/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Consider a planet moving in an e... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a planet moving in an elliptical orbit around the Sun. The work done on the planet by the gravitational force of the Sun (1) is zero in any small part of the orbit (2) is zero in some parts of the orbit (3) is zero in complete revolution (4) is zero in no part of the motion

Short Answer

Expert verified
The work done by the gravitational force is zero in complete revolution. (Option 3)

Step by step solution

01

Understanding Work and Energy

To solve this problem, we need to understand that work is defined as the force acting on an object over the distance the object moves. Mathematically, it is expressed as \( W = \vec{F} \cdot \vec{d} \), where \( \vec{F} \) is the force vector and \( \vec{d} \) is the displacement vector. If these two vectors are perpendicular, then the work done is zero.
02

Gravitational Force Characteristics

The gravitational force from the Sun on the planet acts as a central force, always pointing towards the Sun. As such, it is always perpendicular to the direction of the planet's instantaneous velocity in its elliptical orbit.
03

Gravitational Force and Planet's Path

Since the gravitational force is always perpendicular to the planet's path (velocity vector), the work done by this force over any segment of the orbit is zero. This is a characteristic of conservative forces, where the potential energy is converted to other forms without any net work done over a full cycle.
04

Conclusion on Full Revolution

For an elliptical orbit, the gravitational force does not perform net work on the planet over one complete orbit. Since work is the result of force along a path and here the force is perpendicular throughout the path, the net work over a complete orbit is zero.

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Key Concepts

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

Elliptical Orbit
An elliptical orbit is a common path taken by celestial bodies, such as planets. It is not a perfect circle but an elongated curve. This shape results from the gravitational pull of a larger body, like the Sun, onto the smaller body, such as a planet.

In an elliptical orbit, the distance between the two objects changes continuously, which affects the gravitational force experienced by the orbiting body. The planet moves faster when it is closer to the Sun and slower when it is farther away. This variability in speed helps balance the different gravitational pulls at various points in the orbit.

To visualize an elliptical orbit, imagine an oval track. The Sun is located at one of the two focal points of this track, not at the center, affecting how the gravitational force interacts with the planet's motion.
Central Force
A central force is a type of force that always acts along the line connecting the centers of two objects. In the context of a planet orbiting the Sun, the gravitational force is a prime example of a central force.

  • The direction of the central force is always towards or away from the center of the force-exerting body.
  • This force is responsible for creating stable orbits, such as the elliptical orbit of a planet around the Sun.
Because the force always points towards the Sun, it is always perpendicular to the motion of the planet, which ensures that no work is done over a full orbit cycle. This is due to the work-energy principle, which states that work is only done when a force component acts along the direction of motion.
Conservative Forces
Conservative forces are forces that do not dissipate energy. Instead, they can store and convert energy, like gravitational potential energy, without any net loss in a system over time.

  • Examples include gravitational and electrostatic forces.
  • They guarantee the conservation of mechanical energy in a closed system.
In the case of a planet's orbit, gravitational force is a conservative force. It does not perform work over a complete orbit because any energy changes are perfectly balanced by the conversion between kinetic and potential energies. Thus, throughout the elliptical orbit, the planet's total mechanical energy remains constant.
Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. It is given by the formula:

\[ U = -\frac{G M m}{r} \]

where:
  • \( G \) is the gravitational constant,
  • \( M \) is the mass of the larger body,
  • \( m \) is the mass of the smaller body,
  • \( r \) is the distance between the centers of the two bodies.
As a planet moves in its elliptical orbit, its distance from the Sun changes, causing its gravitational potential energy to vary. However, any increase in potential energy when the planet moves away from the Sun results in a decrease in kinetic energy (and vice versa) so that their sum—the total mechanical energy—remains constant. Understanding gravitational potential energy is key to understanding why no net work is done in a complete orbit.

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Most popular questions from this chapter

Two satellites of the same mass are launched in the same orbit around the earth so as to rotate opposite to each other. If they collide inelastically and stick together as wreckage. the total energy of the system just after collision is (1) \(-\frac{2 G M m}{r}\) (2) \(-\frac{G M m}{r}\) (3) \(\frac{G M m}{2 r}\) (4) \(\frac{G M m}{4 r}\)

A particle of mass \(m\) is thrown in a parabolic path towards earth having mass \(M\) and radius \(R\). It comes to a greatest approach distance \(R / 2\) from surface of earth. Magnitude of total energy change required to make the particle moving in an elliptical path having eccentricity \(1 / 2\) with the greatest approach point either aphelion or perihelion are (1) \(\frac{1}{3} \frac{G M m}{R}\) (2) \(\frac{1}{6} \frac{G M m}{R}\) (3) \(\frac{1}{2} \frac{G M m}{R}\) (4) \(\frac{G M m}{R}\)

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) \(m^{-\frac{1}{2}}\)

Two astronauts have deserted their spaceship in a region of space far from the gravitational attraction of any other body. Each has a mass of \(100 \mathrm{~kg}\) and they are \(100 \mathrm{~m}\) apart. They are initially at rest relative to one another. How long will it be before the gravitational attraction brings them \(1 \mathrm{~cm}\) closer together? (1) \(2.52\) days (2) \(1.41\) days (3) \(0.70\) days (4) \(1.41 \mathrm{~s}\)

\- Consider two solid uniform spherical objects of the same density \(\rho .\) One has radius \(R\) and the other has radius \(2 R\) They are in outer space where the gravitational fields from other objects are negligible. If they are arranged with their surface touching, what is the contact force between the objects due to their traditional attraction? (1) \(G \pi^{2} R^{4}\) (2) \(\frac{128}{81} G \pi^{2} R^{4} \rho^{2}\) (3) \(\frac{128}{81} G \pi^{2}\) (4) \(\frac{128}{87} \pi^{2} R^{2} G\)

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