Imagine a light planet revolving around a very massive star in a circular orbit of radius $R$ with a period of revolution $T$. If the gravitational force of attraction between the planet and the star is proportional to $R^{-5/2}$, then,
$T^2 \propto R^2$
$T^2 \propto R^{7/2}$
$T^2 \propto R^{3/2}$
$T^2 \propto R^{3.75}$
In order to shift a body of mass $m$ from a circular orbit of radius $3R$ to a higher radius $5R$ around the earth, the work done is
The height at which the weight of a body becomes $\frac{1}{9} ^{th}$ its weight on the surface of earth (radius of earth is $R$)
A tunnel is dug along a diameter of the earth. If $M_e$ and $R_e$ are the mass and radius of the earth respectively. Then the force on a particle of mass $'m'$ placed in the tunnel at a distance $r$ from the centre is
The condition for a uniform spherical mass m of radius r to be a black hole is [ $G$ = gravitational constant and $g$ = acceleration due to gravity]
Figure shows the orbit of a planet $P$ round the sun $S.$ $AB$ and $CD$ are the minor and major axes of the ellipse.
If $t_1$ is the time taken by the planet to travel along $ACB$ and $t_2$ the time along $BDA,$ then