According to Kepler’s law the time period of a satellite varies with its radius as
${T^2} \propto {R^3}$
${T^3} \propto {R^2}$
${T^2} \propto (\frac{1}{R^3})$
${T^3} \propto (\frac{1}{R^2})$
The mean radius of earth is $R$, and its angular speed on its axis is $\omega$. What will be the radius of orbit of a geostationary satellite?
A geostationary satellite is revolving around the earth. To make it escape from gravitational field of earth, is velocity must be increased ........ $\%$
Starting from the centre of the earth having radius $R,$ the variation of $g$ (acceleration due to gravity) is shown by
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 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