A GPS Satellite revolces around the earth ina geostationary orbit. ( It always s

Question

A GPS Satellite revolces around the earth ina geostationary orbit. ( It always seems to be in the same place in the sky)

a) Use newtons law of gravitation to show how the period of rotation of any satellite dpends on the distance from the center of the earth.

b) Calculate the distance from the geostationary satellite to the center of the earth, as well as the distance to the surface of the earth.

c)Now suppose that we launch a small rocket from the satellite and we want it to rise to 100,000,000 meters above the surface of the earth before it stops and returns to the earth. what speed must we impart to the rocket?

Explanation / Answer

a)

Here , let the time period is T ,

now, for rotation of satellite

Gm1 * M /r^2 = m1 * (2pi/T)^2 * r

G*M/r^3 = 4 pi^2 /T^2

T = sqrt(4pi^2 * r^3 /(G*M))

where T is the time period of the satellite.

Please post other questions in seperate posts

b)

Now, for geostationary satellite ,

24 * 3600 = sqrt(4 *3.141^2 * r^3/(6.673 *10^-11 * 5.98 *10^24))

solving for r

r = 4.226 *10^7 m

the distance of satellite from the centre of earth is 4.226 *10^7 m

distance from surface = 4.226 *10^7 - 6371 *10^3

distance from surface = 3.589 *10^7 m

the distance of satellite from surface is 3.589 *10^7 m

c)

Here ,

let the speed is u m/s

using conservation of energy ,

- G * M * m /R + 0.5 * m * u^2 = -G*M*m/r

-6.673 *10^-11 * 5.98 *10^24 / (6371 *10^3) + 0.5 * u^2 = -6.673 *10^-11 * 5.98 *10^24 / (6371 *10^3 + 100 *10^6)

solving for u

u = 10852 m/s

the speed of satellite is 10852 m/s

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