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Chapter 5: Problem 14

In case of an orbiting satellite, if the radius of orbit is decreased (1) its KE decreases (2) its PE decreases (3) its ME decreases (4) its speed decreases

Short Answer

Expert verified
The correct option is (3) its ME decreases.

Step by step solution

01

Understand Satellite's Energy Types

For an orbiting satellite, its total mechanical energy (ME) is the sum of its kinetic energy (KE) and potential energy (PE). KE increases as speed increases, and PE is negative and depends on the radius of the orbit: \[ KE = \frac{1}{2} mv^2 \] \[ PE = -\frac{GMm}{r} \] where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, \( m \) is the mass of the satellite, and \( r \) is the radius of the orbit.
02

Effect of Decreasing Orbit Radius on Potential Energy

As the radius \( r \) decreases, since PE is inversely proportional to \( r \), the magnitude of PE \(-\frac{GMm}{r}\) becomes less negative, indicating an increase in PE (decrease of the negative value). Hence, option (2) is incorrect as PE actually 'decreases' in terms of absolute value.
03

Effect of Decreasing Orbit Radius on Kinetic Energy

When the radius \( r \) decreases, the speed \( v \) of the satellite increases due to the gravitational force pulling it closer. Therefore, the kinetic energy \( KE = \frac{1}{2} mv^2 \) increases, making option (1) incorrect.
04

Consideration of Mechanical Energy

The mechanical energy \( ME = KE + PE \) is negative and becomes more negative as the orbit is lowered, meaning more energy is needed to escape the gravitational pull. The fact that the more bound system has less energy (more negative ME) makes option (3) correct as ME actually decreases.
05

Assessing the Effect on Speed

A decreased radius results in an increased gravitational pull and thus an increase in the satellite's speed. Consequently, option (4) is incorrect since the speed increases instead of decreasing.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy (KE) is the energy of motion. For an object like a satellite, it depends on its mass and speed. The formula to calculate kinetic energy is given by:
  • \( KE = \frac{1}{2} mv^2 \)
This shows that a satellite's kinetic energy increases as its speed increases.

When the radius of the satellite's orbit becomes smaller, the satellite speeds up. This is due to gravitational forces pulling it closer to the massive celestial body around which it orbits. Increasing speed leads to an increase in kinetic energy. Thus, if you decrease the orbit's radius, the kinetic energy actually increases, not decreases.
This concept explains why a closer orbit results in higher speeds, defying the common misconception that the kinetic energy might decrease.
Potential Energy
Potential energy (PE) in gravitational contexts indicates the energy stored in an object due to its position in a gravitational field. For a satellite, potential energy is gravitational and depends on the radius of its orbit:
  • \( PE = -\frac{GMm}{r} \)
Here, \( G \) is the gravitational constant, \( M \) is the mass of the planet or star, \( m \) is the mass of the satellite, and \( r \) is the orbital radius.

This formula shows that potential energy is inversely proportional to the radius \( r \). As the satellite gets closer, or the radius decreases, the potential energy becomes less negative, which means, in absolute terms, it increases.When talking about potential energy 'decreasing,' it's essential to understand that it refers to the negativity becoming smaller. Thus, potential energy becomes more positive as a satellite moves to a closer orbit.
Mechanical Energy
Mechanical energy (ME) is the sum of kinetic energy and potential energy:
  • \( ME = KE + PE \)
For satellites, mechanical energy is typically negative because of the way potential energy operates within gravitational fields.

When a satellite's orbit decreases, both kinetic energy and the absolute magnitude of potential energy change, but the potential energy change dominates. As a result, the total mechanical energy becomes more negative, indicating that the satellite is more strongly bound in its gravitational orbit.This decrease in mechanical energy reflects the gain in kinetic energy and the loss in potential energy in absolute terms. Thus, contrary to what might seem intuitive, the system's mechanical energy decreases when the orbit lowers in radius.

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Most popular questions from this chapter

A point \(P\) is on the axis of a fixed ring of mass \(M\) and radius \(R\), at a distance \(2 R\) from the centre \(O\). A small particle starts from \(P\) and reaches \(O\) under the gravitational attraction only. Its speed at \(O\) will be (1) zero (2) \(\sqrt{\frac{2 G M}{R}}\) (3) \(\sqrt{\frac{2 G M}{R}(\sqrt{5}-1)}\) (4) \(\sqrt{\frac{2 G M}{R}\left(1-\frac{1}{\sqrt{5}}\right)}\)

Which of the following are correct? (1) An astronaut going from the earth to the Moon will experience weightlessness once. (2) When a thin uniform spherical shell gradually shrinks maintaining its shape, the gravitational potential at its centre decreases. (3) In the case of a spherical shell, the plot of \(V\) versus \(r\) is continuous. (4) In the case of a spherical shell, the plot of gravitational field intensity \(I\) versus \(r\) is continuous.

A double star is a system of two stars of masses \(m\) and \(2 m\), rotating about their centre of mass only under their mutual gravitational attraction. If \(r\) is the separation between these two stars then their time period of rotation about their centre of mass will be proportional to (1) \(r^{\frac{3}{2}}\) (2) \(r\) (3) \(m \frac{1}{2}\) (4) \(m^{-\frac{1}{2}}\)

Which of the following statements are true for a particle on the surface of the earth? (1) The linear speed is minimum at the equator. (2) The angular speed is maximum at the equator. (3) The linear speed is minimum at the poles. (4) The angular speed is \(7.3 \times 10^{-5} \mathrm{rad} \mathrm{s}^{-1}\) at the equator.

If \(g\) is the acceleration due to gravity on the earth's surface. the change in the potential energy of an object of mass \(m\) raised from the surface of the earth to a height equal to the radius \(R\) of the earth is (1) \(\frac{m g R}{2}\) (2) \(2 \mathrm{mg} R\) (3) \(m g R\) (4) \(-m g R\)
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