The value of acceleration due to gravity at E | vedclass

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(C) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R^2} = 9.8\, m\,s^{-2}$.At an altitude $h$,the acceleration

Vedclass · Accepted Answer

$2.6 	imes 10^6\, m$

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(C) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R^2} = 9.8\, m\,s^{-2}$.At an altitude $h$,the acceleration due to gravity $g'$ is given by $g' = \frac{GM}{(R+h)^2}$.We are given $g' = 4.9\, m\,s^{-2}$,which is $\frac{g}{2}$.So,$\frac{GM}{(R+h)^2} = \frac{1}{2} \frac{GM}{R^2}$.Taking the reciprocal and square root on both sides: $\frac{R+h}{R} = \sqrt{2}$.$1 + \frac{h}{R} = 1.414$.$\frac{h}{R} = 0.414$.$h = 0.414 	imes R = 0.414 	imes 6.4 	imes 10^6\, m$.$h \approx 2.65 	imes 10^6\, m$.

The value of acceleration due to gravity at Earth's surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$,is close to: (Radius of Earth $= 6.4 \times 10^6\, m$)

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