<p>I'm not sure where in the above, the eccentricity of 0.0679 came from, maybe he meant 0.0167. Also, the Earths orbit does not come that close to the sun as the diagram shows. The orbit is near circular, with an eccentricity of 0.01671123, which means a and b (which isn't show in the above) are very close values to one another. The Focus points can be given be $F_1 = (-ae, 0)$ and $F_2 = (ae, 0)$ and eccentricity being $e = \sqrt{1-(b^2/a^2)}$. Use Pythagoras to figure the rest out.</p>

CC BY-SA 4.0
2018-05-08T02:50:29.800Z
I've seen the numbers before for high and low, so I didn't even question that a source wasn't given.    Here's a common source for those two numbers:    https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_(eccentricity)
userLTK
CC BY-SA 4.0
2018-05-09T04:56:12.823Z
The Milankovitch hypothesis is a suggestion on previous states of the Earth's orbits, but not today. I don't personally know much about the validity of the hypothesis. [Wikipedia](https://en.wikipedia.org/wiki/Earth) has the eccentricity document as 0.0167086, which is close to the value I calculated. These two number are a few significant digits apart, but close enough. I would probably use their number, not the one I calculated. I'm sure a professional astronomer made the calculation with much better data than I have.
Byron
CC BY-SA 4.0
2018-05-09T14:58:03.123Z
I think Milankovich cycles are much more than a hypothesis.  They're well established.   The math gets a bit tricky, but future and past eccentricities can be calculated easily enough with today's computers and perturbation theory mathematics.   More details here:   https://en.wikipedia.org/wiki/Perturbation_(astronomy)#Periodic_nature and http://www.scholarpedia.org/article/Stability_of_the_solar_system#The_problem_of_the_eccentricities  As for the precision of the numbers, I wouldn't bet my savings on any number Wikipedia provides, but they're probably in the ballpark.
userLTK
CC BY-SA 4.0
2018-05-10T20:13:16.007Z
I don't know. The Solar System's periods are stable on human terms. If you go out millions of years, there is no telling what happens. Though, I would point out that Charles Hapgood came up with a theory relative to the the icecaps shifting and Einstein rebuked the hypothesis. There have been many other hypothesis made about climate, most recently, CO2 emission are the cause of all climate change. I prefer to live in the here and now.
Byron
CC BY-SA 4.0
2024-11-17T19:04:22.823Z
The Earth's orbital eccentricity is perturbed by the periodic influence of other planets in an entirely predictable way many millions of years into the future.
ProfRob
CC BY-SA 4.0
2024-12-05T02:01:41.233Z
He updated the question. Check the history.
Byron
CC BY-SA 4.0
2018-05-07T19:58:12.360Z
2018-05-08T13:29:23.830Z
2018-05-08T13:29:23.830Z
Byron
<p>I like the math of ellipses cause there's enough parts to keep it interesting, but it's not super complex either.   The answer to your question as you put it, is remarkably simple.  I've bolded the letters that represent distances for easier reading.</p>
<p>Eccentricity or <span class="math-container">$e = c/a$</span>, where <span class="math-container">$a$</span> is the semi major axis and <span class="math-container">$c$</span> is the distance between the focal point (sun's location) and the center of the ellipse. <span class="math-container">$ea$</span> is the distance between the center point and the focal point.</p>
<p>So, the answer to your question is simply <span class="math-container">$a$</span> plus <span class="math-container">$ea$</span> and <span class="math-container">$a$</span> minus <span class="math-container">$ea$</span>.   For the Earth and sun, the semi major axis or <span class="math-container">$a$</span>, is <span class="math-container">$149,600,000\:\mathrm{km}$</span>, then simply add and subtract <span class="math-container">$ea$</span> to both.</p>
<p>When <span class="math-container">$e = .0679$</span>, <span class="math-container">$ea = 10.15\:\mathrm{million\:km}$</span>. So <span class="math-container">$149,600,000 \pm 10,150,000$</span>. <span class="math-container">$159,750,000$</span> and <span class="math-container">$139,450,000\:\mathrm{km}$</span> at furthest and closest.</p>
<p>When <span class="math-container">$e = .000055$</span>, <span class="math-container">$ea = .008\:\mathrm{million\:km}$</span>, so <span class="math-container">$149,680,000$</span> and <span class="math-container">$149,520,000$</span> at most circular.</p>
<p>Simple picture</p>
<p><a href="https://i.sstatic.net/V6UFb.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/V6UFb.gif" alt="enter image description here" /></a></p>
<p>More detailed picture</p>
<p><a href="https://i.sstatic.net/995Rd.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/995Rd.gif" alt="enter image description here" /></a></p>
<p>Maths and sources of pictures, <a href="http://facweb.bhc.edu/academics/science/harwoodr/geol102/study/Eccentricity.htm" rel="nofollow noreferrer">here</a> and <a href="http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html" rel="nofollow noreferrer">here</a></p>
<p>footnote:  I hope this wasn't a homework question.</p>
<p><strong>UPDATE</strong></p>
<p>Eccentricity is one of the Milankovitch cycles which plays a role in the rise and fall of massive glaciars.   Low eccentricity tends to warm the Earth and high eccentricity tends to cool it.   These effects are relativey weak, but weak effects, combined with feedbacks or other cycles can be enough to trigger or lead to the recession of massive glaciers.</p>
<p>Earth has, most recently been on a roughly 100,000 year glaciation cycle which lines up with the eccentricity cycle.   This was confusing to researchers in this area, because the eccentricity cycle is weaker than the axial tilt or precession cycles, which operate on a roughly 41,000 and 26,000 year periods.</p>
<p>You can see more detail on this under the 100,000 year problem.</p>
<p>The simple reason why high eccentricity tends to cool the Earth goes back to Kepler's laws of equal areas.   Earth moves more slowly in it's orbit when it's more distant from the sun, and in moving more slowly, it spends more time further from the Sun than closer to it.   This leads to a small (I think it's considerably less than 1%), but very real decrease in the solar energy Earth gets.   So - high eccentricity means less solar energy per year.  That tends towards forming glaciers.</p>
<p>Despite this effect being small, it's thought that the eccentricity, when it lines up with the other milankovitch cycles, is enough to trigger the end of a glaciation period.</p>
<p><a href="https://en.wikipedia.org/wiki/100,000-year_problem" rel="nofollow noreferrer">100,000 year problem</a></p>
<p>(note) - the Wiki article says 21,000 and 40,000 years in the first paragraph, but later corrects that to 41,000 in the 2nd paragraph.   21,000 years refers to precession, which varies between 21,000 and 29,000 years - I said 26,000 above as an approximate average.</p>
<p>It also mentions a 400,000 year cycle, which is also eccentricity.  Eccentricity follows both a 100,000 and a 400,000 year cycle which, well, more than one planet affects Earth's orbit, so it makes sense if you think about it that way.)    We're entering a low point in the 400,000 year cycle which generally means less variation and a longer, but less dramatic interglacial.   110,000 years ago - the last interglacial, the Eemian, was quite dramatic in it's spikes.</p>

