50 points! a 9.0 x 10^3 kg satellite orbits the earth at the distance of 2.56 x 10^7 m from earth’s surface. what is its period? showing work is required! !

From Kepler's 3rd law, time period of satellite orbiting around planet in a circular orbit is given by,
T² = 4π²r³/GM 
M= mass of the Earth = 6×10²⁴ Kg. 
∴ T² = 4×3.14²×(2.56×10⁷)³/6.67×10⁻¹¹×6×10²⁴ 
   T² = 1653339719
∴ T = 40661.28 seconds.

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