Mastering Physics Solutions: Orbital Speed of a Satellite

Orbital Speed of a Satellite

Part A = At a distance greater than r
Part B = Tm is greater than Te

Solution Below:

Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed v at a distance r from the center of the earth. The second satellite travels at a speed that is less than v.

Part A

At what distance from the center of the earth does the second satellite orbit?

  • 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

Now assume that a satellite of mass m is orbiting the earth at a distance r from the center of the earth with speed Ve. An identical satellite is orbiting the moon at the same distance with a speed Vm . How does the time Tm it takes the satellite circling the moon to make one revolution compare to the time Te 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

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