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
${v_0}\, = \,\sqrt 2 {v_e}$
${v_0}\, = \,{v_e}$
${v_0}\, = \,\frac{{{v_0}}}{2}$
${v_e}\, = \,\sqrt 2 {v_0}$
Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to
A geostationary satellite is revolving around the earth. To make it escape from gravitational field of earth, is velocity must be increased ........ $\%$
The weight of a body on the surface of the earth is $63\, N$. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ? (in $N$)
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,
The masses and radii of the earth and the moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are distance $d$ apart. The minimum speed with which particle of mass $m$ should be projected from a point midway between the two centres so as to escape to infinity is