular motions, r(t) and θ1(t), can be calculated without the inverse-cube force; afterwards, its effect can be calculated by multiplying the angular speed of the particle

<math>

\omega\_{2} = \frac{d\theta\_{2}}{dt} = k \frac{d\theta\_{1}}{dt} = k \omega\_{1}. </math>

## Mathematical statement

Consider a particle moving under an arbitrary central force F1(r) whose magnitude depends only on the distance r between the particle and a fixed center. Since the motion of a particle under a central force always lies in a plane, the position of the particle can be described by polar coordinates (r, θ1), the radius and angle of the particle relative to the center of force (Figure 1). Both of these coordinates, r(t) and θ1(t), change with time t as the particle moves.

Imagine a second particle with the same mass m and with the same radial motion r(t), but one whose angular speed is k times faster than that of the first particle. In other words, the azimuthal angles of the two particles are related by the equation θ2(t) = k θ1(t). Newton showed that the motion of the second particle can be produced by adding an inverse-cube central force to whatever force F1(r) acts on the first particle

<math>

F_2(r) - F_1(r) = \frac{L_1^2}{mr^3} \left( 1 - k^2 \right) </math>

where L1 is the magnitude of the first particle's angular momentum, which is a constant of motion (conserved) for central forces.

If k2 is greater than one, F2 − F1 is a negative number; thus, the added inverse-cube force is attractive, as observed in the green planet of Figures 1–4 and 9. By contrast, if k2 is less than one, F2−F1 is a positive number; the added inverse-cube force is repulsive, as observed in the green planet of Figures 5 and 10, and in the red planet of Figures 4 and 5.

### Alteration of the particle path

The addition of such an inverse-cube force also changes the path followed by the particle. The path of the particle ignores the time dependencies of the radial and angular motions, such as r(t) and θ1(t); rather, it relates the radius and angle variables to one another. For this purpose, the angle variable is unrestricted and can increase indefinitely as the particle revolves around the central point multiple times. For example, if the particle revolves twice about the central point and returns to its starting position, its final angle is not the same as its initial angle; rather, it has increased by 2×360° = 720°. Formally, the angle variable is defined as the integral of the angular speed

<math>

\theta_1 \equiv \int \omega_1(t)\\, dt. </math>

A similar definition holds for θ2, the angle of the second particle.

If the path of the first particle is described in the form r = g(θ1), the path of the second particle is given by the function r = g(θ2/k), since θ2 = k θ1. For example, let the path of the first particle be an ellipse

<math>

\frac{1}{r} = A + B \cos \theta_1 </math>

where A and B are constants; then, the path of the second particle is given by

<math>

\frac{1}{r} = A + B \cos \left( \frac{\theta_2}{k} \right). </math>

## Orbital precession

If k is close, but not equal, to one, the second orbit resembles the first, but revolves gradually about the center of force; this is known as orbital precession (Figure 3). If k is greater than one, the orbit precesses in the same direction as the orbit (Figure 3); if k is less than one, the orbit precesses in the opposite direction.

Although the orbit in Figure 3 may seem to rotate uniformly, i.e., at a constant angular speed, this is true only for circular orbits. If the orbit rotates at an angular speed Ω, the angular speed of the second particle is faster or slower than that of the first particle by Ω; in other words, the angular speeds would satisfy the equation ω2 = ω1 + Ω. However, Newton's theorem of revolving orbits states that the angular speeds are related by multiplication: ω2 = kω1, where k is a constant. Combining these two equations shows that the angular speed of the precession equals Ω = (k − 1)ω1. Hence, Ω is constant only if ω1 is constant. According to the conservation of angular momentum, ω1 changes with the radius r

<math>

\omega\_{1} = \frac{L\_{1}}{m r^{2}}; </math>

where m and L1 are the first particle's mass and angular momentum, respectively, both of which are constant. Hence, ω1 is constant only if the radius r is constant, i.e., when the orbit is a circle. However, in that case, the orbit does not change as it precesses.

## Illustrative example: Cotes's spirals

The simplest illustration of Newton's theorem occurs when there is no initial force, i.e., F1(r) = 0. In this case, the first particle is stationary or travels in a straight line. If it travels in a straight line that does not pass through the origin (yellow line in Figure 6) the equation for such a line may be written in the polar coordinates (r, θ1) as

<math>

\frac{1}{r} = \frac{1}{b} \cos\\ (\theta_1 - \theta_0) </math>

where θ0 is the angle at which the distance is minimized (Figure 6). The distance r begins at infinity (when θ1 – θ0 = −90°), and decreases gradually until θ1 – θ0 = 0°, when the distance reaches a minimum, then gradually increases again to infinity at θ1 – θ0 = 90°. The minimum distance b is the impact parameter, which is defined as the length of the perpendicular from the fixed center to the line of motion. The same radial motion is possible when an inverse-cube central force is added.

