At what approximate height from the earth's surface, acceleration due to gravity will be half the value of g? [radius of earth $R_{e} = 6.4 \times 10^{6} m]$

3200 km

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2500 km

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2650 km

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3050 km

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Solution

The correct option is C 2650 km
$\frac{g}{2} = g {(\frac{R}{R + h})}^{2}$
$\Rightarrow \frac{R}{R + h} = \frac{1}{\sqrt{2}}$
$\Rightarrow R + h = \sqrt{2} R$
$\Rightarrow h = (\sqrt{2} - 1) R = 0.414 R$
$= 0.414 \times 6400 k m$
$= 2649.6 k m$
$= 2650 k m$

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Calculate acceleration due to gravity at a height of $2000 k m$ above the Earth's surface. Mass of Earth = $6 \times 10^{24} k g$ , radius of Earth = $6.4 \times 10^{6}$ m and (Gravitational constant, G $= 6.7 \times 10^{- 11} N m^{2} k g^{- 2}$ )

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Given mass of earth is 6 × $10^{24}$ kg and mean radius of earth is 6.4 × $10^{6}$ m. Calculate the value of acceleration due to gravity (g) on the surface of the earth.

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At what approximate height from the earth's surface, acceleration due to gravity will be half the value of g? [radius of earth Re=6.4×106m]

The correct option is C 2650 kmg2=g(RR+h)2 ⇒RR+h=1√2 ⇒R+h=√2R ⇒h=(√2−1)R=0.414R =0.414×6400 km =2649.6 km =2650 km

At what approximate height from the earth's surface, acceleration due to gravity will be half the value of g? [radius of earth $R_{e} = 6.4 \times 10^{6} m]$

The correct option is C 2650 km
$\frac{g}{2} = g {(\frac{R}{R + h})}^{2}$
$\Rightarrow \frac{R}{R + h} = \frac{1}{\sqrt{2}}$
$\Rightarrow R + h = \sqrt{2} R$
$\Rightarrow h = (\sqrt{2} - 1) R = 0.414 R$
$= 0.414 \times 6400 k m$
$= 2649.6 k m$
$= 2650 k m$