<blockquote>
<p>If you divide 333,000 by 10,000, you get 33.3, meaning the Sun should be yanking on an object placed at L1 with more than thirty times the force as the Earth is....</p>
</blockquote>
<p>That's not how the Lagrange points are defined.</p>
<p>There is a point somewhere between the Sun and the Earth where the gravitational acceleration of some object toward the Sun is exactly equal in magnitude but opposite in direction to the gravitation of that object toward the Earth. Denoting <span class="math-container">$G$</span> as the Newtonian gravitational constant, <span class="math-container">$M$</span> as the mass of the Sun, <span class="math-container">$m$</span> as the mass of the Earth, <span class="math-container">$R$</span> as the distance between the Earth and the Sun, and <span class="math-container">$r$</span> as the distance between the Earth and the point in question, this is given by
<span class="math-container">$$\frac{G M}{(R-r)^2} = \frac{G m}{r^2}$$</span>
The result is that <span class="math-container">$r$</span> is about 258810 km, or about 2/3 of the distance between the Earth and the Moon.</p>
<p>Note that this means the gravitational acceleration of the Moon toward the Sun is more than twice the gravitational acceleration of the Moon toward the Earth. (Some use this fact as an argument that the Moon orbits the Sun rather than the Earth. Force is not a good metric for determining whether body <span class="math-container">$c$</span> orbits body <span class="math-container">$a$</span> or body <span class="math-container">$b$</span>. The Moon orbits the Earth.)</p>
<p>I'll look at the circular restricted three body problem. The three body problem in general looks at how three bodies subject to mutual gravitational force (and nothing else) interact. The restricted three body problem limits the investigation to cases where one of the three bodies has such a tiny mass compared to the other two that this third body has essentially no effect on the behavior of the two primary bodies. The circular restricted three body problem (CR3TB for short) restricts things even more: The two primary bodies are assumed to be in circular orbits about one another in the circular restricted three body problem.</p>
<p>It can be helpful to look at things from the perspective of a rotating frame of reference to understand the concept of the Lagrange points. This frame has its origin at the center of mass of the two primary objects, has its axis of rotation the same as the axis of rotation of the two bodies' orbital angular momentum axis, and and rotates at the same rate at which the two bodies orbit one another.</p>
<p>The two primary bodies have fixed locations from the perspective of this rotating frame. This raises an interesting question: Are there any points at which the third body (the body with negligible mass) can be placed that will also have a fixed location from the perspective of this rotating frame? The answer is yes. There are five such points, These are the Lagrange points. All five lie on thee orbital plane of the two primary bodies. Three are collinear with the two primary bodies. The other two form equilateral triangles with the two primary bodies.</p>
<p>I'll look at L<sub>1</sub>, the collinear Lagrange point that lies between the two primary bodies. I'll denote</p>
<ul>
<li><span class="math-container">$M$</span> as the mass of the larger primary body,</li>
<li><span class="math-container">$m$</span> as the mass of the smaller primary body,</li>
<li><span class="math-container">$R$</span> as the distance between the two primary bodies,</li>
<li><span class="math-container">$d$</span> as the distance between the larger primary body and the center of mass of the two primary bodies,</li>
<li><span class="math-container">$\omega$</span> as the angular velocity of the orbit of the two primary bodies about one another, and</li>
<li><span class="math-container">$r$</span> as the distance between the smaller primary body and the L<sub>1</sub> point.</li>
</ul>
<p>This is a uniformly rotating frame, so points in it will be subject to centrifugal and Coriolis accelerations. The Coriolis acceleration  vanishes for a stationary third body, leaving three accelerations that must sum to zero to find the location of that stationary point:
<span class="math-container">$$(R-d-r)\,\omega^2 - \frac{GM}{(R-r)^2} + \frac{Gm}{r^2} = 0 \tag{1}$$</span>
From Newtonian mechanics, <span class="math-container">$d=R\frac m{M+m}$</span>. Denoting <span class="math-container">$\frac{m}{M+m}$</span> as <span class="math-container">$\mu$</span> results in <span class="math-container">$d=R\mu$</span>. Using the fact that <span class="math-container">$\omega^2=\frac{G(M+m)}{R^3}$</span> (also from Newtonian mechanics) and denoting <span class="math-container">$x=\frac{r}{R}$</span>, equation (1) can be rewritten as
<span class="math-container">$$(1-\mu-x)\,(1-x)^2\,x^2 - (1-\mu)\,x^2 + \mu\,(1-x)^2 = 0\tag{2}$$</span></p>
<p>In the case of the Sun-Earth L<sub>1</sub> point (or more precisely, the Sun - Earth-Moon barycenter L<sub>1</sub> point), this results in <span class="math-container">$x\approx 0.0100109943$</span>. Given the scaling, this is value of <span class="math-container">$x$</span> is in astronomical units. Conversion to metric yields <span class="math-container">$r$</span> is approximately 1.497623 million kilometers, or about 930580 miles.</p>
<blockquote>
<p>P.S.: Is L2 exactly as far from us as L1? Or just approximately?</p>
</blockquote>
<p>Just approximately. The equation equivalent to equation (2) for the L<sub>2</sub> point is
<span class="math-container">$$(1-\mu+x)\,(1+x)^2\,x^2 - (1-\mu)\,x^2 - \mu\,(1+x)^2 = 0\tag{3}$$</span>
The real solution is <span class="math-container">$x\approx0.0100782578$</span>, a bit larger than the solution for the L<sub>1</sub> point. Equation (3), like equation (2), is a quintic equation. Both equations have four complex roots and one real root for <span class="math-container">$0&lt;\mu&lt;1$</span>. The real solutions are very close to one another. However, the Sun-Earth L<sub>2</sub> point is a bit further from the Earth than is the Sun-Earth L<sub>1</sub> point.</p>
<p>Quintic equations are nasty. Like many quintics, equations (2) and (3) can only be solved by numerical methods. There is a cubic equation that yields a common approximate value for the distance from the Earth to both the the Sun-Earth L<sub>2</sub> point and the Sun-Earth L<sub>1</sub> point.</p>

