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The angular momentum of the earth revolving round the sun is proportional to \(r^{n}\), where \(r\) is the distance between the centres of earth and the sun. The value of \(n\) is (a) 1 (b) \(-2\) (c) \(-1\) (d) \(\frac{1}{2}\)

Short Answer

Expert verified
The value of \(n\) is \(\frac{1}{2}\) (option d).

Step by step solution

01

Understand Angular Momentum Formula

Angular momentum \(L\) of an object in orbit is given by \(L = mrv\), where \(m\) is mass, \(r\) is radius, and \(v\) is velocity. Earth's angular momentum around the sun can be expressed as \(L = mrv\).
02

Relate Velocity with Gravitational Force

The gravitational force \(F_g\) provides the necessary centripetal force for earth's orbit. Using \(F_g = \frac{GMm}{r^2} = \frac{m v^2}{r}\), we can solve for velocity, \(v = \sqrt{\frac{GM}{r}}\).
03

Substitute Velocity in Angular Momentum Formula

Substitute \(v = \sqrt{\frac{GM}{r}}\) into \(L = mrv\) resulting in \(L = mr \sqrt{\frac{GM}{r}} = m\sqrt{GMr}\).
04

Simplify the Expression

Simplify further by factoring terms to get \(L = m\sqrt{G} \cdot M^{1/2} \cdot r^{1/2}\).
05

Determine the Power of r

From the expression \(L = m\sqrt{G} \cdot M^{1/2} \cdot r^{1/2}\), it is observed that angular momentum \(L\) is proportional to \(r^{1/2}\). Therefore, \(n = \frac{1}{2}\).

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Key Concepts

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

Gravitational Force
The gravitational force is a fundamental force that acts between two bodies with mass. It is described by Newton's law of gravitation, which states that every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between those two points.

Mathematically, this force is expressed as:
\[ F_g = \frac{G M m}{r^2} \]Here,
  • \(G\) is the gravitational constant,
  • \(M\) and \(m\) are the masses of the two objects,
  • \(r\) is the distance between the centers of the two objects.
Gravitational force is crucial in understanding orbital motion, as it dictates the forces that hold planets in orbit around the sun. Without it, celestial mechanics would not lead to stable orbits, and planets such as Earth would not follow the paths they do today.
Centripetal Force
When an object follows a curved path, it experiences a force called the centripetal force. This force acts inward, toward the center of the circle or path the object is following. In the context of orbital mechanics, the centripetal force required to keep an object like Earth in orbit is provided by gravity.

For an object in circular motion, this force is given by:
\[ F_c = \frac{m v^2}{r} \]Where,
  • \(m\) is the mass of the object,
  • \(v\) is the velocity of the object,
  • \(r\) is the radius of the path.
The link between gravitational and centripetal forces becomes evident since the gravitational force provides the necessary centripetal force to keep Earth in its almost circular orbit around the sun.
Orbital Mechanics
Orbital mechanics is the study of the motions of objects in space, governed by the laws of physics, particularly Newton's laws of motion and universal gravitation. In orbital mechanics, we often analyze how bodies like planets, moons, and satellites move in the vastness of space.

A key aspect of orbital mechanics is understanding how gravitational forces affect these motions. For instance, the stable orbit of Earth around the sun involves a balance between the gravitational pull of the sun and the inertia of Earth's motion.

During orbit:
  • Gravitational force acts as the centripetal force keeping celestial bodies in their paths.
  • The velocity of an orbiting object, such as a planet, determines how it balances gravitational forces to stay in orbit, following a predictable elliptical or circular path based on initial conditions.
The equilibrium between gravitational and centripetal forces implies that the gravitational force is fully accounted for by the centripetal effect to keep a planet continuing in its orbit without spiraling inwards or drifting away, an essential component in maintaining orbital stability in our solar system.

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Most popular questions from this chapter

A system of binary stars of masses \(m_{A}\) and \(m_{B}\) are moving in circular orbits of radius \(r_{A}\) and \(r_{B}\) respectively. If \(T_{A}\) and \(T_{B}\) are the time periods of masses \(m_{A}\) and \(m_{B}\) respectively, then (a) \(\frac{T_{A}}{T_{B}}=\left(\frac{r_{A}}{r_{B}}\right)^{3 / 2}\) (b) \(T_{A}>T_{B}\) (if \(\left.r_{A}>r_{B}\right)\) (c) \(T_{A}>T_{B}\) (if \(\left.m_{A}>m_{B}\right)\) (d) \(T_{A}=T_{B}\)

If suddenly the gravitational force of attraction between earth and a satellite revolving around it becomes zero then the satellite will (a) continue to move in its orbit with same velocity (b) move tangentially to the original orbit with the same velocity (c) become stationary in its orbit (d) move towards the earth

If earth rotates \(n\) time faster than its present speed \(\omega\) about its axis in order that the bodies lying on the equator of earth just fly off into the space, then the value of \(n\) is equal to (take radius of earth \(R\) ) (a) \(\omega \sqrt{g / R}\) (b) \(\frac{1}{\omega} \sqrt{\frac{g}{R}}\) (c) \(\omega \sqrt{2 g / R}\) (d) \(\omega \sqrt{g / 2 R}\)

The radius (in \(\mathrm{km}\) ) to which the present radius of the earth \((R=6400 \mathrm{~km})\) to the compressed so that the escape velocity is increased 10 times is (a) \(6.4\) (b) 64 (c) 640 (d) 4800

At a given place where acceleration due to gravity is \(g \mathrm{~m} / \mathrm{s}^{2}\) a sphere of lead of density \(\rho \mathrm{kg} / \mathrm{m}^{3}\) is gently released in a column of non-viscous liquid of density \(\sigma \mathrm{kg} / \mathrm{m}^{3}\). If \(\rho>\sigma\), then the acceleration of the sphere is (a) \(g\) (b) zero (c) \(\left(\frac{\rho-\sigma}{\rho}\right) g\) (d) \(\frac{\sigma}{\rho} g\)

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