Question

If $R$ is the radius of the Earth then the height above the Earth's surface at which the acceleration due to gravity decreases by $20\%$ is

• A. $\left(\frac{\sqrt{5}}{2}-1\right)R$
• B. $\left(\frac{\sqrt{5}}{2}+1\right)R$
• C. $\left(5\sqrt{2}-1\right)R$
• D. $\left(5\sqrt{2}+1\right)R$

Answer 1

Answer: (A)

If $g$ and $g_h$ denote the accelerations due to gravity at the Earths surface and at the height $h$ above the Earths surface respectively, then
$\frac{g_h}{g}=\frac{R^2}{{\left(R+h\right)}^2}$
But according to the question
$g_h=g-\frac{20}{100}$
$g=\frac{80}{100}$
$g=\frac{4}{5}g$
$\therefore \frac{4}{5}=\frac{R^2}{{\left(R+h\right)}^2}$ or $\sqrt{\frac{4}{5}}=\frac{R}{\left(R+h\right)}$
or $\frac{2}{\sqrt{5}}=\frac{R}{R+h}$ or $2R+2h=\sqrt{5}R$
or $2h=\left(\sqrt{5}-2\right)R$ or
$h=\frac{\left(\sqrt{5}-2\right)R}{2}$
$=\left(\frac{\sqrt{5}}{2}-1\right)R$