Explanation
Acceleration due to gravity at a height $$'h'$$ is given by:-
$$'g'=g\left ( \dfrac{R}{R+H} \right )^{2}$$
where $$g\rightarrow $$ on the surface of the earth.
$$R\rightarrow $$ Radius of earth.
As $$g'=g/9,$$ we get :
$$\Rightarrow \dfrac{g}{9}=g\left ( \dfrac{R}{R+H} \right )^{2}$$
$$\Rightarrow \dfrac{R}{R+H}=\dfrac{1}{3}\Rightarrow h=2R$$
Explanation:
The product of weight and the height to which it has been elevated above the surface of the earth is gravitational potential energy.
The expression of gravitational potential energy is defined as,
$$U=mgh$$
The variation of g with depth is given by
$${{g}^{'}}=\dfrac{g}{1-\dfrac{d}{R}}\,$$ where d is depth
At a depth half way to the centre of the earth $$d=\dfrac{R}{2}$$
$$ {{g}^{'}}=\dfrac{g}{1-\dfrac{R}{2R}} $$
$$ g'=\dfrac{g}{2} $$
On multiplying both sides by m
$$ mg'=\dfrac{mg}{2} $$
$$ {{W}^{'}}=\dfrac{W}{2} $$
At the depth half way, weight becomes half.
At what altitude (h) above the earth's surface would the acceleration due to gravity be one fourth of its value at the earths surface?
Please disable the adBlock and continue. Thank you.