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Angular momentum of a planet of mass \(m\) orbiting around sun is \(J\), areal velocity of its radius vector will be (a) \(\frac{1}{2} m J\) (b) \(\frac{J}{2 m}\) (c) \(\frac{m}{2 J}\) (d) \(\frac{1}{2 m J}\)

Short Answer

Expert verified
The areal velocity of the radius vector of the planet is \(\frac{J}{2m}\). Hence, the correct answer is option (b).

Step by step solution

01

Understanding Kepler's Second Law

Kepler's second law, also known as the law of equal areas, states that a line segment joining a planet and the sun will sweep out equal areas in equal amounts of time. The implication of this law is the conservation of angular momentum and it further implies that the areal velocity of a planet is a constant.
02

Relate Angular Momentum and Areal Velocity

From Kepler's second law, we can derive that the areal velocity \(v\) is half the angular momentum \(J\) divided by the mass of the planet. Mathematically, this relationship can be expressed as \(v = \frac{J}{2m}\).
03

Select the Correct Answer

By comparing the derived expression for areal velocity to the given options, it is clear that the correct answer is \(\frac{J}{2 m}\).

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Key Concepts

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

Areal Velocity
Areal velocity is an intriguing concept that ties into the motion of celestial bodies, particularly planets orbiting a star. Think of a planet orbiting the sun, with an imaginary line connecting the center of the planet to the center of the sun. As time progresses, this line sweeps out a section of the orbit, forming an area. Areal velocity refers to the rate at which this area is swept out.

Kepler's Second Law, sometimes called the "Law of Equal Areas," highlights the unique characteristic of areal velocity: as a planet moves around its sun, it sweeps out equal areas in equal time intervals. This means that even though the speed of the planet changes as it orbits, the rate of area swept (the areal velocity) remains constant. This concept emphasizes the conservation of angular momentum in celestial mechanics.

To calculate areal velocity, you can use the formula \( v = \frac{J}{2m} \), where \( J \) is the angular momentum and \( m \) is the mass of the planet. This relationship signifies that the area swept over time is directly related to the angular momentum of the planet's orbit, making it a crucial aspect of understanding planetary motion.
Angular Momentum
Angular momentum is an essential concept in physics, especially when examining objects in rotational motion. For planets orbiting a star, like Earth around the Sun, angular momentum is particularly significant. In simple terms, angular momentum is the quantity of rotation of an object and depends on its mass, shape, and rotational velocity.

For an orbiting planet, angular momentum \( J \) can be expressed mathematically as the product of the planet's mass \( m \), its velocity \( v \), and its distance from the sun. Also, angular momentum is conserved when no external torque acts on the system. This means that the total angular momentum of a closed system remains constant over time. In the context of planetary motion, this conservation explains why suns and planets interact in a stable, predictable manner.

Understanding angular momentum is crucial for grasping more profound concepts related to orbital mechanics and how they apply universally, from small asteroids to magnificent galaxies. Without the conservation of angular momentum, the elegant and intricate dance of celestial bodies would not be possible.
Planetary Motion
The motion of planets around a star, such as the motion of the planets in our solar system around the sun, is a beautiful demonstration of celestial mechanics. The laws that govern these motions are known as Kepler's Laws of Planetary Motion. These laws describe how planets move in elliptical orbits with the sun at one focus, how they sweep equal areas in equal times (areal velocity), and how the square of the orbital period relates to the cube of the semi-major axis of the orbit.

Kepler’s efforts laid the groundwork for our current understanding of elliptical orbits, which are paths that planets follow. It shines light on how speed varies, such as moving faster when closer to the sun and slower when farther away, yet still maintaining a consistent areal velocity.
  • Elliptical orbits mean that the distance between the planet and the star varies.
  • Planets move faster when closest to the sun (perihelion) and slower when farthest from the sun (aphelion).
  • The gravitational force between a planet and star is central to planetary motion, providing the centripetal force necessary to keep them in orbit.
These fundamental principles shape our understanding not only of our solar system but of the universe as a whole. The planetary motion characteristics observed by Kepler remain a cornerstone in astronomy, defining how we study and interpret the movements of celestial bodies.

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Most popular questions from this chapter

A planet moves around the sun. At a given point \(P\), it is closed from the sun at a distance \(d_{1}\) and has a speed \(v_{1}\). At another point \(Q\). when it is farthest from the sun at a distance \(d_{2}\), its speed will be [MP PMT 1987\(]\) (a) \(\frac{d_{1}^{2} v_{1}}{d_{2}^{2}}\) (b) \(\frac{d_{2} v_{1}}{d_{1}}\) (c) \(\frac{d_{1} v_{1}}{d_{2}}\) (d) \(\frac{d_{2}^{2} v_{1}}{d_{1}^{2}}\)

Potential energy of a satellite having mass ' \(m\) ' and rotating at a height of \(6.4 \times 10^{6} \mathrm{~m}\) from the earth centre is [AIIMS 2000; CBSE PMT 2001; BHU 2001] (a) \(-0.5 m g R_{e}\) (b) \(-m g R_{e}\) (c) \(-2 m g R_{e}\) (d) \(4 m g R_{e}\)

One project after deviation from its path, starts moving round the earth in a circular path at radius equal to nine times the radius at earth \(R\), its time period will be (a) \(2 \pi \sqrt{\frac{R}{g}}\) (b) \(27 \times 2 \pi \sqrt{\frac{R}{g}}\) (c) \(\pi \sqrt{\frac{R}{g}}\) (d) \(8 \times 2 \pi \sqrt{\frac{R}{g}}\)

The gravitational potential due to the earth at infinite distance from it is zero. Let the gravitational potential at a point \(P\) be \(-5 J / \mathrm{kg}\). Suppose, we arbitrarily assume the gravitational potential at infinity to be \(+10 J / k g\), then the gravitational potential at \(P\) will be (a) \(-5 \mathrm{~J} / \mathrm{kg}\) (b) \(+5 J / k g\) (c) \(-15 J / k g\) (d) \(+15 \mathrm{~J} / \mathrm{kg}\)

A ball is dropped from a spacecraft revolving around the earth at a height of \(120 \mathrm{~km}\). What will happen to the ball (a) It will continue to move with velocity \(v\) along the original orbit of spacecraft (b) If will move with the same speed tangentially to the spacecraft (c) It will fall down to the earth gradually (d) It will go very far in the space

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