If the period of revolution of an artificial satellite above the earth's surface be T and the density of earth be p, then prove that p `T^(2)` is a universal constant. Also calculate the value of this constant. Given G=`6.67xx10^(-11)m^(3) kg^(-1)s^(-2)`

If the period of revolution of an artificial satellite just above the earth's surface is T and the density of earth is p, then pT^(2) is

If the period of revolution of an artificial satellite just above the earth be T second and the density of earth be rho , kg//m^(3) then (G = 6.67 xx 10^(-11) m^(3)//kg . "second"^(2) )

If a satellites is revolving close to a planet of density rho with period T , show that the quantity rho T_(2) is a universal constant.

Taking moon's period of revolution about earth as 20 days and neglecting the effect of the sun and the effect other planet on its motion, calculate its distance from the earth. Given G=6.67xx10^(-11)Nm^(-2)kg^(-2) and mass of the earth =6xx10^(24)kg .

A satellite revolves around a planet of mean density 4.38 xx 10^(3)kg//m^(3) , in an orbit close to its surface. Calculate the time period of the satellite. Take, G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .

A synchronous satellite goes around the earth one in every 24 h. What is the radius of orbit of the synchronous satellite in terms of the earth's radius ? (Given: Mass of the earth , M_(E)=5.98xx10^(24) kg, radius of the earth, R_(E)=6.37xx10^(6)m , universal constant of gravitational , G=6.67xx10^(-11)Nm^(2)kg^(-2) )

A satellite of mass m is revolving close to surface of a plant of density d with time period T . The value of universal gravitational constant on planet is given by

Give the order of the following : Gravitational constant ( 6.67xx10^(-11)N-m^(2)//kg^(2) )

A satellite is orbiting just above the surface of a planet of average density D with period T . If G is the universal gravitational constant, the quantity (3pi)/G is equal to