The ratio of binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around the earth at a height $^{'} h^{'}$ above the earth's surface is ( $R =$ radius of the earth)

\frac{2 (R + h)}{R}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{R + h}{2 R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R + h}{R}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{R}{R + h}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{2 (R + h)}{R}$
Binding energy of satellite $E_{s} = \frac{G M m}{2 (R + h)}$
Binding energy of body at rest $E_{r} = \frac{G M m}{R}$
$\frac{E_{r}}{E_{s}} = \frac{2 (R + r)}{R}$

Suggest Corrections

Similar questions

The ratio of the energy required to raise a satellite upto a height h above the surface of earth to that the kinetic energy of the satellite into the orbit there is (R= radius of earth)

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 stellite in terms of

g_{0}

(value of acceleration due to gravity at the earth's surface) is

A satellite of mass $m$ is revolving around the earth at height $R$ (radius of the earth) from the earth's surface. Its potential energy will be:

Q. Periodic time (T) of a satellite revolving above Earth’s surface at a height equal to the radius of Earth (R), is

g = 9.8 m / s^{2}

Q. What will be velocity of a satellite revolving around the earth at a height

h

above surface of earth if radius of earth is

R .

Join BYJU'S Learning Program

The ratio of binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around the earth at a height ′h′ above the earth's surface is (R= radius of the earth)

The correct option is A 2(R+h)RBinding energy of satellite Es=GMm2(R+h)Binding energy of body at rest Er=GMmRErEs=2(R+r)R

The ratio of binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of same mass revolving around the earth at a height $^{'} h^{'}$ above the earth's surface is ( $R =$ radius of the earth)

The correct option is A $\frac{2 (R + h)}{R}$
Binding energy of satellite $E_{s} = \frac{G M m}{2 (R + h)}$
Binding energy of body at rest $E_{r} = \frac{G M m}{R}$
$\frac{E_{r}}{E_{s}} = \frac{2 (R + r)}{R}$