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LuckyWell [14K]
3 years ago
8

A satellite of mass 5600 kg orbits the Earth and has a period of 6200 s.

Physics
1 answer:
Troyanec [42]3 years ago
8 0

Answer:

(a)  Radius of orbit will be =7.32\times10^6m

(b) Earth gravitational force will be =4.18\times 10^4N

(C) Height will be 0.92\times 10^6m

Explanation:

We have given

Mass of the earth, M=6\times 10^{24}kg

Mass of the satellite, m = 5600 kg

Radius of earth, R=6.4\times 10^6m

Time period T = 6200 sec

We know that \omega =\frac{2\pi }{T}=\frac{2\times 3.14}{6200}=0.00101rad/sec

Now

(a) We know that \omega ^2=\frac{GM}{R^3}

R^3=\frac{GM}{\omega ^2}  

R^3=\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{0.00101 ^2}

R^3=3.92\times 10^{20}

Radius of the orbit R=7.32\times 10^6m

(b)

Force F=\frac{GMm}{R^2}=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 5600}{(7.32\times 10^6)^2}=4.18\times 10^4N

(c)

Altitude h=radius\ of\ orbit-radius\ of\ earth=7.32\times 10^6-6.4\times 10^6=0.92\times 10^6m

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