Two masses m₁ and m₂ are at rest a infinity. ir relative velocity of approach due to gravitation

English

Gujarati

Hindi

7.Gravitation

medium

Two masses $m_1$ and $m_2$ are at rest a infinity. Find their relative velocity of approach due to gravitational attraction when their separation is $d$

A

$\sqrt {\frac{{G\left( {{m_1}\, + \,{m_2}} \right)}}{{2d}}} $

B

$\sqrt {\frac{{2G\left( {{m_1}\, + \,{m_2}} \right)}}{{d}}} $

C

$\sqrt {\frac{{G\left( {{m_1}\, + \,{m_2}} \right)}}{{8d}}} $

D

$81\sqrt {\frac{{G\left( {{m_1}\, + \,{m_2}} \right)}}{d}} $

Solution

By using $\frac{1}{2} \mu\left(V_{\mathrm{rel}}\right)^{2}=\frac{\mathrm{G} \mathrm{m}_{1} \mathrm{m}_{2}}{\mathrm{d}}$

Standard 11

Physics

Share

Similar Questions

Energy required to move a body of mass $m$ from an orbit of radius $2R$ to $3R$ is

medium

(AIEEE-2002)

A satellite is revolving round the earth with orbital speed $v_0$. If it stops suddenly, the speed with which it will strike the surface of earth would be : ($v_e =$ escape velocity of a particle on earth's surface)

hard

An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance $R$ to $2R$ from the centre of the earth. What is the gain in its potential energy ?

medium

A mass $m$ , travelling at speed $V_0$ in a straight line from far away is deflected when it passes near a black hole of mass $M$ which is at a perpendicular distance $R$ from the original line of flight. $a$ , the distance of closest approach between the mass and the black hole is given by the relation

hard

A satellite of mass $m$, initially at rest on the earth, is launched into a circular orbit at a height equal to the radius of the earth. The minimum energy required is

hard