Question

At what height above the surface of the earth$,$ the value of $g$ becomes $64\%$ of its value on the surface is the earth? (Radius of the earth is $6400km$ )

• A. $1600km$
• B. $2650km$
• C. $3200km$
• D. $9038km$

Answer 1

The correct option is A $1600km$

Let the force of gravitational on an object of mass $m$ at an altitude $h$ from earth's surface be equal to $F.$

Radius of earth $=R_e=6400km$

and mass of earth be $M_e.$

$F=G\frac{M_{e}m}{{\left(R_e+h\right)}^2}=m{g}^{\prime }$

So$,$ $g^{\prime }=\frac{G{M}_e}{{\left(R_e+h\right)}^2}$

$g=\frac{G{M}_e}{{R_e}^2}$

$\frac{g^{\prime }}{g}={\left(\frac{R_e}{R_e+h}\right)}^2=\frac{64}{100}$

$\frac{R_e}{R_e+h}=0.8$

So the altitude $=h=\frac{R_e}{4}=1600km$

Hence,

option $\left(A\right)$ is correct answer.