An inverse-cube central force F2(r) has the form

<math>

F_2(r) = \frac{\mu}{r^3} </math>

where the numerator μ may be positive (repulsive) or negative (attractive). If such an inverse-cube force is introduced, Newton's theorem says that the corresponding solutions have a shape called Cotes's spirals. These are curves defined by the equation

<math>

\frac{1}{r} = \frac{1}{b} \cos\\ \left(\frac{\theta_2 - \theta_0}{k} \right) </math>

where the constant k equals

<math>

k^2 = 1 - \frac{m \mu}{L_1^2} </math>

When the right-hand side of the equation is a positive real number, the solution corresponds to an epispiral. When the argument θ1 – θ0 equals ±90°×k, the cosine goes to zero and the radius goes to infinity. Thus, when k is less than one, the range of allowed angles becomes small and the force is repulsive (red curve on right in Figure 7). On the other hand, when k is greater than one, the range of allowed angles increases, corresponding to an attractive force (green, cyan and blue curves on left in Figure 7); the orbit of the particle can even wrap around the center several times. The possible values of the parameter k may range from zero to infinity, which corresponds to values of μ ranging from negative infinity up to the positive upper limit, L12/m. Thus, for all attractive inverse-cube forces (negative μ) there is a corresponding epispiral orbit, as for some repulsive ones (μ \< L12/m), as illustrated in Figure 7. Stronger repulsive forces correspond to a faster linear motion.

One of the other solution types is given in terms of the hyperbolic cosine:

<math>

\frac{1}{r} = \frac{1}{b} \cosh\\ \left(\frac{\theta_0 - \theta_2}{\lambda} \right) </math>

where the constant λ satisfies

<math>

\lambda^2 = \frac{m \mu}{L_1^{2}} - 1 </math>

This form of Cotes's spirals corresponds to one of the two Poinsot's spirals (Figure 8). The possible values of λ range from zero to infinity, which corresponds to values of μ greater than the positive number L12/m. Thus, Poinsot spiral motion only occurs for repulsive inverse-cube central forces, and applies in the case that L is not too large for the given μ.

Taking the limit of k or λ going to zero yields the third form of a Cotes's spiral, the so-called reciprocal spiral or hyperbolic spiral, as a solution

<math>

\frac{1}{r} = A \theta_2 + \varepsilon </math>

where A and ε are arbitrary constants. Such curves result when the strength μ of the repulsive force exactly balances the angular momentum-mass term

<math>

\mu = \frac{L\_{1}^{2}}{m} </math>

## Closed orbits and inverse-cube central forces

Two types of central forces—those that increase linearly with distance, F = Cr, such as Hooke's law, and inverse-square forces, F = C/r2, such as Newton's law of universal gravitation and Coulomb's law—have a very unusual property. A particle moving under either type of force always returns to its starting place with its initial velocity, provided that it lacks sufficient energy to move out to infinity. In other words, the path of a bound particle is always closed and its motion repeats indefinitely, no matter what its initial position or velocity. As shown by Bertrand's theorem, this property is not true for other types of forces; in general, a particle will not return to its starting point with the same velocity.

However, Newton's theorem shows that an inverse-cubic force may be applied to a particle moving under a linear or inverse-square force such that its orbit remains closed, provided that k equals a rational number. (A number is called "rational" if it can be written as a fraction m/n, where m and n are integers.) In such cases, the addition of the inverse-cubic force causes the particle to complete m rotations about the center of force in the same time that the original particle completes n rotations. This method for producing closed orbits does not violate Bertrand's theorem, because the added inverse-cubic force depends on the initial velocity of the particle.

Harmonic and subharmonic orbits are special types of such closed orbits. A closed trajectory is called a harmonic orbit if k is an integer, i.e., if n = 1 in the formula k = m/n. For example, if k = 3 (green planet in Figures 1 and 4, green orbit in Figure 9), the resulting orbit is the third harmonic of the original orbit. Conversely, the closed trajectory is called a subharmonic orbit if k is the inverse of an integer, i.e., if m = 1 in the formula k = m/n. For example, if k = 1/3 (green planet in Figure 5, green orbit in Figure 10), the resulting orbit is called the third subharmonic of the original orbit. Although such orbits are unlikely to occur in nature, they are helpful for illustrating Newton's theorem.

