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 -

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 - (A) (mg0 R2/2(R - h)) (B) -(mg0 R2/2(R + h)) (C) (2mg0 R2/2(R + h)) (D) - (2 mg0 R2/(R + h))

Tardigrade Admin · Accepted Answer

T.E = - (GMm/2r) ∴ g0 = gsurface = (GM/R2) GM = g0R2 T.E = - (g0R2m/2(R+h))

Q. 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 -

$\frac{m g _{0} R ^{2}}{2 ( R - h )}$

$- \frac{m g _{0} R ^{2}}{2 ( R + h )}$

$\frac{2 m g _{0} R ^{2}}{2 ( R + h )}$

$- \frac{2 m g _{0} R ^{2}}{( R + h )}$

$T . E = - \frac{GM m}{2 r}$
$∴ g_{0} = g_{s u r f a ce} = \frac{GM}{R ^{2}}$
$GM = g_{0} R^{2}$
$T . E = - \frac{g _{0} R ^{2} m}{2 ( R + h )}$

Q. 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 -

2(R−h)mg0​R2​

−2(R+h)mg0​R2​

2(R+h)2mg0​R2​

−(R+h)2mg0​R2​