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?

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

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.

Yes, you are correct. In order to solve the equation M(moon)/d(moon)^2 = M(earth)/d(earth)^2, you will need the distance of the Moon from the satellite. Let's call this distance "d(satellite-moon)".

To find the distance of the satellite from the Earth, we can use the fact that the satellite has the same orbital period as the Moon, which is 27.3 days.

We know that the orbital period of a satellite is related to its orbital radius (distance from the center of the Earth) by the formula:

T = 2π√(R^3/GM(earth))

where T is the orbital period, R is the orbital radius, G is the gravitational constant, and M(earth) is the mass of the Earth.

For the Moon, we have:

27.3 days = 2π√(d(moon)^3/GM(earth))

For the satellite, we have:

27.3 days = 2π√(d(satellite-earth)^3/GM(earth))

Since the satellite and the Moon have the same orbital period, we can equate these two equations:

2π√(d(moon)^3/GM(earth)) = 2π√(d(satellite-earth)^3/GM(earth))

Simplifying and rearranging the equation, we get:

d(satellite-earth)^3 = (d(moon)^3)/1

Taking the cube root of both sides, we get:

d(satellite-earth) = (d(moon))/1^(1/3)

Therefore, the distance of the satellite from Earth should be the cube root of the distance of the Moon from the satellite.

Yes, to solve that equation, you will need the distance of the Moon from the satellite. To find this distance, you can use the concept of equilibrium points in the three-body problem. In this case, you are looking for the point of unstable equilibrium between the Earth and the Moon, where the gravitational forces cancel each other out and the satellite can remain at equilibrium.

This point is known as the L1 Lagrangian point, and it lies along the line connecting the centers of mass of the Earth and the Moon. To calculate the distance from the Earth to the L1 point, you can use the equation:

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

In this equation, M(moon) and M(earth) are the masses of the Moon and Earth respectively, and d(moon) and d(earth) are the distances of the Moon and Earth from the L1 point respectively.

To solve this equation, you will need to know the values of the masses of the Moon and Earth. Once you have those values, you can rearrange the equation to solve for the distance of the Earth from the L1 point.