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If v(e) is escape velocity and v(0), is ...

If `v_(e)` is escape velocity and `v_(0)`, is orbital velocity of satellite for orbit close to the earth's surface. Then are related by

A

`v_(o)=sqrt(2)v_(e )`

B

`v_(o)=v_(e )`

C

`v_(e )=(v_(o))/(2)`

D

`v_(e )=sqrt(2)v_(o)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to establish the relationship between escape velocity (\(v_e\)) and orbital velocity (\(v_0\)) for a satellite in a circular orbit close to the Earth's surface. ### Step-by-Step Solution: 1. **Define Orbital Velocity (\(v_0\))**: The orbital velocity for a satellite in a circular orbit close to the Earth's surface can be derived from the gravitational force acting as the centripetal force. The formula for orbital velocity is given by: \[ v_0 = \sqrt{\frac{GM}{r}} \] where \(G\) is the universal gravitational constant, \(M\) is the mass of the Earth, and \(r\) is the distance from the center of the Earth to the satellite. For a satellite close to the Earth's surface, \(r\) is approximately equal to the radius of the Earth (\(R\)): \[ v_0 = \sqrt{\frac{GM}{R}} \] 2. **Define Escape Velocity (\(v_e\))**: The escape velocity is the minimum velocity required for an object to break free from the gravitational attraction of the Earth without any additional propulsion. The formula for escape velocity is given by: \[ v_e = \sqrt{\frac{2GM}{R}} \] 3. **Relate Escape Velocity to Orbital Velocity**: Now, we can relate \(v_e\) to \(v_0\) by substituting the expressions we derived: \[ v_e = \sqrt{\frac{2GM}{R}} \quad \text{and} \quad v_0 = \sqrt{\frac{GM}{R}} \] To find the relationship, we can express \(v_e\) in terms of \(v_0\): \[ v_e = \sqrt{2} \cdot \sqrt{\frac{GM}{R}} = \sqrt{2} \cdot v_0 \] 4. **Final Relationship**: Therefore, we conclude that: \[ v_e = \sqrt{2} \cdot v_0 \] This shows that the escape velocity is \(\sqrt{2}\) times the orbital velocity for a satellite in a circular orbit close to the Earth's surface. ### Conclusion: The relationship between escape velocity and orbital velocity for a satellite close to the Earth's surface is: \[ v_e = \sqrt{2} \cdot v_0 \]

To solve the problem, we need to establish the relationship between escape velocity (\(v_e\)) and orbital velocity (\(v_0\)) for a satellite in a circular orbit close to the Earth's surface. ### Step-by-Step Solution: 1. **Define Orbital Velocity (\(v_0\))**: The orbital velocity for a satellite in a circular orbit close to the Earth's surface can be derived from the gravitational force acting as the centripetal force. The formula for orbital velocity is given by: \[ v_0 = \sqrt{\frac{GM}{r}} ...
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