Question

at what height the acceleration due to gravity above the earth surface would be half of its value of the surface of earth radius of earth 6400km​

Answer 1

Given:

Radius of Earth $\sf (R_e)$ = 6400 km

To Find:

Height at which the acceleration due to gravity above the earth surface would be half of its value of the surface of Earth (h)

Answer:

Acceleration due to gravity at Earth surface = g

At a height above Earth surface accelration due to gravity (g') = $\sf \dfrac{g}{2}$

Variation in value of acceleration due gravity with height:

$\boxed{ \boxed{ \bf{g' = \dfrac{g}{ \bigg[1 + \dfrac{h}{R_e} \bigg] ^{2} } }}}$

By substituting values we get:

$\rm \implies \dfrac{ \cancel{g}}{2} = \dfrac{\cancel{g}}{ \bigg[1 + \dfrac{h}{6400} \bigg] ^{2} } \\ \\ \rm \implies \bigg[1 + \dfrac{h}{6400} \bigg] ^{2} = 2 \\ \\ \rm \implies 1 + \dfrac{h}{6400} = \sqrt{2} \\ \\ \rm \implies \dfrac{h}{6400} = \sqrt{2} - 1 \\ \\ \rm \implies h = 6400( \sqrt{2} - 1) \\ \\ \rm \implies h = 6400(1.414 - 1) \\ \\ \rm \implies h = 6400 \times 0.414 \\ \\ \rm \implies h = 2649.6 \: km$

$\therefore$ Height at which the acceleration due to gravity above the earth surface would be half of its value of the surface of Earth (h) = 2649.6 km

Answer 2

g/2 = g/(1 + h/R)²

(1 + h/R)² = 2

1 + h/R = √2

h = R(√2 - 1)