The binding energy of a satellite of mass m in a orbit of radius r is (R = radius of earth, g = acceleration due to gravity)

Question

The binding energy of a satellite of mass m in a orbit of radius r is (R = radius of earth, g = acceleration due to gravity) (A) (mgR2/r) (B) (mgR2/2r) (C) -(mgR2/r) (D) -(mgR2/2r)

Tardigrade Admin · Accepted Answer

The energy required to remove the satellite from its orbit around the earth to infinity is called binding energy of the satellite. It is equal to negative of total mechanical energy of satellite in its orbit.
Thus, binding energy

= - E = \frac{GM m}{2 r}

but,

g = \frac{GM}{R ^{2}} \Rightarrow GM = g R^{2}

=

BE = \frac{g m R ^{2}}{2 r}

Q. The binding energy of a satellite of mass m in a orbit of radius r is (R = radius of earth, g = acceleration due to gravity)

$\frac{m g R ^{2}}{r}$

$\frac{m g R ^{2}}{2 r}$

$- \frac{m g R ^{2}}{r}$

$- \frac{m g R ^{2}}{2 r}$

Q. The binding energy of a satellite of mass m in a orbit of radius r is (R = radius of earth, g = acceleration due to gravity)

rmgR2​

2rmgR2​

−rmgR2​

−2rmgR2​

The energy required to remove the satellite from its orbit around the earth to infinity is called binding energy of the satellite. It is equal to negative of total mechanical energy of satellite in its orbit. Thus, binding energy =−E=2rGMm​ but, g=R2GM​⇒GM=gR2 = BE=2rgmR2​

$\frac{m g R ^{2}}{r}$

$\frac{m g R ^{2}}{2 r}$

$- \frac{m g R ^{2}}{r}$

$- \frac{m g R ^{2}}{2 r}$