Question: 41. The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is :
(1) \frac{5 \mathrm{G} m M}{6 R}
(2) \frac{2 \mathrm{G} m M}{3 R}
(3) \frac{G m M}{2 R}
(4) \frac{\mathrm{G} m M}{3 R}
Answer: Option (1)
Explanation:
The total mechanical energy of a satellite of mass m in a circular orbit of
radius r around the earth is given by E = -\frac{G M m}{2 r}.
The altitude of the satellite is 2R above the earth’s surface,
so the radius of the circular orbit is r = R + 2R = 3R.
Therefore, the total energy of the satellite in the final circular orbit is
E_f = -\frac{G M m}{2 \times 3R} = -\frac{G M m}{6R}.
Initially, the satellite is at rest on the surface of the earth.
Its initial mechanical energy is equal to its gravitational potential energy:
E_i = -\frac{G M m}{R}.
The minimum energy required to launch the satellite is the change in total energy:
\Delta E = E_f - E_i.
Substituting the values,
\Delta E = \left(-\frac{G M m}{6R}\right) - \left(-\frac{G M m}{R}\right).
\Delta E = \frac{G M m}{R}\left(1 - \frac{1}{6}\right) = \frac{5 G M m}{6R}.
Hence, the minimum energy required is \frac{5 \mathrm{G} m M}{6 R},
which corresponds to option (1).