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Gravitation

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Question : 22 of 25
Marks: +1, -0
As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinityto be zero).
Mass of the earth =6.0Ă—1024= 6.0 \times 10^{24} kg, radius = 6400 km.
Solution:  
Gravitational potential at height h from the surface of earth is
V=−GMR+hV = -\frac{G M}{R+h}
=−6.67Ă—10−11Ă—(6Ă—1024)6.4Ă—106+36Ă—106= \frac{-6.67 \times 10^{-11} \times (6 \times 10^{24})}{6.4 \times 10^{6} + 36 \times 10^{6}}
=−9.4Ă—106 J kg−1= -9.4 \times 10^{6} \,\mathrm{J}\,\mathrm{kg}^{-1}
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