Question

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 :

• A. \(\frac{5GmM}{6R}\)
• B. \(\frac{2GmM}{3R}\)
• C. \(\frac{GmM}{2R}\)
• D. \(\frac{GmM}{3R}\)

Answer 1

The Correct Option is A

Minimum Energy Required to Launch a Satellite 

The minimum energy required to launch a satellite into orbit is equal to the work done in moving the satellite from the Earth's surface to its orbit. This energy is calculated based on the change in gravitational potential energy and kinetic energy.

Step 1: Gravitational Potential Energy at the Earth's Surface

The gravitational potential energy \( U_{\text{initial}} \) of a satellite of mass \( m \) at the Earth's surface is:

$$ U_{\text{initial}} = -\frac{G m M}{R} $$

\( G \) = Gravitational constant

\( m \) = Mass of the satellite

\( M \) = Mass of the Earth

\( R \) = Radius of the Earth

Step 2: Gravitational Potential Energy at the Orbit

The altitude of the orbit is \( 2R \), so the distance from the center of the Earth to the satellite is \( 3R \). The gravitational potential energy \( U_{\text{final}} \) at this orbit is:

$$ U_{\text{final}} = -\frac{G m M}{3R} $$

Step 3: Work Done in Moving the Satellite

The work done, which is the minimum energy required to move the satellite, is the change in potential energy:

$$ \Delta U = U_{\text{final}} - U_{\text{initial}} $$

Substituting the values:

$$ \Delta U = -\frac{G m M}{3R} - \left(-\frac{G m M}{R} \right) $$

Rewriting:

$$ \Delta U = \frac{G m M}{R} - \frac{G m M}{3R} $$

$$ \Delta U = \frac{2 G m M}{3R} $$

Step 4: Total Energy of the Satellite in the Orbit

The total mechanical energy of the satellite in orbit is the sum of its potential energy and kinetic energy. The total energy \( E_{\text{total}} \) is given by:

$$ E_{\text{total}} = -\frac{G m M}{6R} $$

Step 5: Minimum Energy Required to Launch the Satellite

The total minimum energy required to launch the satellite is the sum of the work done in lifting the satellite and the total energy required to keep it in orbit:

$$ \text{Minimum Energy} = \Delta U + E_{\text{total}} $$

Substituting the values:

$$ \text{Minimum Energy} = \frac{2G m M}{3R} + \left(-\frac{G m M}{6R} \right) $$

Rewriting with a common denominator:

$$ \text{Minimum Energy} = \frac{4G m M}{6R} - \frac{G m M}{6R} $$

$$ \text{Minimum Energy} = \frac{5G m M}{6R} $$

Conclusion

The minimum energy required to launch the satellite into orbit is:

$$ \frac{5G m M}{6R} $$