## Limit of nearly circular orbits

In Proposition 45 of his Principia, Newton applies his theorem of revolving orbits to develop a method for finding the force laws that govern the motions of planets. Johannes Kepler had noted that the orbits of most planets and the Moon seemed to be ellipses, and the long axis of those ellipses can determined accurately from astronomical measurements. The long axis is defined as the line connecting the positions of minimum and maximum distances to the central point, i.e., the line connecting the two apses. For illustration, the long axis of the planet Mercury is defined as the line through its successive positions of perihelion and aphelion. Over time, the long axis of most orbiting bodies rotates gradually, generally no more than a few degrees per complete revolution, because of gravitational perturbations from other bodies, oblateness in the attracting body, general relativistic effects, and other effects. Newton's method uses this apsidal precession as a sensitive probe of the type of force being applied to the planets.

Newton's theorem describes only the effects of adding an inverse-cube central force. However, Newton extends his theorem to an arbitrary central force F(r) by restricting his attention to orbits that are nearly circular, such as ellipses with low orbital eccentricity (ε ≤ 0.1), which is true of seven of the eight planetary orbits in the solar system. Newton also applied his theorem to the planet Mercury, which has an eccentricity ε of roughly 0.21, and suggested that it may pertain to Halley's comet, whose orbit has an eccentricity of roughly 0.97.

A qualitative justification for this extrapolation of his method has been suggested by Valluri, Wilson and Harper. According to their argument, Newton considered the apsidal precession angle α (the angle between the vectors of successive minimum and maximum distance from the center) to be a smooth, continuous function of the orbital eccentricity ε. For the inverse-square force, α equals 180°; the vectors to the positions of minimum and maximum distances lie on the same line. If α is initially not 180° at low ε (quasi-circular orbits) then, in general, α will equal 180° only for isolated values of ε; a randomly chosen value of ε would be very unlikely to give α = 180°. Therefore, the observed slow rotation of the apsides of planetary orbits suggest that the force of gravity is an inverse-square law.

### Quantitative formula

To simplify the equations, Newton writes F(r) in terms of a new function C(r)

<math>

F(r) = \frac{C(r)}{R r^3} </math>

where R is the average radius of the nearly circular orbit. Newton expands C(r) in a series—now known as a Taylor expansion—in powers of the distance r, one of the first appearances of such a series. By equating the resulting inverse-cube force term with the inverse-cube force for revolving orbits, Newton derives an equivalent angular scaling factor k for nearly circular orbits:

<math>

\frac{1}{k^{2}} = \left( \frac{R}{C} \right) \left. \frac{dC}{dr} \right\|\_{r=R} </math>

In other words, the application of an arbitrary central force F(r) to a nearly circular elliptical orbit can accelerate the angular motion by the factor k without affecting the radial motion significantly. If an elliptical orbit is stationary, the particle rotates about the center of force by 180° as it moves from one end of the long axis to the other (the two apses). Thus, the corresponding apsidal angle α for a general central force equals k×180°, using the general law θ2 = k θ1.

### Examples

Newton illustrates his formula with three examples. In the first two, the central force is a power law, F(r) = rn−3, so C(r) is proportional to rn. The formula above indicates that the angular motion is multiplied by a factor k = 1/√n, so that the apsidal angle α equals 180°/√n.

This angular scaling can be seen in the apsidal precession, i.e., in the gradual rotation of the long axis of the ellipse (Figure 3). As noted above, the orbit as a whole rotates with a mean angular speed Ω=(k−1)ω, where ω equals the mean angular speed of the particle about the stationary ellipse. If the particle requires a time T to move from one apse to the other, this implies that, in the same time, the long axis will rotate by an angle β = ΩT = (k − 1)ωT = (k − 1)×180°. For an inverse-square law such as Newton's law of universal gravitation, where n equals 1, there is no angular scaling (k = 1), the apsidal angle α is 180°, and the elliptical orbit is stationary (Ω = β = 0).

As a final illustration, Newton considers a sum of two power laws

<math>

C(r) \propto a r^m + b r^n </math>

which multiplies the angular speed by a factor

<math>

k = \sqrt{\frac{a + b}{am + bn}} </math>

Newton applies both of these formulae (the power law and sum of two power laws) to examine the apsidal precession of the Moon's orbit.

## Precession of the Moon's orbit

The motion of the Moon can be measured accurately, and is noticeably more complex than that of the planets. The ancient Greek astronomers, Hipparchus and Ptolemy, had noted several periodic variations in the Moon's orbit, such as small oscillations in its orbital eccentricity and the inclination of its orbit to the plane of the ecliptic. These oscillations generally occur on a once-monthly or twice-monthly time-scale. The line of its apses precesses gradually with a period of roughly 8.85 years, while its line of nodes turns a full circle in roughly double that time, 18.6 years. This accounts for the roughly 18-year periodicity of eclipses, the so-called Saros cycle. However, both lines experience small fluctuations in their motion, again on the monthly time-scale.

