Question

The height at which, the acceleration due to gravity decreases by $36\%$ of its value on the surface of the earth, the radius of the earth being $R$, is ?

• A. $\frac{R}{6}$
• B. $\frac{R}{4}$
• C. $\frac{R}{2}$
• D. $\frac{2}{3}R$

Answer 1

The correct option is B $\frac{R}{4}$

The acceleration due to gravity at height $h$ is given by

$g^{\prime }=\frac{g{R}^2}{(R+h{)}^2}$. . . . .(1)

The acceleration due to gravity decreases at height $h$ decreases by $36$%

$g^{\prime }=g-\frac{36g}{100}=\frac{64}{100}=0.64g$. . . . . .(2)

From equation (1) and equation (2), we get

$\frac{g{R}^2}{(R+h{)}^2}=\frac{64}{100}g$

$(\frac{R}{R+h}{)}^2=(\frac{8}{10}{)}^2$

$\frac{R}{R+h}=\frac{8}{10}$

$10R=8R+8h$

$2R=8R$

$h=\frac{R}{4}$

The correct option is B.