A rocket of mass $M$ is launched vertically from the surface of the earth with an initial speed $V$. Assuming the radius of the earth to be $R$ and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is
$\frac{R}{{\left( {\frac{{gR}}{{2{V^2}}} - 1} \right)}}$
$R\left( {\frac{{gR}}{{2{V^2}}} - 1} \right)$
$\frac{R}{{\left( {\frac{{2gR}}{{{V^2}}} - 1} \right)}}$
$R\left( {\frac{{2gR}}{{{V^2}}} - 1} \right)$
The escape velocity of a body from earth's surface is $v_e$ . The escape velocity of the same body from a height equal to $R$ from the earth's surface will be
Assume that a tunnel is dug through earth from North pole to south pole and that the earth is a non-rotating, uniform sphere of density $\rho $. The gravitational force on a particle of mass $m$ dropped into the tunnel when it reaches a distance $r$ from the centre of earth is
The magnitudes of gravitational field at distances $r_1$ and $r_2$ from the centre of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively. Then-
The orbital velocity of an artificial satellite in a circular orbit very close to earth is $v$. The velocity of a geo-stationary satellite orbiting in circular orbit at an altitude of $3R$ from earth's surface will be
A satellite $S$ is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then