Posted by **Roberto** on Sunday, May 8, 2011 at 1:02am.

A satellite is placed between the Earth and the Moon, along a straight line that connects their centers of mass. The satellite has an orbital period around the Earth that is the same as that of the Moon, 27.3 days. How far away from the Earth should this satellite be placed?

Physics - drwls, Saturday, May 7, 2011 at 9:04pm

What you have here is a three-body problem. You cannot use Kepler's law to solve it.

It turns out that there is a point of unstable equilbrium between the earth and the moon where the gravity forces cancel and the satellite will remain at equilbrium at that location and revolve around the earth at the same angular velocity as the moon. There are also two other points along the earth-moon line (one beyond the moon and another on the opposite side of the earth from the moon) and two others at equilateral triangle points. This set of locations is called the Lagrangian points.

Put the satellite at a point between earth and moon where distances to earth and to the moon are such that

M(moon)/d(moon)^2 = M(earth)/d(earth)^2

But to solve this equation won't I need the distance of the moon from the satllite?

- Physics -
**drwls**, Sunday, May 8, 2011 at 5:43am
No. You solve for

d(moon)/d(earth) =

sqrt[M(moon)/M(earth)] = 0.111,

and also require that

d(earth) + d(moon) = d = 3.84*10^5 km

- Physics (correction) -
**drwls**, Sunday, May 8, 2011 at 6:08am
The point in question is called Lagrangian Point L1. At that location, the gravity forces of earth and moon do NOT cancel. Instead, they act in opposite directions such that the net force provides the centripetal force to keep it revolving at the same angular speed as the moon.

You can find the required equation here:

http://www.ottisoft.com/Activities/Lagrange%20point%20L1.htm

It is 84.9% of the way from earth to moon.

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