Q: What will be the acceleration due to gravity at height h if h >> R. Where R is radius of earth and g is acceleration due to gravity on the surface of earth
(a) $\frac{g}{(1+\frac{h}{R})^2}$
(b) $g(1-\frac{2h}{R})$
(c) $\frac{g}{(1-\frac{h}{R})^2}$
(d) $g(1-\frac{h}{R})$
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Ans: (a)
Sol: Acceleration due to gravity near the surface of earth ,
$\displaystyle g = \frac{G M}{R^2} $
Acceleration due to gravity at the height h above the surface of earth ,
$\displaystyle g’ = \frac{G M}{(R+h)^2} $
On dividing ,
$\displaystyle \frac{g’}{g} = \frac{R^2}{(R+h)^2} $
$\displaystyle \frac{g’}{g} = \frac{R^2}{R^2(1+\frac{h}{R})^2} $
$\displaystyle \frac{g’}{g} = \frac{1}{(1+\frac{h}{R})^2} $
$\displaystyle g’ = \frac{g}{(1+\frac{h}{R})^2} $