Gravitation

JEE Main level, single correct, medium

62 of 223

Question 62 of 223

The weight of a body on the surface of the earth is 100N100N. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

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Answer: (A) 64N64N

The acceleration due to gravity changes as one goes to a height from the surface of the earth. Let the new acceleration due to gravity be g′g'. The formula for acceleration due to gravity at a height is given by g′=gR2(R+h)2g'=\dfrac{g{R}^{2}}{{\left(R+h\right)}^{2}}
The weight of the body on surface is W=mg=100NW=mg=100N
The weight at the given height is W=mg′W=mg'.
⇒W′=100×R2(R+R4)2⇒W′=100×R225R216=64N\Rightarrow W'=100\times \dfrac{{R}^{2}}{{\left(R+\dfrac{R}{4}\right)}^{2}} \Rightarrow W'=100\times \dfrac{{R}^{2}}{\dfrac{25{R}^{2}}{16}}=64N

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