<p>I am assuming that by adding 1 km/s, you mean increasing the tangential speed of the Earth by 1 km/s. This would increase both the kinetic energy and the angular momentum.</p>
<p>This is a relatively small increase in both quantities and it would send the Earth into a slightly more eccentric orbit than it has now. It would not come close to Mars.</p>
<p>Using the vis-viva equation, we can say
<span class="math-container">$$ a^{-1} = \frac{2}{r} - \frac{v^2}{GM_\odot}\ , $$</span>
where <span class="math-container">$a$</span> is the new semi-major axis, <span class="math-container">$r$</span> is the radial separation of the Earth and Sun when the impulse is given and <span class="math-container">$v$</span> is the new speed. From there we know that the aphelion radius is <span class="math-container">$r_a = a(1+e)$</span>, where <span class="math-container">$e$</span> is the new eccentricity.</p>
<p>Exactly what happens depends where in the Earth's orbit, the impulse is given. The biggest effect will be when Earth is at perihelion, increasing <span class="math-container">$v$</span> to 31.3 km/s at <span class="math-container">$r_p= 0.983$</span> au. The vis-viva equation gives <span class="math-container">$a=1.075$</span> au, so that <span class="math-container">$e=0.085$</span> and <span class="math-container">$r_a = 1.167$</span> au. Still some way short of the 1.38 au perihelion of Mars.</p>
<p>A better question might be - how much speed do we need to add to the Earth's orbit to get it to Mars. If you can assume the impulse is given at perihelion and it intercepts Mars at its perhelion (not quite right because they aren't in the same plane), then the equations above suggest <span class="math-container">$&gt; 2.2$</span> km/s is needed.</p>

CC BY-SA 4.0
2021-07-14T14:55:50.063Z
Is there a crazy billionaire considering attaching rocker boosters to Earth as an alternative to going to Mars by rocket ship?  Or maybe as a silly solution to global warming?
Barmar
CC BY-SA 4.0
2021-07-14T15:11:17.997Z
@Barmar Oddly, I was back-of-the-enveloping just that idea (boost orbit to cool the planet) in the shower this morning.  Turns out [SciAm](https://www.scientificamerican.com/article/a-modest-proposal-lets-change-earths-orbit/) actually did the same thing.  It would take about ten trillion times as much energy as we currently produce on the world's electrical grids annually.
J...
CC BY-SA 4.0
2021-07-14T20:21:38.700Z
"but it is significantly more than 1 km/s" [citation needed]
Phil Frost
CC BY-SA 4.0
2021-07-14T21:59:08.733Z
@PhilFrost It looks like it could be as small as 3 km/s if you timed it right. This can be calculated using the equations above.
ProfRob
CC BY-SA 4.0
2021-07-14T22:26:06.450Z
@Barmar Forget billionaires, at that point you may as well just market yourself as a supervillain!
user45266
CC BY-SA 4.0
2021-07-14T22:28:56.830Z
@user45266 I'm still trying to hire my henchpeople (I'm a woke supervillain).
Barmar
CC BY-SA 4.0
2021-07-15T10:17:25.177Z
Attaching rocket boosters to the Moon will be probably safer - and one can still drag the Earth with proper timing.
fraxinus
CC BY-SA 4.0
2021-07-14T08:16:37.363Z
2021-07-15T05:28:12.170Z
2021-07-15T05:28:12.170Z
ProfRob
<p>Using calculations from here: <a href="https://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/orbitalvelocity.html" rel="noreferrer">https://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/orbitalvelocity.html</a></p>
<p>New semi-major axis</p>
<p><span class="math-container">$$ a = \frac{150000000000 \cdot 0.0000000000667 \cdot 2\cdot 10^{30}}{2 \cdot 0.0000000000667 \cdot 2 \cdot 10^{30} - 150000000000 \cdot 30000 \cdot 30000)}\\ = 151.821\,\mathrm{million\,km} $$</span></p>
<p>New orbital period
<span class="math-container">$$ p = \frac{\sqrt{4\cdot\pi^2\cdot a^3/(0.0000000000667\cdot (2\cdot 10^{30} + 5.972 \cdot 10^{24}))}}{60\cdot 60\cdot 24}\\ = 372.465\,\mathrm{days}$$</span></p>