CC BY-SA 4.0
2022-02-06T07:16:59.333Z
I'm upvoting, but perhaps the relevance of the centrifugal force term could be better explained?  I think "the Sun pulls twice as hard as the Earth does on the moon" is a bit of a jarring "no way that's right" kind of fact to anyone who hasn't worked out the math and why there's more terms to consider than that.
zibadawa timmy
CC BY-SA 4.0
2022-02-06T07:27:57.877Z
@zibadawatimmy The centrifugal force is simply a consequence of working in a rotating frame of reference. That the Sun pulls on the Moon more than twice as hard as does the Earth is a fact and was meant to be jarring. I'll update the post.
David Hammen
CC BY-SA 4.0
2022-02-06T10:22:29.533Z
@jpa Ack! Thanks! The gravitational acceleration of the Moon toward the Sun is of course very close to the same as the gravitational acceleration of the Earth toward the Sun. The difference between the two accelerations is oftentimes called the third body effect, and this is rather small. I corrected that statement to *The gravitational acceleration of the Moon toward the Sun is more than twice the gravitational acceleration of the Moon toward the Earth*, which is what I meant to write (and did write in the original version of my answer).
David Hammen
CC BY-SA 4.0
2022-02-06T03:10:09.230Z
2022-02-06T10:21:14.993Z
2022-02-06T10:21:14.993Z
David Hammen
<p>Try to follow my simple logic:</p>
<p>The Sun is almost exactly 333,000 times as massive as Earth, and gravitational strength increases linearly with mass, so the Sun's gravity is about 333,000 times ours.</p>
<p>The distance from the Earth to L1 is about 932,000 miles. And the distance from the Sun to L1 is one AU (just under 93 mil. miles) minus 932,000, or just over 92 million miles.</p>
<p>Gravitational strength drops off with the square of distance. And if the Sun is about 99 times further from L1 than L1 is from Earth. Square that and you get approximately 9,748. Or round up to 10,000.</p>
<p>If you divide 333,000 by 10,000, you get 33.3, meaning the Sun should be yanking on an object placed at L1 with more than thirty times the force as the Earth is.</p>
<p>Where is the fault in my simplified logic?</p>
<p>(P.S.: Is L2 exactly as far from us as L1, or just approximately?)</p>

