What is the average orbital speed about the Earth?

Question

GIVEN: A satellite is about 2.2e4 km from the (center of the) Earth.

Answer 1

You need to calculate the force of gravity on the satellite, and then the orbital velocity corresponding to a centripetal force equal to the force of gravity.

The force of gravity is GMm/r^2, where G is the gravitational constant, 6.67 x 10^-11 N-m^2/kg^2, M is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit, which was given. You were not given the mass of the Earth, but you will need it: 5.97 x 10^24 kg. You were also not given the mass of the satellite, but you'll find that you don't need it.

The centripetal force on the satellite is equal to (mv^2)/r, where m is again the mass of the satellite, v is the velocity you are trying to find, and r is again the orbital radius.

Setting these two values equal gives you GMm/r^2 = (mv^2)/r. You can cancel m from both sides, and multiply through by r, leaving you GM/r = v^2 ==> v = sqrt(GM/r), where you were given r and G and M are known constants.

Answer 2

The force of gravity, and therefore the g factor in W = mg, diminishes as the square of the distance from the center of the Earth. Thus, the weight at r = 2.2 X 10^4 km would be w = mg (R^2/r^2) where g = 9.81 m/sec^2 at Earth's radius R ~ 6,000 km.

The satellite's weight in orbit (w) is offset by the centrifugal force F = mv^2/r [See source]; so that w = mg(R/r)^2 = mv^2/r = F. Then v^2 = gR^2/r so that v = R sqrt(g/r) is the tangential velocity of the satellite in orbit at r radius from the center of the Earth. You can do the math. (Note G in F = GmM/r^2 is not needed, nor are the masses germane.)

Answer 3

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