Question

At what height above the surface of the earth value of acceleration due to gravity is reduced to one fourth of its value on the surface of the earth?

Answer 1

at R/2 if R is radious of earth....please mark as brainlist

Answer 2

To find,

The height above the surface of the earth at which the gravity is reduced to 1/4th of its value on the surface of the earth.

Solution,

according to the question

$g_{h}$ = 1/4 $g$ (where $g_{h}$ is gravity at height h)

at a certain height above the surface of the earth gravity is given by,

$g_{h}$ = $g$ $(R/R+h)^{2}$  

where,

$R$ → the radius of the earth

$h$ → height above the earth's surface

⇒ 1/4 $g$ = $g$ $(R/R+h)^{2}$

⇒ 1/2 = $R/R+h$

⇒2$R$ = $R$ + $h$

⇒ R = h

∴ At a height R above the surface of the earth, the value of the gravity reduces to 1/4 times the gravity at the surface.