Given: G=6.67259*10^ -11 Nm^ 2 /kg^ 2 The acceleration of gravity on the surface of a planet of radius R = 7040l is 11.1 m/s ^ 2 What is the period of a satellite in circu lar orbit h = 15488kn above the surface? Answer in units of s

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  1. Given: G = 6.67259 × 10−11 Nm2/kg2The acceleration of gravity on the surface of a planet of radius R = 3430 km is 4.43
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  2. Given: G = 6.67259 × 10−11 Nm2/kg2The acceleration of gravity on the surface of a planet of radius R = 3430 km is 4.43
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