A top-secret U.S. military Satellite is in geostationary around the Earth. Satel

Question

A top-secret U.S. military Satellite is in geostationary around the Earth. Satellites in geostationary orbit always to be at a fixed location in the sky relative to the planet surface.

Given:

ME (Mass of Earth) = 5.97x10^(24)kg

RE (Radius of Earth) = 6.37x10^(6)m

G= 6.67x10^(-11) m^(3)/kgs^(2)

a) Calculate the altitude h of the satellite.

b) Calculate the orbital speed of the satellite in geostationary orbit.

c) Consiter the satellite in geostationary orbit at a constant speed. Is it accelerating? If YES, calculate the geostationary satellite in magnitude of the acceleration.

d) The satellite enters surveillance-mode by switching to a low altitude orbit of radius r2. Calculate the orbital period T2 if r2=1/4 r1

e)Did the kenetic energy of the satellite rise or decreade by switching to a lower orbit?

Explanation / Answer

a) Let r is the radius of the orbit.
we know, T^2 = 4*pi*r^3/(G*Me) (Here T is time periode and r is radius of orbit)

so, r = (T^2*G*Me/(4*pi))^1/3

= ((1*24*60*60)^2*6.67*10^-11*5.98*10^24/(4*pi^2))^(1/3)

= 4.225*10^7 m

h = r - Re

= 4.225*10^7 - 6.37*10^6

= 3.59*10^7 m or 35900 km
b)

Vo = 2*pi*r/T

= 2*pi*4.225*10^7/(24*60*60)

= 3073 m/s

c) yes.

g = G*M/r^2

= 6.67*10^-11*5.97*10^24/(4.225*10^7)^2

= 0.223 m/s^2

d)

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

T2 = T1*(1/4)^(2/3)

= T1*0.397

= 24*0.397

= 9.52 hours

e) kinetic enrgy rises. Because it has to move with increasing speed.

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