The acceleration due to gravity at height h above the earth if h R (radius of earth) is given by

(B) g^{\prime}=g\left(1-\frac{2 h^2}{R^2}\right)

(C) g^{\prime}=g\left(1-\frac{h}{2 R}\right)

(D) g^{\prime}=g\left(1-\frac{h^2}{2 R^2}\right)

JEE Mains · Physics · STD 11 - Gravitation

The acceleration due to gravity at height \(h\) above the earth if \(h \ll R\) (radius of earth) is given by

Pick an option to check your answer.

Verified Solution

Correct Answer

(A) \(g^{\prime}=g\left(1-\frac{2 h}{R}\right)\)

Detailed explanation

For point outside the surface of earth

\(g =\frac{ GM }{ r ^2}\)

I = distance from center of earth

\(\Rightarrow g ( h )=\frac{ GM }{( R + h )^2} \Rightarrow g ( h )=\frac{ GM }{ R ^2\left(1+\frac{ h }{ R }\right)^2}\)

\(\Rightarrow g ( h )=\frac{ GM }{ R ^2}\left(1+\frac{ h }{ R }\right)^{-2}\)

\(\text { If } h \ll R ,\left(1+\frac{ h }{ R }\right)^{-2} \approx 1-\frac{2 h }{ R }\)

\(\Rightarrow g ( h )=\frac{ GM }{ R ^2}\left(1-\frac{2 h }{ R }\right)\)

\(\Rightarrow g ( h )= g _{\text {surface }}\left(1-\frac{2 h }{ R }\right), \frac{ GM }{ R ^2}= g _{\text {surface }}\)

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