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
$a = R{\left( {1 + \frac{{2GM}}{{aV_0^2}}} \right)^{1/2}}$
$a = R{\left( {1 + \frac{{aV_0^2}}{{2GM}}} \right)^{1/2}}$
$a = R{\left( {1 + \frac{{GM}}{{2aV_0^2}}} \right)^{ - 1/2}}$
$a = R{\left( {1 + \frac{{2GM}}{{aV_0^2}}} \right)^{ - 1/2}}$
If the distance between the centres of Earth and Moon is $D$ and mass of Earth is $81\, times$ that of Moon. At what distance from the centre of Earth gravitational field will be zero?
If the radius of earth shrinks by $1.5 \%$ (mass remaining same), then the value of gravitational acceleration changes by ......... $\%$
The value of $g$ at the surface of earth is $9.8 \,m / s ^2$. Then the value of ' $g$ ' at a place $480 \,km$ above the surface of the earth will be nearly .......... $m / s ^2$ (radius of the earth is $6400 \,km$ )
Two masses $m_1$ and $m_2$ start to move towards each other due to mutual gravitational force. If distance covered by $m_1$ is $x$, then the distance covered by $m_2$ is
What should be the angular speed of the earth, so that a body lying on the equator may appear weightlessness $(g = 10\,m/s^2, R = 6400\,km)$