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Q. The orbital speed of an artificial satellite very close to the surface of the earth is $ {{v}_{o}} $ . Then the orbital speed of another artificial satellite at a height equal to the three times the radius of the earth is:

KEAMKEAM 2001

Solution:

$ {{v}_{o}}=\sqrt{\frac{GM}{R}} $
$ v_{o}^{}=\sqrt{\frac{GM}{R+3R}}=\sqrt{\frac{GM}{4R}} $
$ \Rightarrow $ $ v_{o}^{}=\frac{1}{2}{{v}_{o}}=0.5\,{{v}_{o}} $