CC BY-SA 4.0
2021-07-14T15:26:45.440Z
With the input precision you use, you can only give the results at most as accurate as 150 million km and 370 days. Everything else is pretending an accuracy your input values don't warrant. 88.59394756% of all statistics give an over-precise accuracy.
planetmaker
CC BY-SA 4.0
2021-07-15T02:40:45.510Z
Instead of writing many zeros, you can use [scientific notation](https://en.wikipedia.org/wiki/Scientific_notation?wprov=sfla1) and write numbers in form $m × 10^n$. Also, instead of using dots to show multiplication, use cross sign "×".
Nilay Ghosh
CC BY-SA 4.0
2021-07-15T05:47:10.430Z
@NilayGhosh, curious, why is cross sign better than the dot for multiplication in this case?
justhalf
CC BY-SA 4.0
2021-07-15T06:29:37.503Z
@justhalf Cross sign is generally use in scientific notation (see above). Also dots may sometimes be confusing with the decimal points. I usually prefer cross than dots.
Nilay Ghosh
CC BY-SA 4.0
2021-07-15T08:37:04.483Z
@NilayGhosh I tend to disagree with using the cross sign for scientific notation, that's a a very cultural thing (depending on country, scientific community etc.). For instance, some fields prefer to write $m E n = m \cdot 10^n$, e.g. $1.23{\rm e}45$ for $1.23 \cdot 10^{45}$.
B--rian
CC BY-SA 4.0
2021-07-15T08:58:44.780Z
Where I grew up, dots were used for regular multiplication in school and uni (maths courses); the cross was reserved for the vector cross product. That's also a country where we use "," to separate decimals, not "." as in the English world.
AnoE
CC BY-SA 4.0
2021-07-14T15:10:29.647Z
2021-07-14T15:24:30.940Z
2021-07-14T15:24:30.940Z
Ivan Borsuk
<p>How would Earth's orbit be affected if we (hypothetically) added <span class="math-container">$1 {\rm km/s}$</span> to its orbital velocity? Would Earth reach close to Mars' orbit?
Could Earth get  gravitational assist from Mars and go to outer solar system?</p>

CC BY-SA 4.0
2021-07-14T07:40:12.120Z
Welcome to astronomy SE! I edited your question slightly, trying to make it a bit clearer. I also added two relevant tags.
B--rian
CC BY-SA 4.0
2021-07-14T14:32:41.237Z
Earth, ~30km/s. Venus ~35km/s, Mars ~24km/s. If you added 1km/sec instantaneously then Earth's orbit would be eccentric, rising above our normal distance from the sun towards Mars at apoapsis and returning to our nominal orbital distance at periapsis. If you wanted to keep a roughly circular orbit you would need to **slow down** first, allowing the Earth to drop closer to the sun. You could then **slow down** again to circularize and we would end up in a slightly tighter orbit, closer to Venus and 1km/s faster. So which of these is your question?
J...
CC BY-SA 4.0
2021-07-14T15:37:29.840Z
For comparison, Earth gains about 1km/s of orbital velocity every year as it goes from aphelion to perihelion, and loses it on the other half of the year.
notovny
CC BY-SA 4.0
2021-07-14T16:01:07.610Z
This could happen if the Earth was hit by a lone planet entering the solar system. The Earth could even be "marbled" out of the system. It could also be sent to the Sun. Both scenarios are not too beneficient for people and Nature. If Nature provided such a hitting loner it would be "bye-bye" for mankind.
Deschele Schilder
CC BY-SA 4.0
2021-07-14T18:03:04.450Z
You need anywhere from 4.23 to 4.89 km/s of delta V to get to Mars from Earth.  So 1 km/s is a decent change, but not enough to get near Mars.  May I suggest using [this tool](https://trajbrowser.arc.nasa.gov/traj_browser.php?maxMag=25&maxOCC=4&chk_target_list=on&target_list=mars%0D%0A&mission_class=oneway&mission_type=rendezvous&LD1=2010&LD2=2040&maxDT=4.0&DTunit=yrs&maxDV=12.0&min=DV&wdw_width=-1&submit=Search#a_load_results) to select an  efficient date for this hypothetical velocity gain?
Wyck
CC BY-SA 4.0
2021-07-14T21:56:26.373Z
change orbit from 1Au almost-circular to 1au x 1.1466au ellipse. use this toy to play with the numbers: https://www.omnicalculator.com/physics/orbital-velocity
PcMan
<gravity><mathematics><hypothetical><earths-orbit>
What if Earth gained 1 km/s orbital velocity?
Sasidhar
# What if Earth gained 1 km/s orbital velocity?