In 1673, Jeremiah Horrocks published a reasonably accurate model of the Moon's motion in which the Moon was assumed to follow a precessing elliptical orbit. A sufficiently accurate and simple method for predicting the Moon's motion would have solved the navigational problem of determining a ship's longitude; in Newton's time, the goal was to predict the Moon's position to 2' (two arc-minutes), which would correspond to a 1° error in terrestrial longitude. Horrocks' model predicted the lunar position with errors no more than 10 arc-minutes; for comparison, the diameter of the Moon is roughly 30 arc-minutes.

Newton used his theorem of revolving orbits in two ways to account for the apsidal precession of the Moon. First, he showed that the Moon's observed apsidal precession could be accounted for by changing the force law of gravity from an inverse-square law to a power law in which the exponent was 2 + 4/243 (roughly 2.0165)

<math>

F(r) = - \frac{GMm}{r^{2 + 4/243}} </math>

In 1894, Asaph Hall adopted this approach of modifying the exponent in the inverse-square law slightly to explain an anomalous orbital precession of the planet Mercury, which had been observed in 1859 by Urbain Le Verrier. Ironically, Hall's theory was ruled out by careful astronomical observations of the Moon. The currently accepted explanation for this precession involves the theory of general relativity, which (to first approximation) adds an inverse-quartic force, i.e., one that varies as the inverse fourth power of distance.

As a second approach to explaining the Moon's precession, Newton suggested that the perturbing influence of the Sun on the Moon's motion might be approximately equivalent to an additional linear force

<math>

F(r) = \frac{A}{r^{2}} + B r </math>

The first term corresponds to the gravitational attraction between the Moon and the Earth, where r is the Moon's distance from the Earth. The second term, so Newton reasoned, might represent the average perturbing force of the Sun's gravity of the Earth-Moon system. Such a force law could also result if the Earth were surrounded by a spherical dust cloud of uniform density. Using the formula for k for nearly circular orbits, and estimates of A and B, Newton showed that this force law could not account for the Moon's precession, since the predicted apsidal angle α was (≈ 180.76°) rather than the observed α (≈ 181.525°). For every revolution, the long axis would rotate 1.5°, roughly half of the observed 3.0°

## Generalization

Isaac Newton first published his theorem in 1687, as Propositions 43–45 of Book I of his Philosophiæ Naturalis Principia Mathematica. However, as astrophysicist Subrahmanyan Chandrasekhar noted in his 1995 commentary on Newton's Principia, the theorem remained largely unknown and undeveloped for over three centuries.

The first generalization of Newton's theorem was discovered by Mahomed and Vawda in 2000. As Newton did, they assumed that the angular motion of the second particle was k times faster than that of the first particle, θ2 = k θ1. In contrast to Newton, however, Mahomed and Vawda did not require that the radial motion of the two particles be the same, r1 = r2. Rather, they required that the inverse radii be related by a linear equation

<math>

\frac{1}{r\_{2}(t)} = \frac{a}{r\_{1}(t)} + b </math>

This transformation of the variables changes the path of the particle. If the path of the first particle is written r1 = g(θ1), the second particle's path can be written as

<math>

\frac{a r_2}{1 - b r_2} = g\left( \frac{\theta_2}{k} \right) </math>

If the motion of the first particle is produced by a central force F1(r), Mahomed and Vawda showed that the motion of the second particle can be produced by the following force

<math>

F_2(r_2) = \frac{a^3}{\left( 1 - b r_2 \right)^2} F\_{1}\left( \frac{a r_2}{1 - b r_2} \right) + \frac{L^2}{mr^3} \left( 1 - k^2 \right) - \frac{bL^2}{mr^2} </math>

According to this equation, the second force F2(r) is obtained by scaling the first force and changing its argument, as well as by adding inverse-square and inverse-cube central forces.

For comparison, Newton's theorem of revolving orbits corresponds to the case a = 1 and b = 0, so that r1 = r2. In this case, the original force is not scaled, and its argument is unchanged; the inverse-cube force is added, but the inverse-square term is not. Also, the path of the second particle is r2 = g(θ2/k), consistent with the formula given above.

## Derivations

### Newton's derivation

Newton's derivation is found in Section IX of his Principia, specifically Propositions 43–45. His derivations of these Propositions are based largely on geometry.

