Question

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

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

Answer 1

Answer: (C)

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
$g^{\prime }=\frac{GM}{(R+h{)}^2}$
$\frac{g}{g^{\prime }}={\left(\frac{R+h}{R}\right)}^2$
$={\left(1+\frac{h}{R}\right)}^2$
$\frac{g^{\prime }}{g}={\left(1+\frac{h}{R}\right)}^{-2}$
$=\left(1-\frac{2h}{R}\right)$
but $g^{\prime }\frac{g}{2}$ (given)
$\frac{g/2}{h}=1-\frac{2h}{R}$
$\frac{2h}{R}=\frac{1}{2}$
$h=\frac{R}{4}$