(Asked by: username_0 on 2021-07-14)

How would Earth's orbit be affected if we (hypothetically) added $1 {\rm km/s}$ to its orbital velocity? Would Earth reach close to Mars' orbit?
Could Earth get gravitational assist from Mars and go to outer solar system?

**Tags:** gravity, mathematics, hypothetical, earths-orbit
> **Comments:**
> **username_1**: Welcome to astronomy SE! I edited your question slightly, trying to make it a bit clearer. I also added two relevant tags.  
> **username_2**: Earth, ~30km/s. Venus ~35km/s, Mars ~24km/s. If you added 1km/sec instantaneously then Earth's orbit would be eccentric, rising above our normal distance from the sun towards Mars at apoapsis and returning to our nominal orbital distance at periapsis. If you wanted to keep a roughly circular orbit you would need to **slow down** first, allowing the Earth to drop closer to the sun. You could then **slow down** again to circularize and we would end up in a slightly tighter orbit, closer to Venus and 1km/s faster. So which of these is your question?  
> **username_3**: For comparison, Earth gains about 1km/s of orbital velocity every year as it goes from aphelion to perihelion, and loses it on the other half of the year.  
> **username_4**: This could happen if the Earth was hit by a lone planet entering the solar system. The Earth could even be "marbled" out of the system. It could also be sent to the Sun. Both scenarios are not too beneficient for people and Nature. If Nature provided such a hitting loner it would be "bye-bye" for mankind.  
> **username_5**: You need anywhere from 4.23 to 4.89 km/s of delta V to get to Mars from Earth.  So 1 km/s is a decent change, but not enough to get near Mars.  May I suggest using [this tool](https://trajbrowser.arc.nasa.gov/traj_browser.php?maxMag=25&maxOCC=4&chk_target_list=on&target_list=mars%0D%0A&mission_class=oneway&mission_type=rendezvous&LD1=2010&LD2=2040&maxDT=4.0&DTunit=yrs&maxDV=12.0&min=DV&wdw_width=-1&submit=Search#a_load_results) to select an  efficient date for this hypothetical velocity gain?  
> **username_6**: change orbit from 1Au almost-circular to 1au x 1.1466au ellipse. use this toy to play with the numbers: https://www.omnicalculator.com/physics/orbital-velocity  

---

## 2 answers

### Answer by username_7 on 2021-07-14. Score: 13

I am assuming that by adding 1 km/s, you mean increasing the tangential speed of the Earth by 1 km/s. This would increase both the kinetic energy and the angular momentum.

This is a relatively small increase in both quantities and it would send the Earth into a slightly more eccentric orbit than it has now. It would not come close to Mars.

Using the vis-viva equation, we can say
$$ a^{-1} = \frac{2}{r} - \frac{v^2}{GM\_\odot}\ , $$
where $a$ is the new semi-major axis, $r$ is the radial separation of the Earth and Sun when the impulse is given and $v$ is the new speed. From there we know that the aphelion radius is $r\_a = a(1+e)$, where $e$ is the new eccentricity.

