7
$\begingroup$

Is there any method to find a point or plane in space where the gravitational field of the Moon and Earth are the same?

And if so, what happens to a spacecraft if it passes through that point or crosses the plane (in engine shut-off condition with zero acceleration)?

$\endgroup$
9
  • 3
    $\begingroup$ I would imagine all Lunar spacecraft have passed that point. $\endgroup$ May 25 at 17:43
  • 6
    $\begingroup$ You may enjoy what-if.xkcd.com/157 $\endgroup$
    – PM 2Ring
    May 25 at 18:56
  • 15
    $\begingroup$ My man rediscovers Lagrange points $\endgroup$
    – Topcode
    May 26 at 1:50
  • 2
    $\begingroup$ @OrganicMarble They didn't pass through it. The trajectories from Earth to lunar orbit and from lunar orbit to the Earth passed well away from that point, but not because bad things happen. The goal was to get to the Moon. That point is rather useless, but it is harmless. $\endgroup$ May 26 at 12:30
  • 1
    $\begingroup$ @Topcode The point the OP was asking for is not one of the Lagrange points. See PM2Ring's answer. $\endgroup$ May 26 at 12:32

5 Answers 5

14
$\begingroup$

Because the gravitational field is a field, there's a couple ways this question could be interpreted:

  1. The magnitude of the gravitational accelerations due to the Earth and Moon are equal
  2. The gravitational accelerations due to the Earth and Moon are equal in magnitude and direction (they contribute the same amount of acceleration)
  3. The gravitational accelerations due to the Earth and Moon are equal in magnitude and opposing in direction (they cancel out)

I'm going to point out that each of these interpretations gives a different answer. I'm going to assume you meant the last one.

This point in space is called EML-1, or the Earth-Moon Lagrange point 1. Here's an image I stole from Wikipedia/Wikimedia that shows where you can find it.

enter image description here

If you were to coast an object through this point, it would accelerate very little, as the gravitational accelerations of the Earth and Moon cancel out. It would still accelerate a little bit due to the gravities of the Sun, Jupiter, and literally everything else in the observable universe.

This and the other Lagrange points are locations where apparent accelerations due to gravity from the Earth, gravity from the Moon, and the rotation of the whole Earth-Moon system due to the moon orbiting cancel out. If you were able to go to one of these and stop moving (relative to the rotating Earth-Moon system) you would stay put. Well, you'd stay put for a while--eventually those little accelerations from the Sun, Jupiter, etc. (even the lumpiness of mountains on the moon) would perturb your stable orbit & you'd drift off. This is worse at EML-1, -2, and -3; EML-4 and -5 are more stable.

Worth noting: all/most two-body gravitational systems have a set of associated Lagrange points, including the Earth-Sun system. Sometimes, we send things to them.

$\endgroup$
9
  • 4
    $\begingroup$ Due to the eccentricities of the orbits, the distances to the Earth-Moon Lagrange points vary a little. Here's a weekly plot for EML-1 for last year, using data from JPL Horizons. i.stack.imgur.com/xswb4.png $\endgroup$
    – PM 2Ring
    May 25 at 18:36
  • 1
    $\begingroup$ have you worked out the simple math to prove gravity force is indeed equal and opposite? My understanding of Lagrange points is that they relate to angular velocity matching... not gravity force equaled. $\endgroup$
    – BradV
    May 25 at 19:27
  • $\begingroup$ the question pertained to gravity force alone, not centrifugal force balancing against gravity. $\endgroup$
    – BradV
    May 25 at 20:09
  • 1
    $\begingroup$ yes @Pm 2Ring... in a sense that is true...but was not the question. Stand-alone gravity forces exist without motion or momentum. The EML1 point just happens to coexist on the same line between Earth and Moon. $\endgroup$
    – BradV
    May 25 at 20:27
  • 5
    $\begingroup$ The Lagrange points are calculated from the perspective of a rotating frame and therefore include the fictitious centrifugal acceleration. I don't think this is what the OP was after. However, the Lagrange points are much, much more useful than the useless point in space that the OP asked for. $\endgroup$ May 26 at 12:24
8
$\begingroup$

The point you seek is near the Earth-Moon Lagrange L1 point, but not identical to it. I'll call your point the centre of gravity; note that it is not equal to the centre of mass. In this answer, I'll calculate the location of the centre of gravity and compare it to the location of the L1 point.

