According to Kepler’s law the time period of a satellite varies with its radius as
${T^2} \propto {R^3}$
${T^3} \propto {R^2}$
${T^2} \propto (\frac{1}{R^3})$
${T^3} \propto (\frac{1}{R^2})$
When a body is taken from pole to the equator its weight
In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be $(g = 10\,m{s^{ - 2}}$ and radius of earth is $6400 \,kms)$
A geo-stationary satellite is orbiting the earth at a height of $6 R$ above the surface of earth, $R$ being the radius of earth. The time period of another satellite at a height of $2.5 R$ from the surface of earth is
The orbital angular momentum of a satellite revolving at a distance $r$ from the centre is $L.$ If the distance is increased to $4r$ then the new angular momentum will be
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