Question

The acceleration due to gravity becomes $\left(\frac{g}{2}\right)$ g = acceleration due to gravity on the surface of the earth) at a height equal to

• A. 4R
• B. R/4
• C. 2R
• D. R/2

Answer 1

Answer: (B)

The acceleration due to gravity $g=\frac{GM}{R^2}$ At a height h above the earth's surface, the acceleration due to gravity is $g^{\prime }=\frac{GM}{{(R+h)}^2}$ $\therefore$ $\frac{g}{g^{\prime }}={\left(\frac{R+h}{R}\right)}^2={\left(1+\frac{h}{R}\right)}^2$ $\frac{g^{\prime }}{g}={\left(1+\frac{h}{R}\right)}^{-2}$ $=\left(1-\frac{2h}{R}\right)$ But $g^{\prime }=\frac{g}{2}$ (given) $\therefore$ $\frac{g/2}{g}=1-\frac{2h}{R}$ $\frac{2h}{R}=\frac{1}{2}$ $h=\frac{R}{4}$