CC BY-SA 3.0
2016-10-20T23:08:32.507Z
I guess in a sense it's homework for my job as a guide. I'm putting together a presentation about ice ages, and I'm talking about Milankovitch Cycles, but I wanted to include this info just so people could conceptualize the variances better.
Fred
CC BY-SA 3.0
2016-10-21T12:45:48.097Z
Converted your bold math text to latex, feel free to undo if you want.
zephyr
CC BY-SA 4.0
2024-11-17T13:18:24.973Z
Dumb Q, but is $a$ constant as $e$ varies in Milankovitch cycles?  I guess so, but worth stating, with a reference.
ProfRob
CC BY-SA 4.0
2024-11-17T18:20:31.293Z
@ProfRob: I can’t remember the source, but I did read somewhere that a is indeed constant as e varies. However, I don’t understand *why* that would be…
Pierre Paquette
CC BY-SA 4.0
2024-11-17T18:56:44.137Z
@PierrePaquette conservation of energy
ProfRob
CC BY-SA 4.0
2024-11-18T03:24:12.960Z
@ProfRob care to elaborate?
Pierre Paquette
CC BY-SA 4.0
2024-11-18T06:36:01.287Z
@PierrePaquette the energy of an orbit depends on its semi major axis, not its eccentricity. You can't change the semi major axis without transferring energy.
ProfRob
CC BY-SA 4.0
2024-12-01T19:29:30.657Z
@Fred  comment from 8 years ago, but I just saw it today due to seeing another comment or vote up.

