Question

If the period of the moon is 28 days, what is the period (in hours ("hr")) of a 62.5 kg satellite with an orbital radius of 44.0 x103 km around the Earth?

Answer 1

Using force balance on satellite

Fg = Fc

G*M*m/r^2 = m*V^2/r

V = sqrt (G*M/r)

Time period is given by

T = 2*pi/w = 2*pi*r/v

T = 2*pi*sqrt (r^3/GM)

Now Given that Period of moon = 28 days = 2419200 sec

Now we can see that Time period is directly proportional to square root of r^3, where r is orbital radius

T2/T1 = (r2/r1)^(3/2)

r1 = Orbital radius of moon = 3.85*10^8 m

r2 = 44*10^3 km = 4.4*10^7 m

T1 = 28 days = 2419200 sec

So

T2 = T1*(r2/r1)^(3/2)

T2 = 2419200*(4.4*10^7/(3.85*10^8))

T2 = 276480 sec = 76.8 hrs

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