Gravitation

JEE Main level, single correct, medium

89 of 223

Question 89 of 223

The value of acceleration due to gravity at Earth's surface is 9.8ms−29.8m{s}^{-2} . The altitude above its surface at which the acceleration due to gravity decreases to 4.9ms−24.9m{s}^{-2}, is close to: (Radius of earth =6.4×106m=6.4\times {10}^{6}m )

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Answer: (B) 2.6×106m2.6\times {10}^{6}m

Acceleration due to gravity g=GMR2g=\dfrac{GM}{{R}^{2}} .......(1)
⇒g2=GM(R+h)2\Rightarrow \dfrac{g}{2}=\dfrac{GM}{{(R+h)}^{2}} ............(2)
From (1) and (2),
GM2R2=GM(R+h)2\dfrac{GM}{2{R}^{2}}=\dfrac{GM}{{(R+h)}^{2}}
R+h=2RR+h=\sqrt{2}R
h=0.414×6.4×106h=0.414\times 6.4\times {10}^{6}
=2.6×106m=2.6\times {10}^{6}\text{m}

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