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 10.0 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?

h = m

Part B

You observe another satellite directly overhead and traveling east to west. This satellite is again overhead in 10.0 hours. How far is this satellite's orbit above the surface of the earth?

h= m

Explanation / Answer

Mass of earth (m) = 5.98 e24 kg
Graviattional constant (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 10 hours ( 36,000 sec ) = 36,000 * 7.2921 e-5 = 2.625 radians

Rotation by satellite to catch up = ( 2 * pi ) + 2.625 = 8.905 radians

Sattelite rotation rate required = 8.905 / 36,000 = 2.4736 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 = 1.175 e7 meters (satellite orbital radius)

Satellite altitude = r - R = 0.5379 meters (ANSWER)

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

Earths rotation in 10 hours = ( 2 * pi ) * ( 36,000 / 86,164 ) = 2.624 radians

Rotation of satellite = ( 2 * pi ) - 2.624 = 3.656 radians

Rotation rate of satellite = 3.656 / 36,000 = 10.156 e-5 rad / sec

r = cube root ( ( G * m ) / ( rad / sec ² ) )
r = 3.37 e7 meters

Altitude = r - R = 2.735 e7 meters (ANSWER)

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