# How do you find the equilibrium point between two planets using the formula, fG=fG?

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Elise Share
Feb 10, 2018

See explanation below

#### Explanation:

Say you have an object with mass "m" and you want to place it somewhere between two planets where the first planet's gravitational pull is equal to the second's. All you need to do is use Newton's law of universal gravitation.

$F = G \cdot \frac{M \cdot m}{r} ^ 2$ ,

where:

F the gravitational force
G the gravitational constant
M the mass of the first object
m the mass of the second object
r the distance between the two objects (measured from their centers of mass)

Using this equation, we get that the force acted on the object by the first planet (mass " ${M}_{1}$ ") is equal to

${F}_{1} = G \cdot \frac{{M}_{1} \cdot m}{{r}_{1}} ^ 2$

${r}_{1}$ is the distance between the centers of mass of the first planet and the object.

Similarly, for the force acted on the object by the second planet we get

${F}_{2} = G \cdot \frac{{M}_{2} \cdot m}{{r}_{2}} ^ 2$

Since we want to find the equilibrium point between the two planets, we get that

${F}_{1} = {F}_{2}$

Or

$G \cdot \frac{{M}_{1} \cdot m}{{r}_{1}} ^ 2 = G \cdot \frac{{M}_{2} \cdot m}{{r}_{2}} ^ 2$

Simplify and you get

${M}_{1} / {\left({r}_{1}\right)}^{2} = {M}_{2} / {\left({r}_{2}\right)}^{2}$

If you know the relationship between the masses (e.g. ${M}_{1}$ being 4 times greater than ${M}_{2}$), you can easily find the equilibrium point.

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