Proposition 43; Problem 30

It is required to make a body move in a curve that revolves about the center of force in the same manner as another body in the same curve at rest.

Newton's derivation of Proposition 43 depends on his Proposition 2, derived earlier in the Principia. Proposition 2 provides a geometrical test for whether the net force acting on a point mass (a particle) is a central force. Newton showed that a force is central if and only if the particle sweeps out equal areas in equal times as measured from the center.

Newton's derivation begins with a particle moving under an arbitrary central force F1(r); the motion of this particle under this force is described by its radius r(t) from the center as a function of time, and also its angle θ1(t). In an infinitesimal time dt, the particle sweeps out an approximate right triangle whose area is

<math>

dA_1 = \frac{1}{2} r^2 d\theta_1 </math>

Since the force acting on the particle is assumed to be a central force, the particle sweeps out equal angles in equal times, by Newton's Proposition 2. Expressed another way, the rate of sweeping out area is constant

<math>

\frac{dA_1}{dt} = \frac{1}{2} r^2 \frac{d\theta_1}{dt} = \mathrm{constant} </math>

This constant areal velocity can be calculated as follows. At the apapsis and periapsis, the positions of closest and furthest distance from the attracting center, the velocity and radius vectors are perpendicular; therefore, the angular momentum L1 per mass m of the particle (written as h1) can be related to the rate of sweeping out areas

<math>

h_1 = \frac{L_1}{m} = r v_1 = r^2 \frac{d\theta_1}{dt} = 2 \frac{dA_1}{dt} </math>

Now consider a second particle whose orbit is identical in its radius, but whose angular variation is multiplied by a constant factor k

<math>

\theta_2(t) = k \theta_1(t)\\,\\! </math>

The areal velocity of the second particle equals that of the first particle multiplied by the same factor k

<math>

h_2 = 2 \frac{dA_2}{dt} = r^2 \frac{d\theta_2}{dt} = k r^2 \frac{d\theta_1}{dt} = 2 k \frac{dA_1}{dt} = k h_1 </math>

Since k is a constant, the second particle also sweeps out equal areas in equal times. Therefore, by Proposition 2, the second particle is also acted upon by a central force F2(r). This is the conclusion of Proposition 43.

Proposition 44

The difference of the forces, by which two bodies may be made to move equally, one in a fixed, the other in the same orbit revolving, varies inversely as the cube of their common altitudes.

To find the magnitude of F2(r) from the original central force F1(r), Newton calculated their difference F2(r) − F1(r) using geometry and the definition of centripetal acceleration. In Proposition 44 of his Principia, he showed that the difference is proportional to the inverse cube of the radius, specifically by the formula given above, which Newtons writes in terms of the two constant areal velocities, h1 and h2

<math>

F_2(r) - F_1(r) = m \frac{h_1^2 - h_2^2}{r^3} </math>

Proposition 45; Problem 31

To find the motion of the apsides in orbits approaching very near to circles.

In this Proposition, Newton derives the consequences of his theorem of revolving orbits in the limit of nearly circular orbits. This approximation is generally valid for planetary orbits and the orbit of the Moon about the Earth. This approximation also allows Newton to consider a great variety of central force laws, not merely inverse-square and inverse-cube force laws.

### Modern derivation

Modern derivations of Newton's theorem have been published by Whittaker (1937) and Chandrasekhar (1995). By assumption, the second angular speed is k times faster than the first

<math>

\omega\_{2} = \frac{d\theta\_{2}}{dt} = k \frac{d\theta\_{1}}{dt} = k \omega\_{1} </math>

Since the two radii have the same behavior with time, r(t), the conserved angular momenta are related by the same factor k

<math>

L\_{2} = m r^{2} \omega\_{2} = m r^{2} k \omega\_{1} = k L\_{1} \\,\\! </math>

The equation of motion for a radius r of a particle of mass m moving in a central potential V(r) is given by Lagrange's equations

<math>

m\frac{d^2 r}{dt^2} - mr \omega^2 = m\frac{d^2 r}{dt^2} - \frac{L^2}{mr^3} = F(r) </math>

Applying the general formula to the two orbits yields the equation

<math>

m\frac{d^2 r}{dt^2} = F_1(r) + \frac{L_1^2}{mr^{3}} = F_2(r) + \frac{L_2^2}{mr^3} = F_2(r) + \frac{k^2 L_1^2}{mr^3} </math>

which can be re-arranged to the form

<math>

F\_{2}(r) = F_1(r) + \frac{L_1^2}{mr^3} \left( 1 - k^2 \right) </math>

This equation relating the two radial forces can be understood qualitatively as follows. The difference in angular speeds (or equivalently, in angular momenta) causes a difference in the centripetal force requirement; to offset this, the radial force must be altered with an inverse-cube force.

