Question

a ball is dropped from a height half of the earth's radius. find the value of g at this point.​

Answer 1

We will use formula:

g=GMe/(R+h)^2

Answer 2

The acceleration due to gravity at height half of the earth's radius is $4.4m/s^2$.

Explanation:

The acceleration due to gravity at height h above the Earth surface,

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

Here m is the mass of Earth and R is the radius of the Earth.

As per question,

$h=\frac{R}{2}$,

g becomes,

$g`=\frac{Gm}{(R+R/2)^2}$

OR,

$g`=\frac{4}{9} \frac{Gm}{R^2}$

$g`=\frac{4}{9}g$             (Because, $g=\frac{Gm}{R^2} =9.8m/s^2$ at earth surface.)

So,

$g`=0.444\times9.8m/s^2=4.36=4.4m/s^2$

Thus, acceleration due to gravity at height half of the earth's radius is $4.4m/s^2$.

#Learn More: acceleration due to gravity.