The radius of a planet is R_(1) and a satellite revolve round of in a circule of radius R_(2). The time period of revolution is T. The acceleration due to the gravitation of the planet at its surface is

The radius of a planet is R_(1) and a satellite revolve round of in a

1.	The radius of a planet is R_(1) and a satellite revolve round of in a circule of radius R_(2). The time period of revolution is T. The acceleration due to the gravitation of the planet at its surface is
Answer» Solution :Orbital velocity of the SATELLITE `V_(0) = SQRT(g^(1)R_(2))` Where `g^(1) = g((R_(1))/(R_(2)))^(2)` and angular velocity `omega = (V_(0))/(R_(2)) = sqrt((g^(1))/(R_(2))) = sqrt((g(R_(1))^(2))/(R_(2)^(3)))` But time PERIOD of REVOLUTION `T = (2pi)/(omega) = 2pisqrt((R_(2)^(3))/(gR_(1)^(2)))` (or) `T^(2) = 4pi^(2)(R_(2)^(3))/(gR_(1)^(2))` (or) `g = (4pi^(2))/(T^(2)) (R_(2)^(3))/(R_(1)^(2))`