Newton's theorem can be expressed equivalently in terms of potential energy, which is defined for central forces

$F(r) = -\frac{dV}{dr}$

The radial force equation can be written in terms of the two potential energies

<math>

\- \frac{dV_2}{dr} = - \frac{dV_1}{dr} + \frac{L_1^2}{mr^3} \left( 1 - k^2 \right) </math>

Integrating with respect to the distance r, Newtons's theorem states that a k-fold change in angular speed results from adding an inverse-square potential energy to any given potential energy V1(r)

<math>

V_2(r) = V_1(r) + \frac{L_1^2}{2mr^2} \left( 1 - k^2 \right) </math>

## Newton’s Geometric Proof from the Principia

### Simplified Geometric Proof of Proposition 44

Although Newton states that the problem was to be solved by Proposition 6, he does not use it explicitly. In the following, simplified proof, Proposition 6 is used to show how the result is derived.

Newton's detailed proof follows that, and finally Proposition 6 is appended, as it is not well-known.

Proposition 44 uses Proposition 6 to prove a result about revolving orbits. In the propositions following Proposition 6 in Section 2 of the Principia, he applies it to specific curves, for example, conic sections. In the case of Proposition 44, it is applied to any orbit, under the action of an arbitrary force directed towards a fixed point, to produce a corresponding revolving orbit.

In Fig. 1, MN is part of that orbit. At point P, the body moves to Q under the action of a force directed towards S, as before. The force, F(SP) is defined at each point P on the curve.

In Fig. 2, the corresponding part of the revolving orbit is mn with s as its centre of force. Assume that initially, the body in the static orbit starts out at right angles to the radius with speed V. The body in the revolving orbit must also start at right angles and assume its speed is v. In the case shown in Fig. 1, $v > V$ and the force is directed towards S. The argument applies equally if $v < V$. Also, the force may be directed away from the centre.

Let SA be the initial direction of the static orbit, and sa, that of the revolving orbit. If after a certain time the bodies in the respective orbits are at P and p, then the ratios of the angles $\frac{asp}{ASP} = \frac{psq}{PSQ} = \frac{v}{V}$; the ratios of the areas $\frac{asp}{ASP} = \frac{psq}{PSQ} = \frac{v}{V}$; and the radii $SP = sp$, $SQ = sq$, $SA = sa$.

The figure pryx and the arc py in Fig. 2 are the figure PRQT and the arc PQ in Fig. 1, expanded linearly in the horizontal direction in the ratio $\frac{v}{V}$, so that $PU = pu$, $UT = ux$, and $qt = \frac{v}{V}QT$. The straight lines qt and QT should really be circular arcs with centres s and S and radii sq and SQ respectively. In the limit, their ratio becomes $\frac{v}{V}$, whether they are straight lines or arcs.

Since in the limit the forces are parallel to SP and sp, if the same force acted on the body in Fig. 2 as in Fig. 1, the body would arrive at y, since ry = RQ. The difference in horizontal speed does not affect the vertical distances. Newton refers to Corollary 2 of the Laws of Motion, where the motion of the bodies is resolved into a component in the radial direction acted on by the whole force, and the other component transverse to it, acted on by no force.

However, the distance from y to the centre, s, is now greater than SQ, so an additional force is required to move the body to q such that sq = SQ. The extra force is represented by yq, and f is proportional to ry + yq, just as F is to RQ.

$RQ = SP - PU - ST$, $rq = sp - pu - st = RQ + ST - st$.

The difference, $ST - st = yq$, can be found as follows:

$ST^2 = SQ^2 - QT^2$, $st^2 = sq^2 - qt^2$, so $(ST - st)(ST + st) = qt^2 - QT^2$.

And in the limit, as QT and qt approach zero, $ST + st$ becomes equal to $SQ + sq$ or 2SP so

$yq*2SP = \frac{QT^2(v^2 - V^2)}{V^2}$.

Therefore, $rq = RQ + ST - st = RQ + \frac{QT^2(v^2 - V^2) }{2SP*V^2}$.

Since from Proposition 6 (Fig.1 and see below), the force is $\frac{k*QR}{(SP^2*QT^2) }$. Divide by $\frac {(SP^2*QT^2)}{k}$, where k is constant, to obtain the forces $f(sp) = F(sp) + \frac{k(v^2 - V^2) }{2sp^3*V^2}$.