CC BY-SA 4.0
2022-02-06T03:01:08.077Z
[At the Lagrange points, the gravitational forces of the two large bodies **and the centrifugal force** balance each other.](https://en.wikipedia.org/wiki/Lagrange_point)
Keith McClary
CC BY-SA 4.0
2022-02-06T09:47:08.220Z
You may like my Lagrange point diagrams here: https://space.stackexchange.com/a/57679/38535
PM 2Ring
CC BY-SA 4.0
2022-02-07T13:33:43.697Z
You must be *really* worried about $L_2$, where the sun's force and earth's force *reinforce* each other.
Lee Mosher
CC BY-SA 4.0
2022-02-07T21:55:00.397Z
@LeeMosher excellent point!
uhoh
CC BY-SA 4.0
2022-02-22T23:51:33.277Z
Lee Mosher, I hadn't thought of that... for some reason.... Until just now... D'Oh!!
Kurt Hikes
<orbit><gravity><orbital-mechanics><orbital-elements><lagrange-point>
Why is the L1 point (Lagrange) almost 1 million miles from Earth? Shouldn't it be closer to us?
Kurt Hikes
# Why is the L1 point (Lagrange) almost 1 million miles from Earth? Shouldn't it be closer to us?

(Asked by: username_0 on 2022-02-06)

Try to follow my simple logic:

The Sun is almost exactly 333,000 times as massive as Earth, and gravitational strength increases linearly with mass, so the Sun's gravity is about 333,000 times ours.

The distance from the Earth to L1 is about 932,000 miles. And the distance from the Sun to L1 is one AU (just under 93 mil. miles) minus 932,000, or just over 92 million miles.

Gravitational strength drops off with the square of distance. And if the Sun is about 99 times further from L1 than L1 is from Earth. Square that and you get approximately 9,748. Or round up to 10,000.

If you divide 333,000 by 10,000, you get 33.3, meaning the Sun should be yanking on an object placed at L1 with more than thirty times the force as the Earth is.

Where is the fault in my simplified logic?

(P.S.: Is L2 exactly as far from us as L1, or just approximately?)

**Tags:** orbit, gravity, orbital-mechanics, orbital-elements, lagrange-point

> **Comments:**
> **username_1**: [At the Lagrange points, the gravitational forces of the two large bodies **and the centrifugal force** balance each other.](https://en.wikipedia.org/wiki/Lagrange_point)  
> **username_2**: You may like my Lagrange point diagrams here: https://space.stackexchange.com/a/57679/38535  
> **username_3**: You must be *really* worried about $L_2$, where the sun's force and earth's force *reinforce* each other.  
> **username_4**: @LeeMosher excellent point!  
> **username_0**: username_3, I hadn't thought of that... for some reason.... Until just now... D'Oh!!  

---

## 1 answers

### Answer by username_5 on 2022-02-06. Score: 20 | Accepted answer

> If you divide 333,000 by 10,000, you get 33.3, meaning the Sun should be yanking on an object placed at L1 with more than thirty times the force as the Earth is....

That's not how the Lagrange points are defined.

There is a point somewhere between the Sun and the Earth where the gravitational acceleration of some object toward the Sun is exactly equal in magnitude but opposite in direction to the gravitation of that object toward the Earth. Denoting $G$ as the Newtonian gravitational constant, $M$ as the mass of the Sun, $m$ as the mass of the Earth, $R$ as the distance between the Earth and the Sun, and $r$ as the distance between the Earth and the point in question, this is given by
$$\frac{G M}{(R-r)^2} = \frac{G m}{r^2}$$
The result is that $r$ is about 258810 km, or about 2/3 of the distance between the Earth and the Moon.

Note that this means the gravitational acceleration of the Moon toward the Sun is more than twice the gravitational acceleration of the Moon toward the Earth. (Some use this fact as an argument that the Moon orbits the Sun rather than the Earth. Force is not a good metric for determining whether body $c$ orbits body $a$ or body $b$. The Moon orbits the Earth.)

I'll look at the circular restricted three body problem. The three body problem in general looks at how three bodies subject to mutual gravitational force (and nothing else) interact. The restricted three body problem limits the investigation to cases where one of the three bodies has such a tiny mass compared to the other two that this third body has essentially no effect on the behavior of the two primary bodies. The circular restricted three body problem (CR3TB for short) restricts things even more: The two primary bodies are assumed to be in circular orbits about one another in the circular restricted three body problem.

