Question

A satellite of mass 350 kg is in a circular orbit around the Earth at an altitude equal to the Earth's mean radius.

(a) Find the satellite's orbital speed.

m/s

(b) What is the period of its revolution?

min

(c) Calculate the gravitational force acting on it.

N

Answer #1

Part (a)

Consider gravitational force law and second law of motion.

Grav Force, = G.M.m/r^{2}

centripetal force = m.v^{2}/r

Gravitational force = centripetal force

We get

*v*^{2} = *G.M/r*

Here r = r_{e} + r_{e }( mean radius
of Earth)

v^{2} =
6.67x10^{-11}x5.98x10^{24}/2x6.37x10^{6}

*v* = **5.60x10 ^{3}
m/s**

(b)

Calculate time period of revolution.

Time period, T = 2.π.r/v

= 2×3.14 x 2x6.37x10^{6}/5.63x10^{3}

= **1.42x10 ^{4}**

**= 2236.67min**

(c)

Calculate gravitational force acting on it.

F' = GM.m/r^{2}

= 350
x(5.60x10^{3})^{2}/2x6.37x10^{6}

= **861.5N**

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