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
$\frac{R}{{\left( {\frac{{gR}}{{2{V^2}}} - 1} \right)}}$
${R\left( {\frac{{gR}}{{2{V^2}}} - 1} \right)}$
$\frac{R}{{\left( {\frac{{2gR}}{{{V^2}}} - 1} \right)}}$
${R\left( {\frac{{2gR}}{{{V^2}}} - 1} \right)}$
If the gravitational acceleration at surface of Earth is $g$ , then increase in potential energy in lifting an object of mass $m$ to a height equal to half of radius of earth from surface will be
A body of mass $m$ is kept at a small height $h$ above the ground. If the radius of the earth is $R$ and its mass is $M$, the potential energy of the body and earth system (with $h = \infty $ being the reference position ) is
A geo-stationary satellite is orbiting the earth at a height of $5R$ above surface of the earth, $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of earth is
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)
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