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
$a = R{\left( {1 + \frac{{2GM}}{{aV_0^2}}} \right)^{1/2}}$
$a = R{\left( {1 + \frac{{aV_0^2}}{{2GM}}} \right)^{1/2}}$
$a = R{\left( {1 + \frac{{GM}}{{2aV_0^2}}} \right)^{ - 1/2}}$
$a = R{\left( {1 + \frac{{2GM}}{{aV_0^2}}} \right)^{ - 1/2}}$
Asatellite is launched into a circular orbit of radius $R$ around the earth. A second satellite is launched into an orbit of radius $1.02\,R.$ The period of second satellite is larger than the first one by approximately ........ $\%$
The rotation of the earth having $R$ radius about its axis speeds up to a value such that a man at latitude angle $60^o$ feels weightlessness. The duration of the day in such a case is.
The Earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$, where $v$ is its escape velocity from the surface of the Earth. the value of $f$ is
If $v_e$ is escape velocity and $v_0$ is orbital velocity of satellite for orbit close to the earth's surface. Then these are related by
A satellite is launched into a circular orbit of radius $R$ around earth, while a second satellite is launched into a circular orbit of radius $1.02\, {R}$. The percentage difference in the time periods of the two satellites is -