I'm adding a paragraph at the end to address how eccentricity plays a role in glaciation periods (which people often call ice ages).  Glaciation periods and interglacials are the correct terms)
userLTK
CC BY-SA 4.0
2016-10-20T22:47:50.670Z
2024-12-01T19:43:05.873Z
2024-12-01T19:43:05.873Z
userLTK
<p>I'm putting together a presentation on Milankovitch Cycles, and I would like to bring up the difference in distance to the Sun (km) at aphelion and perihelion, when the Earth's orbit is at maximum and minimum eccentricity. The numbers I've found are that the maximum eccentricity of Earth's orbit is 0.0679, and the minimum is 0.000055.</p>

CC BY-SA 4.0
2018-05-08T02:51:10.860Z
Source for high and low eccentricity:   https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_(eccentricity)
userLTK
<orbit><earth>
Distance to the Sun at perihelion and aphelion at maximum and minimum eccentricity
Fred
# Distance to the Sun at perihelion and aphelion at maximum and minimum eccentricity

(Asked by: username_0 on 2016-10-20)

I'm putting together a presentation on Milankovitch Cycles, and I would like to bring up the difference in distance to the Sun (km) at aphelion and perihelion, when the Earth's orbit is at maximum and minimum eccentricity. The numbers I've found are that the maximum eccentricity of Earth's orbit is 0.0679, and the minimum is 0.000055.

**Tags:** orbit, earth
> **Comments:**
> **username_1**: Source for high and low eccentricity:   https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_(eccentricity)  

---

## 2 answers

### Answer by username_1 on 2016-10-20. Score: 4 | Accepted answer

I like the math of ellipses cause there's enough parts to keep it interesting, but it's not super complex either. The answer to your question as you put it, is remarkably simple. I've bolded the letters that represent distances for easier reading.

Eccentricity or $e = c/a$, where $a$ is the semi major axis and $c$ is the distance between the focal point (sun's location) and the center of the ellipse. $ea$ is the distance between the center point and the focal point.

So, the answer to your question is simply $a$ plus $ea$ and $a$ minus $ea$. For the Earth and sun, the semi major axis or $a$, is $149,600,000\:\mathrm{km}$, then simply add and subtract $ea$ to both.

When $e = .0679$, $ea = 10.15\:\mathrm{million\:km}$. So $149,600,000 \pm 10,150,000$. $159,750,000$ and $139,450,000\:\mathrm{km}$ at furthest and closest.

When $e = .000055$, $ea = .008\:\mathrm{million\:km}$, so $149,680,000$ and $149,520,000$ at most circular.

Simple picture

[![enter image description here](https://i.sstatic.net/V6UFb.gif)](https://i.sstatic.net/V6UFb.gif)

More detailed picture

[![enter image description here](https://i.sstatic.net/995Rd.gif)](https://i.sstatic.net/995Rd.gif)

Maths and sources of pictures, [here](http://facweb.bhc.edu/academics/science/harwoodr/geol102/study/Eccentricity.htm) and [here](http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html)

footnote: I hope this wasn't a homework question.

**UPDATE**

Eccentricity is one of the Milankovitch cycles which plays a role in the rise and fall of massive glaciars. Low eccentricity tends to warm the Earth and high eccentricity tends to cool it. These effects are relativey weak, but weak effects, combined with feedbacks or other cycles can be enough to trigger or lead to the recession of massive glaciers.

Earth has, most recently been on a roughly 100,000 year glaciation cycle which lines up with the eccentricity cycle. This was confusing to researchers in this area, because the eccentricity cycle is weaker than the axial tilt or precession cycles, which operate on a roughly 41,000 and 26,000 year periods.

You can see more detail on this under the 100,000 year problem.

The simple reason why high eccentricity tends to cool the Earth goes back to Kepler's laws of equal areas. Earth moves more slowly in it's orbit when it's more distant from the sun, and in moving more slowly, it spends more time further from the Sun than closer to it. This leads to a small (I think it's considerably less than 1%), but very real decrease in the solar energy Earth gets. So - high eccentricity means less solar energy per year. That tends towards forming glaciers.

Despite this effect being small, it's thought that the eccentricity, when it lines up with the other milankovitch cycles, is enough to trigger the end of a glaciation period.

