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 :
(1) V11:V10 (2) V10: V11(V means root)
(3) 10:11
(4) 11 : 10​

Answer 1

Soltuion : So option 1 is correct $v_{1}: v_{2}$ = $\sqrt{11} : \sqrt{10}$

By using formula

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

So at the surface of the earth

T.E=$\frac{1}{2}$ $mv^{2}$ + ( - $\frac{GMm}{R}$

Here V = Velocity attained by the particle at surface.

So for Case 1 : From infinity

=> T.E at infinity is zero .

Now ,Apply the Formula of Conservation of energy -

0=$\frac{1}{2}$ $mv_{1} ^{2}$ + ( - $\frac{GMm}{R}$ )

=> v1 = $\sqrt{\frac{2Gm}{R} }$

ii) So For Case 2 :

At h= 10 R

$T.E_{H}$= 0+ ( - $\frac{GMm}{11R}$ )

( asr=h + R   =11 R )

Now Apply conservation of energy:-

 ( - $\frac{GMm}{11R}$ ) =$\frac{1}{2}$ $m_{2} ^{2}$ +(- $\frac{GMm}{R}$ )

=> $v^{2}$ = $\sqrt{\frac{10}{11}\frac{2GM}{R}}$

Therefore , $v_{1}: v_{2}$ = $\sqrt{11} : \sqrt{10}$