Question

If the acceleration due to gravity at a height $h$ from the surface of the earth is $96\%$ less than its value on the surface, then $h$ is (where $R$ is radius of the earth).

• A. $.5\ast R$
• B. $2R$
• C. $3R$
• D. $4R$

Answer 1

The correct option is D $4R$

the gravitational acceleration at a height $h$ above earth's surface is calculated as:

$g_h=\frac{GM}{(R+h{)}^2}$

where,

$M$ is the mass of the earth

$R$ is the radius of the earth

$G$ is the universal Gravitational constant

at the surface, $g=\frac{GM}{R^2}$

Rearranging $g_h$ using the above equations as follows

$\therefore g_h=\frac{GM}{R^2}\times \frac{R^2}{(R+h{)}^2}=g{\left(\frac{R}{R+h}\right)}^2$

In the given question, $g_h$ is $96\%$ less than its value on the surface $⟹100-96=4\%$ of its value on surface.

$\therefore g_h=0.04g$

$\therefore g{\left(\frac{R}{R+h}\right)}^2=0.04g$

$\therefore \frac{R}{R+h}=\sqrt{0.04}=0.2=\frac{1}{5}$

$\therefore 5R=R+h$

$\therefore h=5R-R=4R$

The gravitational acceleration is $96\%$ less than its value on the surface of the earth at a height of $4R$, where $R$ is the radius of the earth.