A1,160 kg satellite orbits earth with a tangential speed of 7,446 m/s. if the satellite experiences a centripetal force of 8,955 n, what is the height of the satellite above the surface of earth? recall that earth’s radius is 6.38 × 106 m and earth’s mass is 5.97 × 1024 kg.

8.02×10⁵ m

Explanation:

Equation for centripetal force:

F = mv²/r

Solving for r:

r = mv²/F

Given:

F = 8955 N

m = 1160 kg

v = 7446 m/s

r = (1160 kg) (7446 m/s)² / 8955 N

r = 7.182×10⁶ m

The height above the surface is:

h = 7.182×10⁶ m − 6.38×10⁶ m

h = 0.802×10⁶ m

h = 8.02×10⁵ m

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