These both situation is truely shown in graph (D) A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g_0 , the value of acceleration due to gravity at the earth's surface, is

These both situation is truely shown in graph (D) A satellite of mass

1.	These both situation is truely shown in graph (D) A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g_0 , the value of acceleration due to gravity at the earth's surface, is
Answer» `(2mg_0R^2)/(R+h)` `-(2mg_0R^2)/(R+h)` `(mg_0R^2)/(2(R+h))` `-(mg_0R^2)/(2(R+h))` SOLUTION :`implies` TOTAL ENERGY of satellites revoleve AROUND earth . `U = -(GM_em)/(2r)` `:. U = - (GM_e_m)/(2(R+h)) ""...(1) "" [ :. r = R + h]` Now `g_0 = (GM_e)/R^2` `:. GM_e = g_0R^2` In equ. (1) `GM_e = g_0 R^2` `U = -(mg_0R^2)/(2(R+h))`