In Fig. 3, at the initial point A of the static curve, draw the tangent AR, which is perpendicular to SA, and the circle AQD, which just touches the curve at A. Let ρ be the radius of that circle. Since angle SAR is a right angle, the centre of the circle lies on SA. From the property of a circle: $QT^2 = AT(2\rho - AT) = RQ(2\rho - RQ)$, and in the limit as Q approaches A, this becomes $\rho =  QT^2 / 2RQ$.

Hence, $F(SA) = \frac{k}{(2\rho*SA^2) }$.

And since F(SA) is given, this determines the constant k. However, Newton wants the force at A to be of the form $cV^2 / SA^2$, where c is a constant, so that $f(sp) = F(sp) + \frac{c\rho (v^2 - V^2) }{sp^3}$, where $c = \frac{F(SA) * SA^2}{V^2}$.

The expression for f(sp) above is the same as Newton's in Corollary 4 of Proposition 44, except that he uses different letters. He writes $\frac{G}{F} = \frac{v}{V}$ (where G and F are not necessarily equal to v and V respectively), and uses the letter “V” for the constant corresponding to “c”, and the letter “X” for the function F(sp).

The above geometric proof shows very clearly where the additional force arises from to make the orbit revolve with respect to the static orbit.

### Newton’s Proof of Proposition 44

Newton's proof is complicated, in view of the simplicity of the above proof. As an example, his proof requires some deciphering, as the following sentence shows:

“And therefore, if with centre C and any radius CP or Cp a circular sector is described equal to the total area VPC which the body P revolving in an immobile orbit has described in any time by a radius drawn to the centre, the difference between the forces by which the body P in an immobile orbit and body p in a mobile orbit revolve will be to the centripetal force by which some body, by a radius drawn to the centre, would have been able to describe that sector uniformly in the same time in which the area VPC was described as G2 - F2 to F2.”

He initially regards the infinitesimal $\tau$ as fixed, then the areas SPQ and spq are proportional to V and v, respectively; therefore, $QT*SP \propto V\tau$ and $qt*sp \propto v\tau$ at each of the points P and p, and $yq = \frac{qt^2 - QT^2}{2sp} \propto \frac{1}{sp^3}$ so the additional force varies inversely as the cube of the radius.

In Fig.1, XQ is a circular arc, with centre S and radius SQ, meeting SP at X. The perpendicular XY meets RQ at Y, and $YQ = \frac{QT^2}{2SP}$.

Let $\phi(SQ)$ be the force required to make a body move in a circle of radius SQ, if it has the same speed as the transverse speed of the body in the static orbit at Q.

$\frac{f(sp) - F(sp)}{\phi(sp)} = \frac{yq}{YQ} = \frac{(qt^2 - QT^2) }{QT^2} = \frac{(v^2 - V^2) }{V^2}$

at every point, P and in particular at the apside, A:

$\frac{f(SA) - F(SA)}{\phi(SA)} = \frac{(v^2 - V^2) }{V^2}$.

But at A, in Fig. 3., the ratio of the force that makes the body follow the static curve, AE, to that required to make it follow the circle, AB, with radius SA, is inversely as the ratio of their radii of curvature, since they are both moving at the same speed, V, perpendicular to SA: $\frac{F(SA)}{\phi(SA)} = \frac{SA}{\rho}$. From the first part of the proof, $f(sp) - F(sp) = \frac{(f(sa) - F(sa))sa^3}{sp^3} = \frac{\rho F(sa)sa^2(v^2 - V^2)}{V^2sp^3}$.

Substituting Newton's expression for F(SA), gives the result obtained previously.

### Newton’s Proof of Proposition 45

“To find the motion of the apsides in orbits approaching circles.”

Proposition 44 was devised expressly to prove this Proposition. Newton wants to investigate the motion of a body in a nearly circular orbit attracted by a force of the form $f(sp) = \gamma*sp^{n - 3}$.

He approximates the static curve by an ellipse with an inverse square force, F(SP), directed to one of the foci, made to revolve by the addition of an inverse cube force, according to Proposition 44.

For the static ellipse, with the force varying inversely as SP squared, $F(SP) = \frac{cV^2}{SP^2}$, since c is defined above so that $F(SA) = \frac{cV^2}{SA^2}$.

With the body in the static orbit starting from the upper apside at A, it will reach the lower apside, the point closest to S, after moving through an angle of 180 degrees. Newton wants a corresponding revolving orbit starting from apside, a, about a point s, with the lower apside shifted by an angle, α, where $\frac{(180 + \alpha)}{180} = \frac{v}{V}$.

The initial speed, V, at A must be just less than that required to make the body move in a circle. Then ρ can be taken as equal to SA or sa. The problem is to determine v from the value of n, so that α can be found, or given α, to find n.

