The value of escape velocity on a certain planet is $2\, km/s$ . Then the value of orbital speed for a satellite orbiting close to its surface is
$12\,km/s$
$1\, km/s$
$\sqrt 2\,km/s$
$2\sqrt 2\, km / s$
What should be the angular speed of the earth, so that a body lying on the equator may appear weightlessness $(g = 10\,m/s^2, R = 6400\,km)$
The orbit of geostationary satellite is circular, the time period of satellite depends on $(i)$ mass of the satellite $(ii)$ mass of the earth $(iii)$ radius of the orbit $(iv)$ height of the satellite from the surface of the earth
According to Kepler’s law the time period of a satellite varies with its radius as
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 period of a satellite, in a circular orbit near an equatorial plane, will not depend on