Question

Suppose you are at the earth's equator and observe a satellite passing directly overhead and moving from west to east in the sky. Exactly 11.0 hours later, you again observe this satellite to be directly overhead. Assume a circular orbit. How far above the earth's surface is the satellite's orbit?

You observe another satellite directly overhead and traveling east to west. This satellite is again overhead in 11.0 hours. How far is this satellite's orbit above the surface of the earth?

Answer 1

First thing, if we are assuming a circular orbit and if it moves from west to east directly overhead you are on the equator. A satellite can move overhead west to east only above the equator. Now in 11 hours the Earth has rotated 11/24 of a revolution so the satellite has made 1 11/24 rev (which equals 35/24 rev) in that 11 hours. For a period of

11(24/35) = 7.54 hrs = 27154 sec.

r^3 = T^2(GM)/4(pi)^2

r = 19.52 x 10^6 m (this is the distance from the center of the Earth)

h = r - R = 19.52 x 10^6 - 6.39 x 10^6 = 13,132 km above the Earth's Surface

For the second satellite in 11 hours you've revolves 11/24 of an orbit and since the satellite is moving in the other direction is has done only 13/24 of a revolution of the Earth. For a period of 11(24/13) = 20.31 hr = 73108 sec

Using the same formula from above

r = 37.78 x 10^6 m

h = r - R = 37.78 x 10^6 - 6.39 x 10^6 = 31391 km above the Earth

[N.B. : G = 6.67 x 10^-11

M = Earth mass = 5.972 x 10^24 kg]