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Earth orbiting satellite will escape if...

Earth orbiting satellite will escape if

A

its speed is increased by 41 %

B

its KE is doubled

C

Both (a) and (b) are correct

D

Both (a) and (b) are wrong

Text Solution

AI Generated Solution

The correct Answer is:
To determine the conditions under which an Earth-orbiting satellite will escape from its orbit, we need to understand the concepts of escape velocity and orbital velocity. ### Step-by-Step Solution: 1. **Understanding Orbital Velocity**: The orbital velocity (\(v_o\)) of a satellite is the speed required to maintain a stable orbit around the Earth. It is given by the formula: \[ v_o = \sqrt{\frac{GM}{R}} \] where \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the distance from the center of the Earth to the satellite. 2. **Understanding Escape Velocity**: The escape velocity (\(v_e\)) is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body without any additional propulsion. It is given by: \[ v_e = \sqrt{2GM/R} \] This means that the escape velocity is \( \sqrt{2} \) times the orbital velocity: \[ v_e = \sqrt{2} \cdot v_o \] 3. **Condition for Escape**: For a satellite to escape the gravitational pull of the Earth, its velocity must be equal to or greater than the escape velocity. This can be expressed as: \[ v \geq v_e \] where \(v\) is the current velocity of the satellite. 4. **Increasing Orbital Velocity**: To escape, the satellite must increase its orbital velocity. The increase required to reach escape velocity can be calculated as: \[ \Delta v = v_e - v_o = \sqrt{2} \cdot v_o - v_o = ( \sqrt{2} - 1 ) \cdot v_o \] This implies that the satellite's speed must be increased by approximately 41% (since \( \sqrt{2} \approx 1.414 \)). 5. **Kinetic Energy Consideration**: The kinetic energy (\(KE\)) of the satellite is given by: \[ KE = \frac{1}{2} mv^2 \] If the velocity increases to escape velocity, the kinetic energy will increase by a factor of 2, since: \[ KE_{escape} = \frac{1}{2} m v_e^2 = \frac{1}{2} m (2 v_o^2) = 2 \cdot KE_{orbital} \] 6. **Conclusion**: Therefore, an Earth-orbiting satellite will escape if its speed is increased by 41% (to reach escape velocity) and this will result in the kinetic energy increasing by a factor of 2. ### Final Answer: An Earth-orbiting satellite will escape if its speed is increased by approximately 41% to reach escape velocity, resulting in a doubling of its kinetic energy.

To determine the conditions under which an Earth-orbiting satellite will escape from its orbit, we need to understand the concepts of escape velocity and orbital velocity. ### Step-by-Step Solution: 1. **Understanding Orbital Velocity**: The orbital velocity (\(v_o\)) of a satellite is the speed required to maintain a stable orbit around the Earth. It is given by the formula: \[ v_o = \sqrt{\frac{GM}{R}} ...
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Knowledge Check

  • A satellite is revolving in a circular orbit at a height 'h' from the earth's surface (radius of earth R). The minimum increase in its orbital velocity required, So that the satellite could escape from the earth's gravitational field, is close to :(Neglect the effect of atomsphere.)

    A
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    D
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    A
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    B
    (i) and (ii)
    C
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    D
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