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Gravitation

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Question : 19 of 25
Marks: +1, -0
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence?
Mass of the satellite = 200 kg; mass of the earth =6.0Ă—1024kg= 6.0 \times 10^{24} \text{kg};
radius of the earth =6.4Ă—106m= 6.4 \times 10^{6} \text{m}; G=6.67Ă—10−11N m2kg−2G = 6.67 \times 10^{-11} \text{N m}^{2} \text{kg}^{-2}.
Solution:  
Total energy of orbiting satellite at a height h
=−GMm(R+h)+12mv2= - \frac{G M m}{(R+h)} + \frac{1}{2} m v^{2}
=−GMm(R+h)+12mGM(R+h)= - \frac{G M m}{(R+h)} + \frac{1}{2} m \frac{G M}{(R+h)}
=−GMm2(R+h)= \frac{-G M m}{2(R+h)}
Energy expended to rocket the satellite out of the earth’s gravitational field =– (total energy of the orbiting satellite)
=GMm2(R+h)= \frac{G M m}{2(R+h)}
=(6.67Ă—10−11)Ă—(6Ă—1024)Ă—2002(6.4Ă—106+4Ă—105)= \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24}) \times 200}{2 (6.4 \times 10^{6} + 4 \times 10^{5})}
=5.9Ă—109J= 5.9 \times 10^{9} \text{J}
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