# Points in space where the gravitational force due to two bodies are equal in magnitude

Consider a two body system say the earth with mass $M$ and the moon with mass $m$ and distance $d$ between them. Thus there is a point in space between the earth and the moon where the force of gravity on an arbitrary object is equal in magnitude but opposite in direction. Similarly on the other side of the moon, there will be a point in space where the force of gravity on an arbitrary object is equal in magnitude but and identical in direction.

Question:

1. What is the locus of all points where the magnitude of the gravitational force on an arbitrary object due to the earth and the moon is equal in magnitude regardless of the direction. Will it be an ellipse with the moon at one focus?

2. What will be the path of motion of an object if it suddenly pops into existence with zero initial velocity on the above locus? Will it revolve around the larger body or will it revolve around the smaller body or will it revolve around their common center of mass or will it float aimlessly in space?

3. How will the above answers change if we consider the two body system to be an isolated system with no external gravitational influence

• Checkout Lagrangian points on wiki. Oct 12, 2016 at 5:09
• Closely related: physics.stackexchange.com/questions/147908/… Oct 12, 2016 at 6:37
• How do you want to stop the two bodies coming together? I ask, because the angular momentum of the Earth-Moon system is an essential component of the real dynamics. Oct 12, 2016 at 7:30
• What do you think? Please show your attempt to answer this question. Oct 12, 2016 at 15:01

Gravitational field is proportional to mass of object and inversely proportional to the square of distance to object.

$$G*M/R^2 = G*m/r^2 => R^2/r^2 = const => R/r = const$$

I remember from electrostatic course that the locus of such points is a sphere (or a plain iff $R=r$). It's not difficult to prove. Let $x, y$ be coordinates of a point, $M$ located as $(0, 0)$, $m$ located at $(0, a)$.

Square of distance to $M$: $x^2+y^2$

Square of distance to $m$: $(x-a)^2+y^2$

Now we have: $(x-a)^2+y^2 = C*x^2 + C*y^2$

And now we have an equation of the required locus of points: $$(C-1)*x^2 + (C-1)*y^2 = a^2 - 2*a*x$$

This must be a circle (because coefficients of $x^2$ and $y^2$ are equal). In 3-D space this would be a sphere. The sphere encloses the smaller body, but the body is not in it's center.

So, your first question was relatively simple. Question 2 seems to be much more difficult. Some clarifications are necessary: are the two massive bodies "fixed" in space or are they moving as well like Moon and Earth? If yes, is the third body large enough to disturb the motion of the first two ones? Each "yes" makes the problem more difficult. But I do not think there is a nice simple answer even in the most simple case (bodies are fixed).

And I do not understand Question 3. Wasn't this system isolated from external forces all the time?

• This ignores the fact that the Earth-Moon system has (and must have) angular momentum. Oct 12, 2016 at 6:38
• Well, yes. But Question 1 asks exactly this: what is the locus of points where magnitudes of gravitational forces are equal. We do not need to know anything about momentum to calculate gravitational forces. Oct 12, 2016 at 6:48

Two bodies (like Earth and moon) can't be fixed if they are influenced by gravity force each other. They must orbit around a Mass Center. In this case this point are the Lagragian ones and they take into account also centrifugal force. If you want to find the position of L1 L2 L3 I've found a beautiful explanation here: (the page is in Italian, but there isn't so much text, so you can translate it simply with a online translator)

http://it.giocandoconlagravita.wikia.com/wiki/I_Punti_di_Lagrange_L1,_L2_e_L3

• well, theoretically they CAN be fixed even if there is a gravitational force between them. Just use long enough rod with zero mass. Of course if would be difficult if masses of objects are huge like in case of Moon and Earth. Oct 12, 2016 at 7:18