Gravitational field of the Moon and Earth

Question

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)?

Answer 1

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.

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.

Answer 2

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.

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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. Here's a live version of the plotting script.

Here are daily distance plots, courtesy of Horizons.

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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.

Answer 3

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

Answer 4

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.

Answer 5

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.