Planet and Satellite

Practice Questions:

Question 1: difficult

The additional kinetic energy to be provided to a satellite of mass m revolving around a planet of mass M, to transfer it from a circular orbit of radius R1 to another of radius R2(R2 > R1) is :

View Answer

The additional kinetic energy required to transfer a satellite from a circular orbit of radius \( R_1 \) to \( R_2 \) is the difference in kinetic energy between the two orbits.

Kinetic energy in a circular orbit is:

\[
K = \frac{GMm}{2R}
\]

The kinetic energy difference is:

\[
\Delta K = \frac{GMm}{2R_1} - \frac{GMm}{2R_2}
\]

Simplifying:

\[
\Delta K = \frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
\]

So, the additional kinetic energy is:

\[
\Delta K = \frac{GMm}{2} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
\]

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Question 2: difficult

A satellite is launched in the equatorial plane in such a way that it can transmit signals upto \(60^\circ\) latitude on the earth. Then the angular velocity of the satellite is :

1. \(\sqrt{\frac{GM}{8R^3}}\)

2. \(\sqrt{\frac{GM}{2R^3}}\)

3. \(\sqrt{\frac{GM}{4R^3}}\)

4. \(\sqrt{\frac{3\sqrt{3}GM}{8R^3}}\)

View Answer

For a transmission coverage up to latitude \(\theta = 60^\circ\), the orbital radius \(r\) satisfies \(\cos\theta = \frac{R}{r}\). This gives \(r = 2R\). The angular velocity is \(\omega = \sqrt{\frac{GM}{r^3}} = \sqrt{\frac{GM}{8R^3}}\).

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