Question

Height at which the value of g becomes $\frac{1}{16}th$ of that of the earth is

• A.
  $4R$
• B.
  $3R$
• C.
  $\frac{3}{2}R$
• D.
  $2R$

Answer 1

The correct option is B

$3R$
Acceleration due to gravity on the surface of the earth, $g=\frac{GM}{R^2}...\left(i\right)$

Here, G = gravitational constant, M = mass of the earth and R = radius of the earth

Acceleration due to gravity at height h from the surface of the earth, $g_h=\frac{GM}{(R+h{)}^2}...\left(ii\right)$

We are given, $g_h=\frac{1}{16}g$

Taking ratio of (ii) and (i):

$\Rightarrow \frac{g_h}{g}=\frac{\frac{GM}{(R+h{)}^2}}{\frac{GM}{R^2}}\Rightarrow \frac{\frac{1}{16}g}{g}=\frac{R^2}{(R+h{)}^2}\Rightarrow \frac{1}{16}={\left(\frac{R}{R+h}\right)}^2\Rightarrow \frac{R}{R+h}=\frac{1}{4}\Rightarrow 4R=R+h\Rightarrow h=3R$


The height at which the value of acceleration due to gravity becomes $\frac{1}{16}th$ of that on the earth is 3R.

Hence, the correct answer is option (b).