Part A Suppose you are at the earth\'s equator and observe a satellite passing d

Question

Part A

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 12.0 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?

(Answer in Meters)

Part B

You observe another satellite directly overhead and traveling east to west. This satellite is again overhead in 12.0 hours. How far is this satellite's orbit above the surface of the earth?

(Answer in Meters)

Explanation / Answer

Earths mass (m) = 5.98 e24 kg
G (a constant) = 6.673 e-11
Earths surface radius (R) = 6.371 e6 meters
r = satellite orbital radius in meters

A.
Earths sidereal rotation rate = 7.2921 e-5 rad/sec

Earths rotation in 12 hours ( 43200 sec ) = 43200 * 7.2921 e-5 = 3.150 radians

Rotation by satellite to catch up = ( 2 * pi ) + 3.150 = 9.433 radians

Sattelite rotation rate required = 9.433 / 43200 = 2.184 e-4 rad / sec

Equation build :
rad/sec = v / r , and v = square root ( G * m / r ), so :

rad /sec = ( square root ( ( G * m ) / r ) ) / r

Transpose to feature r :

r = cube root ( ( G * m ) / ( rad / sec ² ) )

r = 2.03e7 meters (satellite orbital radius)

Satellite altitude = r - R = 1.393e7 meters (ANSWER)

B.
Earths sidereal (360°) rotation time = 23.93446 h = 86,164 seconds

Earths rotation in 12 hours = ( 2 * pi ) * ( 43200 / 86,164 ) = 3.150 radians

Rotation of satellite = ( 2 * pi ) - 3.4127 = 3.133 radians

Rotation rate of satellite = 3.133 / 43200 = 7.25 e-5 rad / sec

r = cube root ( ( G * m ) / ( rad / sec ² ) )
r = 4.233 e7 meters

Altitude = r - R = 3.596 e7 meters (ANSWER)

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