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 16.0 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  

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Part B

You observe another satellite directly overhead and traveling east to west. This satellite is again overhead in 16.0 hours. How far is this satellite's orbit above the surface of the earth?

h =

  m  

Answer 1

r = (T*(GM)^0.5/2pi)^2/3
Mass of Earth = 5.97*10²⁴ kg
Radius of Earth = 6.38*10⁶ meters
G = 6.67*10^-11 N*m²/kg²

Keep in mind the orbital period is given in hours so you'll need to convert to seconds i.e. 16*60*60

ALSO keep in mind that the earth is rotating every 24 hours. So a sattelite going from west to east is going to take less than 16 hours to be directly overhead. i.e. 16/(1+ 16/24) = 9.6 hours for a full revolution

PartA) r = (9.6 x 3600 x ( 6.67*10^-11 x 5.97*10²⁴)^0.5/6.28)^2/3

distance from earth surface = (22.924294 - 6.38) x 10⁶ m = 16.5443 x 10⁶ m

PartB) T = 16/(1 - 16/24) = 48 hr

r = (48 x 3600 x ( 6.67*10^-11 x 5.97*10²⁴)^0.5/6.28)^2/3

distance from earth surface = (67.031 - 6.38) x 10⁶ m = 60.651 x 10⁶ m