Q39 / 50 · Manipal (MET) 2023 · Physics
The acceleration due to gravity becomes $g / 2$ ( $g=$ acceleration due to gravity on the surface of the earth) at a height equals to
- Correct answerA$\frac{R}{4}$
- B$\frac{R}{2}$
- C$\frac{R}{3}$
- D$\frac{R}{5}$
Solution
The acceleration due to gravity,
$g=\frac{G M}{R^2}\ldots$ (i)
At a height $h$ above the surface of the earth, the acceleration due to gravity is
$g^{\prime}=\frac{G M}{(R+h)^2}\ldots$ (ii)
From Eqs. (i) and (ii), we get
$\frac{g}{g^{\prime}}=\left(\frac{R+h}{R}\right)^2=\left(1+\frac{h}{R}\right)^2$
$\Rightarrow \quad \frac{g^{\prime}}{g^{\prime}}=\left(1+\frac{h}{R}\right)^{-2}=\left(1-\frac{2 h}{R}\right)^2$
Here,
$g^{\prime}=\frac{g}{2}$
$\therefore \quad \frac{g / 2}{g}=\left(1-\frac{2 h}{R}\right) \Rightarrow \frac{2 h}{R}=\frac{1}{2}$ or $h=\frac{R}{4}$
Hence, the correct answer is option: 'A'.