Question

A particle falls on earth
(i) from infinity,
(ii) from a height $10$ times the radius of earth. The ratio of the velocities gained on reaching at the earth's surface is.

• A. $\sqrt{11}:\sqrt{10}$
• B. $\sqrt{10}:\sqrt{11}$
• C. $10:11$
• D. $11:10$

Answer 1

The correct option is A $\sqrt{11}:\sqrt{10}$

$T.E=K.E+P.E=\frac{1}{2}m{v}^2+(-\frac{GMm}{r})$

At surface of earth, $T.E_s=\frac{1}{2}m{v}^2+(-\frac{GMm}{R})$ where v is the velocity attain by particle at surface.

For case 1: $T.E$ at infinity is zero.

Now apply conservation of energy-

$0=\frac{1}{2}m{v}_1^2+(-\frac{GMm}{R})⟹v_1=\sqrt{\frac{2GM}{R}}$

For case 2: At $h=10R$, $T.E_h=0+(-\frac{GMm}{11R})$ $(asr=h+R=11R)$

Now apply conservation of energy-

$(-\frac{GMm}{11R})=\frac{1}{2}m{v}_2^2+(-\frac{GMm}{R})⟹v_2=\sqrt{\frac{10}{11}\frac{2GM}{R}}$

Thus $v_1:v_2=\sqrt{11}:\sqrt{10}$