A geostationary satellite is orbiting the earth at a height of $6\,R$ above the surface of earth ($R$ is the radius of earth). The time period of another satellite at a height of $2.5\,R$ from the surface of the earth is :-
$3 \sqrt 2 \,hour$
$6 \sqrt 2\, hour$
$6\, hour$
$72\, hour$
The height at which the weight of a body becomes $1/16^{th}$, its weight on the surface of earth (radius $R$), is
If the radius of the earth were to shrink by $1\%$ its mass remaining the same, the acceleration due to gravity on the earth's surface would
If the change in the value of ' $g$ ' at a height ' $h$ ' above the surface of the earth is same as at a depth $x$ below it, then ( $x$ and $h$ being much smaller than the radius of the earth)
Escape velocity at the surface of earth is $11.2\,km/sec$ . If radius of planet is double that of earth but mean density same as that of earth then the escape velocity will be ........ $km/sec$
A satellite of mass $m$ is at a distance $a$ from $a$ star of mass $M$. The speed of satellite is $u$. Suppose the law of universal gravity is $F = - G\frac{{Mm}}{{{r^{2.1}}}}$ instead of $F = - G\frac{{Mm}}{{{r^2}}}$, find the speed of the statellite when it is at $a$ distance $b$ from the star.