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When energy of a satellite - planet s...

When energy of a satellite - planet system is positive then satellite will

A

Move around planet in circular orbit

B

Move around planet inelliptical orbit

C

Escape out with minimum speed

D

Escape out with speed greater than escape velocity

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The correct Answer is:
To solve the problem of determining the behavior of a satellite-planet system when the energy is positive, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Energy of the Satellite-Planet System**: The total mechanical energy (E) of a satellite in orbit around a planet is given by the formula: \[ E = K + U \] where \( K \) is the kinetic energy and \( U \) is the gravitational potential energy. 2. **Gravitational Potential Energy**: The gravitational potential energy (U) of a satellite of mass \( m \) at a distance \( r \) from the center of a planet of mass \( M \) is given by: \[ U = -\frac{G M m}{r} \] where \( G \) is the gravitational constant. 3. **Kinetic Energy**: The kinetic energy (K) of the satellite is given by: \[ K = \frac{1}{2} mv^2 \] where \( v \) is the orbital speed of the satellite. 4. **Total Energy**: Therefore, the total energy of the satellite-planet system can be expressed as: \[ E = \frac{1}{2} mv^2 - \frac{G M m}{r} \] 5. **Condition for Positive Energy**: For the total energy \( E \) to be positive: \[ \frac{1}{2} mv^2 - \frac{G M m}{r} > 0 \] Rearranging gives: \[ \frac{1}{2} mv^2 > \frac{G M m}{r} \] Dividing by \( m \) (assuming \( m \neq 0 \)): \[ \frac{1}{2} v^2 > \frac{G M}{r} \] Multiplying both sides by 2: \[ v^2 > \frac{2 G M}{r} \] Taking the square root: \[ v > \sqrt{\frac{2 G M}{r}} \] 6. **Conclusion**: This means that for the energy of the satellite-planet system to be positive, the speed of the satellite must be greater than the escape velocity, which is given by: \[ v_{escape} = \sqrt{\frac{2 G M}{r}} \] Therefore, if the energy of the satellite-planet system is positive, the satellite will be moving with a speed greater than the escape velocity, indicating that it will escape the gravitational influence of the planet. ### Final Answer: When the energy of a satellite-planet system is positive, the satellite will escape from the gravitational field of the planet.
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