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 :

Question

Two planets $P_1$ and $P_2$ with equal mass have radii $R_1$ and $R_2$, respectively, where \[ R_2=\frac{R_1}{2} \] The escape speeds of $P_1$ and $P_2$ are $v_1$ and $v_2$, respectively. Then the value of \[ \frac{v_2}{v_1} \] is:

Answer 1

Minimum Energy for Satellite Launch

The minimum energy needed to place a satellite in orbit equals the work done to transfer it from Earth's surface to its orbital path. This calculation considers changes in gravitational potential and kinetic energy.

Step 1: Gravitational Potential Energy at Earth's Surface

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

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

$ G $ denotes the gravitational constant.

$ m $ is the satellite's mass.

$ M $ is the Earth's mass.

$ R $ is the Earth's radius.

Step 2: Gravitational Potential Energy in Orbit

For an orbit at an altitude of $ 2R $, the distance from Earth's center 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 Moving the Satellite

The work done, representing the minimum energy to move the satellite, is the difference in potential energy:

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

Substituting the values yields:

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

This simplifies to:

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

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

Step 4: Total Satellite Energy in Orbit

The satellite's total mechanical energy in orbit is the sum of its potential and kinetic energies. The total energy ($ E_{\text{total}} $) is calculated as:

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

Step 5: Minimum Launch Energy Calculation

The total minimum energy for launching the satellite is the sum of the energy required for lifting it and the energy to maintain its orbit:

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

Substituting the determined values:

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

Using a common denominator, this becomes:

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

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

Conclusion

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

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

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 :

The Correct Option is A

Solution and Explanation

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