At what altitude (h) above the earths surface would the acceleration due to gravity be one fourth of its value at the earths surface? where, R is the radius of the earth.

Question

At what altitude (h) above the earths surface would the acceleration due to gravity be one fourth of its value at the earths surface? where, R is the radius of the earth. (A) h = R (B) h = 4R (C) h = 2R (D) n = 16R

Tardigrade Admin · Accepted Answer

Given, g'=(g/4) We know that is acceleration due to gravity at height h from the surface of the earth g'=g[(R/R +h)]2 Hence (g/4)=g[(R/R+ h)]2 (∴ g'=(g/4)) (R/R +h)=(1/2) R + h =2 R h = R

Q. At what altitude $(h)$ above the earths surface would the acceleration due to gravity be one fourth of its value at the earths surface?
where, $R$ is the radius of the earth.

h = R

h = 4R

h = 2R

n = 16R

Given, $g^{'} = \frac{g}{4}$
We know that is acceleration due to gravity at height $h$ from the surface of the earth
$g^{'} = g [\frac{R}{R + h}]^{2}$
Hence $\frac{g}{4} = g [\frac{R}{R + h}]^{2}$
$(∴ g^{'} = \frac{g}{4})$
$\frac{R}{R + h} = \frac{1}{2}$
$R + h = 2 R$
$h = R$

Q. At what altitude (h) above the earths surface would the acceleration due to gravity be one fourth of its value at the earths surface? where, R is the radius of the earth.