Question

plzzzz explain it.....​

Answer 1

❏ Question:-

Refer to the questionar attachment ,

❏ Solution:-

We know that the escape velocity is ,

$\longrightarrow\boxed{\large{V_e=\dfrac{2GM}{R}}}$

where,

M= mass of the Earth,

R= Radius of Earth,

Now , At a very large distance from the earth surface the body will only posses Kinetic Energy,

Let , the distance at a large distance is = V .

Now it's thrown at speed of n times the escape velocity ,

So velocity at earth surface = $V_o=nV_e$

Hence,

$\therefore K.E._{\text{at a lage distace}}=(P.E.+K.E.)_{\text{at the earth surface}}$

$\implies \frac{1}{2}mV^2=-\dfrac{GMm}{R}+\dfrac{1}{2}m(V_o)^2$

$\implies \frac{1}{2}mV^2=-\dfrac{1}{2}\times\dfrac{2GM}{R}\times m+\dfrac{1}{2}m(nV_e)^2$

$\implies \frac{1}{2}mV^2=-\dfrac{1}{2}\times V^2_e\times m+\dfrac{1}{2}mn^2{V_e}^{2}$

$\implies \frac{1}{2}mV^2=-\dfrac{1}{2}mV^2_e+\dfrac{1}{2}mn^2{V_e}^{2}$

$\implies \cancel{m}\times[ \frac{1}{2}V^2]=\cancel{m}\times[-\dfrac{1}{2}V^2_e+\dfrac{1}{2}n^2{V_e}^{2}]$

$\implies \frac{1}{2}V^2=-\dfrac{1}{2}{V_e}^{2}+\dfrac{1}{2}n^2{V_e}^{2}$

$\implies V^2=-{V_e}^{2}+n^2{V_e}^{2}$

$\implies V^2={V_e}^{2}[n^2-1]$

$\implies V=\sqrt{{V_e}^{2}[n^2-1]}$

$\implies\boxed{\large{\red{ V=V_e\sqrt{n^2-1}}}}$

So , Hence

➩ option (1) .