I'll use the same notation as Wikipedia's Lagrange point article. $M_1$ is the mass of the Earth, $M_2$ is the mass of the Moon, $R$ is the distance between them, and $r$ is the distance from the point of interest to the Moon. (All distances are measured centre to centre).

From Newton's law of universal gravitation, at the centre of gravity (CoG) we have $$\frac{GM_1}{(R-r)^2} = \frac{GM_2}{r^2}$$

Rearranging, $$\left(\frac{r/R}{1-r/R}\right)^2 = \frac{M_2}{M_1}$$

It's convenient here to work with ratios. Let
$q = M_2/M_1$,
$x = r/R$,
$s = 1 - x$.
In other words, we're working in units where the Earth-Moon distance is $1$, the distance from the CoG to the Moon is $x$, and the distance from the CoG to the Earth is $s$.

So $$\left(\frac{x}{s}\right)^2 = q$$ which leads to $$x = \frac{\sqrt q}{1+\sqrt q}$$ and $$s = \frac{1}{1+\sqrt q}$$

Note that when $M_1=M_2$, $q=1$ and $x=s=\frac12$. Also note that these equations are symmetrical: if we swap $x$ & $s$, we get the inverse mass ratio, $1/q$.

For the Earth & Moon, $q \approx 0.0123000369$. That gives
$x=0.099833$
$s=0.900166$
Using $R=384975$ km for the mean Earth-Moon distance,
$x=38433$ km
$s=346541$ km
Please see my answer here for plots of the annual variation in the Earth-Moon and L1 distance.


Wikipedia gives this equation for the L1 point: $$\frac{M_1}{(R-r)^2} - \frac{M_2}{r^2} = \left(\frac{M_1R}{M_1}-r\right)\frac{M_1+M_2}{R^3}$$

That simplifies to $$\frac1{s^2} - \frac{q}{x^2} = s - qx$$ Hence $$q = \frac{s-1/s^2}{x-1/x^2}$$

That leads to a 5th degree equation in $x$, which can't be solved algebraically (in general), although it's easy enough to solve numerically. However, we don't need to solve it to compare it to the centre of gravity.

We get $$q = \left(\frac xs\right)^2 \left(\frac{1-s^3}{1-x^3}\right)$$

Note that the factor on the left is the equation for $q$ for the centre of gravity. The factor on the right is close to $2x$ for small $x$, so it's fairly close to $1$ when $x$ is close to $0.5$.

Here's a plot comparing the L1 and centre of gravity distance. Plot of L1 & Centre of Gravity vs mass ratio Here's a live version of the plotting script.

Here are daily distance plots, courtesy of Horizons.

Distances from Earth

Distances from Moon


Here's a quick hack of my Lagrange potential surface script, originally from this answer. This version also calculates the CoG distance, and plots it as a green dot at the same height as L1.

Interactive 3D Lagrange surface plot.

$\endgroup$
10
  • 1
    $\begingroup$ I'm going to be a bit 'rough around the edges' here so I can get right to the point. After all is said and done (lots more math/ratios/graph) was my 'quick and dirty' answer right or wrong? Is the Anton Hengst answer (accepted as correct) really correct? $\endgroup$
    – BradV
    May 26 at 14:12
  • 1
    $\begingroup$ @BradV I believe that the OP is asking about the centre of gravity, not the Lagrange point, so your answer is more correct than Anton's. But at that small mass ratio, L1 & CoG are very close. $\endgroup$
    – PM 2Ring
    May 26 at 14:28
  • $\begingroup$ regardless of how close the numbers are... Anton guessed, guessed wrongly, and provided 'answer' that sounded good but misled others. $\endgroup$
    – BradV
    May 26 at 14:36
  • 1
    $\begingroup$ @BradV Anton wrote a great answer... to a slightly different question. I don't agree that it's misleading, exactly, since he does mention that the Lagrange points include the centrifugal force as well as the two gravitational forces. $\endgroup$
    – PM 2Ring
    May 26 at 15:39
  • $\begingroup$ I regret I was harsh. This reminds me of a test question in "Design of Machine Elements" course. Question specifics were carefully tailored so if you chose solution path A you got answer 12.345. if you chose solution path B you got answer 12.567. Both answers would pass the smell test (felt right enough) but path A was not really applicable. The point was... details matter. I used this example when mentoring junior engineers and designers. Getting a close enough answer for the wrong reasons can lead to catastrophe. $\endgroup$
    – BradV
    May 26 at 15:48
4
$\begingroup$

I approached my solution by placing a mass of XX between Earth and Moon. Radius from Earth to Moon is $3.84 \cdot 10^{8}\ m$. ($D_e$) Distance from Earth to mass is R - ($D_m$) distance from Moon to mass.

