Changing Gravity

Classical Mechanics Level 2

What is the value of acceleration due to gravity at a height = R 2 = \dfrac R2 from surface of the earth where R R is the radius of earth ? Take the gravitational acceleration on surface of the earth g = 9.8 m s 2 g = 9.8 \dfrac{m}{s^{2}}

0 4.35 m s − 2 ms^{-2} 6.53 m s − 2 ms^{-2} 4.9 m s − 2 ms^{-2}

Avadhoot Sinkar by Avadhoot Sinkar , 22, India

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2 solutions

g h = G M ( R + h ) 2 g_{h} = \frac{GM}{(R+h)^{2}} where G is the universal gravitational constant, R is the radius of the earth, M is the mass of the earth.
∴ g h α 1 ( R + h ) 2 \therefore g_{h} \alpha \frac{1}{(R+h)^{2}}
g R 2 = g 0 R 2 ( 3 R 2 ) 2 = 4 ⋅ 9.8 9 ≈ 4.35 g_{\frac{R}{2}} = g_{0}\frac{R^{2}}{\left(\frac{3R}{2}\right)^{2}} = \dfrac{4\cdot9.8}{9} \approx 4.35

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Avadhoot Sinkar
Feb 19, 2016

The only trap here is that here you cannot directly use the formula g'=g(1-2h/R) which is obtained by binomial approximation when h<<<R. We should use g'=g/(1+h/R)^2 so we get the answer.

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