At what distance above the surface of earth, the gravitational force will be reduced by $10 %$ , if the radius of earth is $6370$ Km.

750 K m

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650 K m

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450 K m

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344 K m

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Solution

The correct option is A $344 K m$

Gravitational force on the earth surface,

$F_{G_{s}} = \frac{G M m}{R^{2}}$

Gravitational force at a height 'h' above,

$F_{G_{h}} = \frac{G M m}{(R + h)^{2}}$

G.T. $F_{G_{h}} = \frac{90}{100} (F_{G_{S}})$

$\Rightarrow \frac{G M m}{(R + h)^{2}} = \frac{9}{10} (\frac{G M m}{R^{2}})$

$\Rightarrow R + h = \sqrt{\frac{10}{9}} (R)$

$\Rightarrow 6370 + h = \frac{3.162}{3} (6370)$

$\Rightarrow h = \frac{0.162 \times 6370}{3} = 344$ km

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At what distance above the surface of earth, the gravitational force will be reduced by 10%, if the radius of earth is 6370 Km.

The correct option is A 344Km Gravitational force on the earth surface, FGs=GMmR2 Gravitational force at a height 'h' above, FGh=GMm(R+h)2 G.T. FGh=90100(FGS) ⇒GMm(R+h)2=910(GMmR2) ⇒R+h=√109(R) ⇒6370+h=3.1623(6370) ⇒h=0.162×63703=344 km

At what distance above the surface of earth, the gravitational force will be reduced by $10 %$ , if the radius of earth is $6370$ Km.