Gravitation

JEE Main level, single correct, medium

116 of 223

Question 116 of 223

The gravitational potential at a point above the surface of earth is −5.12×107Jkg−1-5.12\times {10}^{7}J{kg}^{-1} and the acceleration due to gravity at that point is 6.4ms−26.4m{s}^{-2}. Assume that the mean radius of earth to be 6400km6400km. The height of this point above the earth's surface is:

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Answer: (A) 1600km1600km

Gravitational potential at a distance rr from the centre of the earth is given by, V=−GMrV=\dfrac{-GM}{r}.
Therefore, we can write
−GMERE+h=−5.12×107…(i)-\dfrac{G{M}_{E}}{{R}_{E}+h}=-5.12\times {10}^{7}\ldots \left(i\right)
also acceleration due to gravity at that point will be
GME(RE+h)2=6.4…(ii)\dfrac{G{M}_{E}}{{\left({R}_{E}+h\right)}^{2}}=6.4\ldots \left(ii\right)
By (i) and (ii), we get
−GMERE+hGME(RE+h)2=5.12×1076.4\dfrac{-\dfrac{G{M}_{E}}{{R}_{E}+h}}{\dfrac{G{M}_{E}}{{\left({R}_{E}+h\right)}^{2}}}=\dfrac{5.12\times {10}^{7}}{6.4}
⇒RE+h=80×105m=8000km⇒h=8000km−6400km=1600km\Rightarrow {R}_{E}+h=80\times {10}^{5}m=8000km \Rightarrow h=8000km-6400km=1600km

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