Gravity force on mass from Earth:

$$\frac{G \cdot M_e \cdot XX}{D_e^{2}}$$

Gravity force on mass from the Moon:

$$\frac{G \cdot M_m \cdot XX}{D_m^{2}}$$

after setting the two against each other, cancelling terms $G$ and $XX$, and solving for distance, I come up with a "equal gravity point" at $3.46 \cdot 10^{8}\ m$ from Earth... 90% of distance to Moon. Please check my work!!!

How close is my calculation to EML1 location?

EDIT BradV 5/26/2022 add graphic

distances and diameters are to scale

enter image description here

$\endgroup$
7
  • $\begingroup$ If you use gravity acceleration instead of gravity force you don't need a mass XX. $\endgroup$
    – Uwe
    May 25 at 20:19
  • $\begingroup$ 346,000 km is around the maximum distance from the centre of the Earth to EML-1. See the graph I linked in my 1st comment on Anton's answer. $\endgroup$
    – PM 2Ring
    May 25 at 20:32
  • $\begingroup$ @PM 2Ring my calculation and your distance to L1 sure seem to align well! $\endgroup$
    – BradV
    May 25 at 20:40
  • $\begingroup$ If the formatting I applied is wrong, then my apologies. I did my best but could have messed something up. For example, I now see that Distance from Earth to mass is R - ($D_m$) distance from Moon to mass. probably should have been Distance from Earth to mass is $R - (D_m)$ distance from Moon to mass. Cheers. $\endgroup$
    – user47149
    May 25 at 22:44
  • 1
    $\begingroup$ @PM 2Ring in our side chat you used the term COG (Center of Gravity, better known as CG) and on your COG, L1 plot within your answer regarding the location where earth and moon gravity sum to zero. Since CG is already a defined and widely used engineering term that has a very different meaning, I'd like to propose another term.... the "Zero G-spot" ;-) rimshot/cymbal crash. $\endgroup$
    – BradV
    May 27 at 15:33
2
$\begingroup$

The gravitational field is a vector quantity. Near L1 we can find a point where the gravitational fields of the Earth and Moon are equal in magnitude but they will be opposite in direction. To have them the same magnitude and direction you want a point on the Earth-Moon line but beyond the moon. To find this point we recall that the gravitational field from a body of mass $M$ at a distance $R$ is $\frac {GM}{R^2}$ toward the body. If we let $M$ be the mass of the Earth, $m$ be the mass of the moon, $r$ be the distance to the Moon and $R$ the distance to the Earth, we are asking that $$\frac {GM}{R^2}=\frac {Gm}{r^2}$$. We will use $380\,000\ \mathrm{km}$ as a reasonable distance from the Earth to the moon, so this becomes $$\frac {GM}{(r+380\,000\ \mathrm{km})^2}=\frac {Gm}{r^2}\\ \frac Mm=\frac 1{0.0123}=\frac {(r+380\,000\ \mathrm{km})^2}{r^2}$$ and we find $$r=47\,000\ \mathrm{km}$$ so the point is $47\,000\ \mathrm{km}$ beyond the Moon. Nothing special happens if a spacecraft passes through this point. The gravitational field is what it is and the acceleration is what it is.

$\endgroup$
3
  • $\begingroup$ This is an interesting take on the question!! Super! This is a very interesting matter of semantics and interpretation of what the question really asks! So... Gravity force due to Earth is A. Gravity force due to Moon is also A. SO total gravity force acting on mass is 2A in your interpretation. $\endgroup$
    – BradV
    May 26 at 22:19
  • $\begingroup$ About a spacecraft passing thru the point... correct, nothing special happens. Trajectory would increasingly bend towards Earth/Moon as approaching point and ease as leaving. $\endgroup$
    – BradV
    May 26 at 22:28
  • $\begingroup$ @BradV: yes, I read the question as asking for that situation. $\endgroup$ May 26 at 23:56
0
$\begingroup$

Yes there would be a point where gravitational fields of moon and earth would be same. When the spacecraft passes through this point net Acceleration due to net force because of field being zero. The spacecraft would cross that point because acceleration would be zero means it would be moving with some constant speed at that point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.