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