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» <html><body><p></p>Solution :Orbital velocity of the <a href="https://interviewquestions.tuteehub.com/tag/satellite-1195060" style="font-weight:bold;" target="_blank" title="Click to know more about SATELLITE">SATELLITE</a> `V_(0) = <a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(g^(1)R_(2))` <br/> Where `g^(1) = g((R_(1))/(R_(2)))^(2)` <br/> and angular velocity `omega = (V_(0))/(R_(2)) = sqrt((g^(1))/(R_(2))) = sqrt((g(R_(1))^(2))/(R_(2)^(3)))` <br/> But time <a href="https://interviewquestions.tuteehub.com/tag/period-1151023" style="font-weight:bold;" target="_blank" title="Click to know more about PERIOD">PERIOD</a> of <a href="https://interviewquestions.tuteehub.com/tag/revolution-623158" style="font-weight:bold;" target="_blank" title="Click to know more about REVOLUTION">REVOLUTION</a> `T = (2pi)/(omega) = 2pisqrt((R_(2)^(3))/(gR_(1)^(2)))` <br/> (or) `T^(2) = 4pi^(2)(R_(2)^(3))/(gR_(1)^(2))` (or) `g = (4pi^(2))/(T^(2)) (R_(2)^(3))/(R_(1)^(2))`</body></html>