$f(sp) = F(sp) + \frac{c\rho (v^2 - V^2)}{sp^3} =  \frac{c(sp*V^2 + sa(v^2 - V^2))}{sp^3}$

Letting $sp = sa - X$,

$f(sp) = \frac{c(s*av^2 - XV^2)}{sp^3}$.

Then “by our method of converging series”: $f(sp) = \frac{\gamma*sp^n}{sp^3} = \frac{\gamma(sa - X)^n}{sp^3} = \frac{\gamma*sa^{n - 1}(sa - nX)}{sp^3}$ plus terms in X2 and above which can be ignored because the orbit is almost circular, so X is small compared to sa.

Comparing the 2 expressions for f(sp), $f(sp) = \frac{\gamma*sa^{n-1}(sa - nX)}{sp^3} = \frac{c(sa*v^2 - XV^2)}{sp^3}$ it follows that $\frac{v}{V} = \frac {1}{n ^{1/2}} = \frac{(180 + \alpha)}{180}$.

Also, $f(sp) = \frac{cv^2*sp^{n-3}}{sa^{n-1}}$.

The ratio of the initial forces at a is given by $\frac{f(sa)}{F(sa)} = \frac{1}{n}$.

### Proposition 6 for Proof of Proposition 44, above

In Fig. 1, a body is moving along a specific curve MN acted on by a (centripetal) force, towards the fixed point S. The force depends only of the distance of the point from S. The aim of this proposition is to determine how the force varies with the radius, SP. The method applies equally to the case where the force is centrifugal.

In a small time, $\tau$, the body moves from P to the nearby point Q. Draw QR parallel to SP meeting the tangent at R, and QT perpendicular to SP meeting it at T.

If there was no force present it would have moved along the tangent at P with the speed that it had at P, arriving at the point, R. If the force on the body moving from P to Q was constant in magnitude and parallel to the direction SP, the arc PQ would be parabolic with PR as its tangent and QR would be proportional to that constant force and the square of the time, $\tau$.

Conversely, if instead of arriving at R, the body was deflected to Q, then a constant force parallel to SP, with magnitude: $F \propto \frac{QR}{\tau^2}$ would have caused it to reach Q instead of R.

However, since the direction of the radius from S to points on the arc PQ and also the magnitude of the force towards S will change along PQ, the above relation will not give the exact force at P. If Q is sufficiently close to P, the direction of force will be almost parallel to SP all along PQ and if the force changes little, PQ can be assumed to be approximated by a parabolic arc with the force given as above in terms of QR and $\tau$.

The time, $\tau$ is proportional to the area of the sector SPQ. This is Kepler's Second Law. A proof is demonstrated in Proposition 1, Book 1, in the Principia. Since the arc PQ can be approximated by a straight line, the area of the sector SPQ and the area of the triangle SPQ can be taken as equal, so

$F \propto \frac{QR}{\tau^2} = \frac{k*QR}{(SP^2*QT^2) }$, where k is constant.

Again, this is not exact for finite lengths PQ. The force law is obtained if the limit of the above expression exists as a function of SP, as PQ approaches zero.

In fact, in time $\tau$, the body with no force would have reached a point, W, further from P than R. However, in the limit QW becomes parallel to SP. The point W is ignored in Newton's proof.

Also, Newton describes QR as the versed sine of the arc with P at its centre and length twice QP. Although this is not strictly the same as the QR that he has in the diagram (Fig.1), in the limit, they become equal.

Notes:

This proposition is based on Galileo's analysis of a body following a parabolic trajectory under the action of a constant acceleration. In Proposition 10, he describes it as Galileo's Theorem, and mentions Galileo several other times in relation to it in the Principia. Combining it with Kepler's Second Law gives the simple and elegant method.

In the historically very important case where MN in Fig. 1 was part of an ellipse and S was one of its foci, Newton showed in Proposition 11 that the limit $QR / QT^2$ was constant at each point on the curve, so that the force on the body directed towards the fixed point S varied inversely as the square of the distance SP.

Besides the ellipse with the centre at the focus, Newton also applied Proposition 6 to the hyperbola (Proposition 12), the parabola (Proposition 13), the ellipse with the centre of force at the centre of the ellipse (Proposition 10), the equiangular spiral (Proposition 9), and the circle with the centre of force not coinciding with the centre, and even on the circumference (Proposition 7).

## See also

- Kepler problem
- Laplace–Runge–Lenz vector
- Two-body problem in general relativity
- Newton's theorem about ovals

Theorem in classical mechanics