Question

The height at which the acceleration due to gravity decreases by 36% of its value on the surface of the earth is (Assume radius of the earth is R)

• A. $\frac{R}{4}$
• B. $\frac{R}{2}$
• C. $\frac{R}{6}$
• D. 4R

Answer 1

The correct option is A $\frac{R}{4}$

Decrease by 36% means that 64% of the original value remains which means that g’/g = 0.64.
$g^{\prime }=g{\left[\frac{R}{R+h}\right]}^{2}or\frac{g^{\prime }}{g}=\left[\frac{R}{R+h}\right]$
$\Rightarrow 0.64={\left[\frac{R}{R+h}\right]}^2$
$\Rightarrow \frac{R}{R+h}=0.8$
$\Rightarrow 0.8h=0.2R$
$\Rightarrow h=\frac{R}{4}$