At what height, the value of ' $g$ ' is half that on the surface of the earth of radius $R$ ?

R

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2 R

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0.414 R

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0.75 R

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Solution

The correct option is B $0.414 R$
We have, $g = \frac{G M}{r^{2}}$ ; $r$ is the distance from the centre of Earth.
$\Rightarrow g \propto \frac{1}{r^{2}}$ above the surface of the earth.
$\Rightarrow \frac{g_{s}}{g_{h}} = \frac{r_{h}^{2}}{r_{s}^{2}}$
Since we want the value of $g_{s}$ to be half, we replace $g_{h}$ by $g_{s} / 2$
$\Rightarrow \frac{g_{s}}{g_{s} / 2} = [\frac{R + h}{R}]^{2}$
$\Rightarrow 2 = [\frac{R + h}{R}]^{2}$
$\Rightarrow \frac{R + h}{R} = \sqrt{2}$
$\Rightarrow \frac{h}{R} = 1.414 - 1 = 0.414$
$\Rightarrow h = 0.414 R$

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At what height, the value of 'g' is half that on the surface of the earth of radius R?

The correct option is B 0.414RWe have, g=GMr2; r is the distance from the centre of Earth.⇒g∝1r2 above the surface of the earth.⇒gsgh=r2hr2sSince we want the value of gs to be half, we replace gh by gs/2⇒gsgs/2=[R+hR]2⇒2=[R+hR]2⇒R+hR=√2⇒hR=1.414−1=0.414⇒h=0.414R

At what height, the value of ' $g$ ' is half that on the surface of the earth of radius $R$ ?