Exactly what happens depends where in the Earth's orbit, the impulse is given. The biggest effect will be when Earth is at perihelion, increasing $v$ to 31.3 km/s at $r\_p= 0.983$ au. The vis-viva equation gives $a=1.075$ au, so that $e=0.085$ and $r\_a = 1.167$ au. Still some way short of the 1.38 au perihelion of Mars.

A better question might be - how much speed do we need to add to the Earth's orbit to get it to Mars. If you can assume the impulse is given at perihelion and it intercepts Mars at its perhelion (not quite right because they aren't in the same plane), then the equations above suggest $> 2.2$ km/s is needed.

> **Comments:**
> **username_8**: Is there a crazy billionaire considering attaching rocker boosters to Earth as an alternative to going to Mars by rocket ship?  Or maybe as a silly solution to global warming?  
> **username_2**: @Barmar Oddly, I was back-of-the-enveloping just that idea (boost orbit to cool the planet) in the shower this morning.  Turns out [SciAm](https://www.scientificamerican.com/article/a-modest-proposal-lets-change-earths-orbit/) actually did the same thing.  It would take about ten trillion times as much energy as we currently produce on the world's electrical grids annually.  
> **username_9**: "but it is significantly more than 1 km/s" [citation needed]  
> **username_7**: @PhilFrost It looks like it could be as small as 3 km/s if you timed it right. This can be calculated using the equations above.  
> **username_10**: @Barmar Forget billionaires, at that point you may as well just market yourself as a supervillain!  
> **username_8**: @user45266 I'm still trying to hire my henchpeople (I'm a woke supervillain).  
> **username_11**: Attaching rocket boosters to the Moon will be probably safer - and one can still drag the Earth with proper timing.  

---

### Answer by username_12 on 2021-07-14. Score: 6

Using calculations from here: <https://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/orbitalvelocity.html>

New semi-major axis

$$ a = \frac{150000000000 \cdot 0.0000000000667 \cdot 2\cdot 10^{30}}{2 \cdot 0.0000000000667 \cdot 2 \cdot 10^{30} - 150000000000 \cdot 30000 \cdot 30000)}\\ = 151.821\,\mathrm{million\,km} $$

New orbital period
$$ p = \frac{\sqrt{4\cdot\pi^2\cdot a^3/(0.0000000000667\cdot (2\cdot 10^{30} + 5.972 \cdot 10^{24}))}}{60\cdot 60\cdot 24}\\ = 372.465\,\mathrm{days}$$

> **Comments:**
> **username_13**: With the input precision you use, you can only give the results at most as accurate as 150 million km and 370 days. Everything else is pretending an accuracy your input values don't warrant. 88.59394756% of all statistics give an over-precise accuracy.  
> **username_14**: Instead of writing many zeros, you can use [scientific notation](https://en.wikipedia.org/wiki/Scientific_notation?wprov=sfla1) and write numbers in form $m × 10^n$. Also, instead of using dots to show multiplication, use cross sign "×".  
> **username_15**: @NilayGhosh, curious, why is cross sign better than the dot for multiplication in this case?  
> **username_14**: @justhalf Cross sign is generally use in scientific notation (see above). Also dots may sometimes be confusing with the decimal points. I usually prefer cross than dots.  
> **username_1**: @NilayGhosh I tend to disagree with using the cross sign for scientific notation, that's a a very cultural thing (depending on country, scientific community etc.). For instance, some fields prefer to write $m E n = m \cdot 10^n$, e.g. $1.23{\rm e}45$ for $1.23 \cdot 10^{45}$.  
> **username_16**: Where I grew up, dots were used for regular multiplication in school and uni (maths courses); the cross was reserved for the vector cross product. That's also a country where we use "," to separate decimals, not "." as in the English world.  

---

