**Orbital Speed of a Satellite**

Part A = At a distance greater than r

Part B = Tm is greater than Te

**Solution Below:**

Part A

- At a distance that is less than r
- At a distance equal to r
**At a distance greater than r**

Remember that v = sqrt(GM/r). Therefore, if v decreases, r must increase. In case you forgot, we can derive v = sqrt(GM/r) by equalizing gravitational and centripetal force:

mv^2/r = GmM/(r^2)

v^2/r = GM/(r^2)

v^2 = GM/r

v = sqrt(GM/r)

At a distance greater than r

Part B

_{e}. An identical satellite is orbiting the moon at the same distance with a speed V

_{m}. How does the time T

_{m}it takes the satellite circling the moon to make one revolution compare to the time T

_{e}it takes the satellite orbiting the earth to make one revolution?

- Tm is less than Te
- Tm is equal to Te
**Tm is greater than Te**

Remember that v = 2πr/T, where T is the period of orbit.

Remember that mv^2/r = GmM/(r^2) (See Part A). We can simplify to v = sqrt(GM/r).

So we have:

v = 2πr/T = sqrt(GM/r)

2πr/T = sqrt(GM/r)

GM/r = (2πr)^2/T^2

T^2 * GM/r = (2πr)^2

T^2 = (2πr)^2 * r / GM

T^2 = (2π)^2 * r^3 / GM

T = 2πr * r^(3/2) / sqrt(GM)

We can see that as the mass increases, the orbital period, T, gets smaller. Since the moon has a smaller mass than earth, Tm must be greater than Te. You can also rearrange the above to give you:

T^2 = (4π^2 / GM) * r^3

This is Kepler’s 3rd law of planetary motion. Your class may rearrange the terms slightly, or incorporate a constant, i.e., T^2 = kr^3, where k = (4π^2 / GM).

Tm is greater than Te