[100,000 year problem](https://en.wikipedia.org/wiki/100,000-year_problem)

(note) - the Wiki article says 21,000 and 40,000 years in the first paragraph, but later corrects that to 41,000 in the 2nd paragraph. 21,000 years refers to precession, which varies between 21,000 and 29,000 years - I said 26,000 above as an approximate average.

It also mentions a 400,000 year cycle, which is also eccentricity. Eccentricity follows both a 100,000 and a 400,000 year cycle which, well, more than one planet affects Earth's orbit, so it makes sense if you think about it that way.) We're entering a low point in the 400,000 year cycle which generally means less variation and a longer, but less dramatic interglacial. 110,000 years ago - the last interglacial, the Eemian, was quite dramatic in it's spikes.

> **Comments:**
> **username_0**: I guess in a sense it's homework for my job as a guide. I'm putting together a presentation about ice ages, and I'm talking about Milankovitch Cycles, but I wanted to include this info just so people could conceptualize the variances better.  
> **username_4**: Converted your bold math text to latex, feel free to undo if you want.  
> **username_3**: Dumb Q, but is $a$ constant as $e$ varies in Milankovitch cycles?  I guess so, but worth stating, with a reference.  
> **username_5**: @ProfRob: I can’t remember the source, but I did read somewhere that a is indeed constant as e varies. However, I don’t understand *why* that would be…  
> **username_3**: @PierrePaquette conservation of energy  
> **username_5**: @ProfRob care to elaborate?  
> **username_3**: @PierrePaquette the energy of an orbit depends on its semi major axis, not its eccentricity. You can't change the semi major axis without transferring energy.  
> **username_1**: @Fred  comment from 8 years ago, but I just saw it today due to seeing another comment or vote up.  I'm adding a paragraph at the end to address how eccentricity plays a role in glaciation periods (which people often call ice ages).  Glaciation periods and interglacials are the correct terms)  

---

### Answer by username_2 on 2018-05-07. Score: 0

I'm not sure where in the above, the eccentricity of 0.0679 came from, maybe he meant 0.0167. Also, the Earths orbit does not come that close to the sun as the diagram shows. The orbit is near circular, with an eccentricity of 0.01671123, which means a and b (which isn't show in the above) are very close values to one another. The Focus points can be given be $F\_1 = (-ae, 0)$ and $F\_2 = (ae, 0)$ and eccentricity being $e = \sqrt{1-(b^2/a^2)}$. Use Pythagoras to figure the rest out.

> **Comments:**
> **username_1**: I've seen the numbers before for high and low, so I didn't even question that a source wasn't given.    Here's a common source for those two numbers:    https://en.wikipedia.org/wiki/Milankovitch_cycles#Orbital_shape_(eccentricity)  
> **username_2**: The Milankovitch hypothesis is a suggestion on previous states of the Earth's orbits, but not today. I don't personally know much about the validity of the hypothesis. [Wikipedia](https://en.wikipedia.org/wiki/Earth) has the eccentricity document as 0.0167086, which is close to the value I calculated. These two number are a few significant digits apart, but close enough. I would probably use their number, not the one I calculated. I'm sure a professional astronomer made the calculation with much better data than I have.  
> **username_1**: I think Milankovich cycles are much more than a hypothesis.  They're well established.   The math gets a bit tricky, but future and past eccentricities can be calculated easily enough with today's computers and perturbation theory mathematics.   More details here:   https://en.wikipedia.org/wiki/Perturbation_(astronomy)#Periodic_nature and http://www.scholarpedia.org/article/Stability_of_the_solar_system#The_problem_of_the_eccentricities  As for the precision of the numbers, I wouldn't bet my savings on any number Wikipedia provides, but they're probably in the ballpark.  
> **username_2**: I don't know. The Solar System's periods are stable on human terms. If you go out millions of years, there is no telling what happens. Though, I would point out that Charles Hapgood came up with a theory relative to the the icecaps shifting and Einstein rebuked the hypothesis. There have been many other hypothesis made about climate, most recently, CO2 emission are the cause of all climate change. I prefer to live in the here and now.  
> **username_3**: The Earth's orbital eccentricity is perturbed by the periodic influence of other planets in an entirely predictable way many millions of years into the future.  
> **username_2**: He updated the question. Check the history.  

---