It can be helpful to look at things from the perspective of a rotating frame of reference to understand the concept of the Lagrange points. This frame has its origin at the center of mass of the two primary objects, has its axis of rotation the same as the axis of rotation of the two bodies' orbital angular momentum axis, and and rotates at the same rate at which the two bodies orbit one another.

The two primary bodies have fixed locations from the perspective of this rotating frame. This raises an interesting question: Are there any points at which the third body (the body with negligible mass) can be placed that will also have a fixed location from the perspective of this rotating frame? The answer is yes. There are five such points, These are the Lagrange points. All five lie on thee orbital plane of the two primary bodies. Three are collinear with the two primary bodies. The other two form equilateral triangles with the two primary bodies.

I'll look at L1, the collinear Lagrange point that lies between the two primary bodies. I'll denote

* $M$ as the mass of the larger primary body,
* $m$ as the mass of the smaller primary body,
* $R$ as the distance between the two primary bodies,
* $d$ as the distance between the larger primary body and the center of mass of the two primary bodies,
* $\omega$ as the angular velocity of the orbit of the two primary bodies about one another, and
* $r$ as the distance between the smaller primary body and the L1 point.

This is a uniformly rotating frame, so points in it will be subject to centrifugal and Coriolis accelerations. The Coriolis acceleration vanishes for a stationary third body, leaving three accelerations that must sum to zero to find the location of that stationary point:
$$(R-d-r)\,\omega^2 - \frac{GM}{(R-r)^2} + \frac{Gm}{r^2} = 0 \tag{1}$$
From Newtonian mechanics, $d=R\frac m{M+m}$. Denoting $\frac{m}{M+m}$ as $\mu$ results in $d=R\mu$. Using the fact that $\omega^2=\frac{G(M+m)}{R^3}$ (also from Newtonian mechanics) and denoting $x=\frac{r}{R}$, equation (1) can be rewritten as
$$(1-\mu-x)\,(1-x)^2\,x^2 - (1-\mu)\,x^2 + \mu\,(1-x)^2 = 0\tag{2}$$

In the case of the Sun-Earth L1 point (or more precisely, the Sun - Earth-Moon barycenter L1 point), this results in $x\approx 0.0100109943$. Given the scaling, this is value of $x$ is in astronomical units. Conversion to metric yields $r$ is approximately 1.497623 million kilometers, or about 930580 miles.

> P.S.: Is L2 exactly as far from us as L1? Or just approximately?

Just approximately. The equation equivalent to equation (2) for the L2 point is
$$(1-\mu+x)\,(1+x)^2\,x^2 - (1-\mu)\,x^2 - \mu\,(1+x)^2 = 0\tag{3}$$
The real solution is $x\approx0.0100782578$, a bit larger than the solution for the L1 point. Equation (3), like equation (2), is a quintic equation. Both equations have four complex roots and one real root for $0<\mu<1$. The real solutions are very close to one another. However, the Sun-Earth L2 point is a bit further from the Earth than is the Sun-Earth L1 point.

Quintic equations are nasty. Like many quintics, equations (2) and (3) can only be solved by numerical methods. There is a cubic equation that yields a common approximate value for the distance from the Earth to both the the Sun-Earth L2 point and the Sun-Earth L1 point.

> **Comments:**
> **username_6**: I'm upvoting, but perhaps the relevance of the centrifugal force term could be better explained?  I think "the Sun pulls twice as hard as the Earth does on the moon" is a bit of a jarring "no way that's right" kind of fact to anyone who hasn't worked out the math and why there's more terms to consider than that.  
> **username_5**: @zibadawatimmy The centrifugal force is simply a consequence of working in a rotating frame of reference. That the Sun pulls on the Moon more than twice as hard as does the Earth is a fact and was meant to be jarring. I'll update the post.  
> **username_5**: @jpa Ack! Thanks! The gravitational acceleration of the Moon toward the Sun is of course very close to the same as the gravitational acceleration of the Earth toward the Sun. The difference between the two accelerations is oftentimes called the third body effect, and this is rather small. I corrected that statement to *The gravitational acceleration of the Moon toward the Sun is more than twice the gravitational acceleration of the Moon toward the Earth*, which is what I meant to write (and did write in the original version of my answer).  

---

