Question

10-3. A 639-kg satellite is in a circular orbit about Earth at a height h = 1.16 x 10^7 m above the Earth’s surface. Find (a) the gravitational force (N) acting on the satellite,

(b) the satellite’s speed (m/s) (magnitude of its velocity, not its angular velocity), and (c) the period (h) of its revolution. Caution: The radius of the satellite’s orbit is not just its height above the Earth’s surface. It also includes the radius of the Earth. The mass of the Earth is 5.98 × 1024 kg, and the radius of the Earth is 6.38 × 106 m. Give the answer to part (c) to two significant figures.

Answer 1

a)

F = G M m / ( r + h) ^2

F = 6.67*10^-11* 639*5.98*10^24 / ( 6.38*10^6 + 11.6*10^6)^2

F = 788.4 N

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b)

orbital speed of satellite

Vo^2 = GM / ( r + h)

Vo^2 = 6.67*10^-11* 5.98*10^24 / ( 11.6*10^6 + 6.38*10^6)

Vo = 4710 m/s

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c)

t = 2 pi ( r +h) / vo

t = 2*3.14* ( 6.38 + 11.6)*10^6 / 4710

t = 6.7 h

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