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

Updated On: Jun 20, 2022
  • $\frac{R}{2}$
  • $2R$
  • $\frac{R}{4}$
  • $4R$
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The Correct Option is C

Solution and Explanation

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} $
$h=\frac{R}{4}$
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Concepts Used:

Escape Speed

Escape speed is the minimum speed, which is required by the object to escape from the gravitational influence of a plannet. Escape speed for Earth’s surface is 11,186 m/sec. 

The formula for escape speed is given below:

ve = (2GM / r)1/2 

where ,

ve = Escape Velocity 

G = Universal Gravitational Constant 

M = Mass of the body to be escaped from 

r = Distance from the centre of the mass