Question

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 g₀, the value of acceleration due to gravity at the earth's surface, is

पर्याय

• `- (2mg_0R^2)/(R + h)`

• `(mg_0R^2)/(2(R + h))`

• `- (mg_0R^2)/(2(R + h))`

• `(2mg_0R^2)/(R + h)`

Answer 1

`- (mg_0R^2)/(2(R + h))`

Explanation:

A satellite's total energy is,

T.E. = `(GMm)/(2(R + h))`  ......(i)

∴ Taking the equation (i) and multiplying and dividing it by R²

T.E. = `(GMmR^2)/(2(R + h)R^2)`

∴ T.E. = `(g_0mR^2)/(2(R + h))`  .....`(because g_0 = (GM)/R^2)`