Question

At what height above the surface of the earth, the value of ‘g’ is reduced to half of the value of ‘g’ on the surface of the earth?

Answer 1

Answer:

2651 km

Explanation:

Formula :

$g_{h} = g (1 + \frac{h}{R} )^{-2}$

Procedure :

Assuming that the radius of the Earth = 6400 km,

$\frac{g}{2} = g (1 + \frac{h}{R} )^{-2}$

⇒ $\frac{1}{2}=(1 + \frac{h}{R} )^{-2}$

⇒ 2 = $(1 + \frac{h}{R} )^{2}$

⇒ √2 = $1 + \frac{h}{6400}$

⇒ √2 - 1 = $\frac{h}{6400}$

∴ h = 6400(√2 - 1)

[Actually h = $R_{e}$(√2 - 1) ]

⇒ h = 2651 km Approximately.

2651 km above the surface of the Earth, the value of g would be reduced to half of the value of g on the surface of the Earth.

Derivation of the Formula :

$g_{h} = \frac{GM}{(R + h)^{2}}$

[R + h, from the centre of the Earth]

⇒ $g_{h} = \frac{GM}{[R(1+\frac{h}{R} )]^{2}}$

∴ $g_{h} = \frac{GM}{R^{2}} (1 + \frac{h}{R})^{-2}$

Thanks !