A satellite revolving in a circular equatorial orbit of radius `R = 2.0 xx 10^(4)` km from west to east appears over a certain point at the equator every `11.6 h`. From these data, calculate the mass of the earth. `(G = 6.67 xx 10^(-11) N m^(2))`

A satellite revolving in a circular equatorial orbit of radius R = 2xx10^(4) km from west to east appears over a certain point at the equator every t = 11.6 hours. Calculate the mass of the earth. The gravitational contant G= 6.67 xx 10^(-11) SI units. [Hint : omega_(app) =omega_("earth") or omega_("real") = 2pi ((1)/(T)+(1)/(t)) ]

A satelite revolving in a circular equatorial orbit from west to east appears over a certain point on the equator every 8 hours therefore it's period is

A satellite is revolving in the circular equatorial orbit of radius R=2xx10^(4)km from east to west. Calculate the interval after which it will appear at the same equatorial town. Given that the radius of the earth =6400km and g (acceleration due to gravity) =10ms^(-2)

A satellite has to revolve round the earth in a circular orbit of radius 8 xx 10^(3) km. The velocity of projection of the satellite in this orbit will be -

A satellite of mass 2000 kg is revolving around the earth in a circular orbit of radius 4xx10^(4) km. The gravitational force exerted by the satellite on the earth is ________. [Mass of the earth = 6xx10^(24)kg, G=6.67xx10^(-11)Nm^(2)//kg^(2) ]

An artificial satellite is revolving in a circular orbit at a height of 600 km from the earth's surface. If its time of revolution be 100 minutes, calculate the acceleration due to gravity acting on it towards the earth. Radius of earth = 6.4 xx 10^(6) m

A satellite whose mass is M , is revolving in circular orbit of radius r around the earth. Time of revolution of satellite is