/*! 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 54 In planetary motion the areal ve... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 54

In planetary motion the areal velocity of position vector of a planet depends on angular velocity \((\omega)\) and the distance of the planet from sun \((r)\). If so the correct relation for areal velocity is (a) \(\frac{d A}{d t} \propto \omega r\) (b) \(\frac{d A}{d t} \propto \omega^{2} r\) (c) \(\frac{d A}{d t} \propto \omega r^{2}\) (d) \(\frac{d A}{d t} \propto \sqrt{\omega r}\)

Short Answer

Expert verified
The correct relation for areal velocity is \(dA/dt \propto 蠅 r^{2}\).

Step by step solution

01

Identify the basic concept

Areal velocity in the context of planetary motion is given by the area swept out per unit time by the line joining the planet to the sun. Considering a small time duration \(dt\), the planet moves a small distance \(ds\) along its elliptical path, making a very small angle \(d\theta\) at sun. The corresponding small area swept is thus given by \(dA = 0.5 * r * r * d\theta\). The rate of change of area with respect to time, \(dA/dt\), is called the areal velocity.
02

Deducing the relation

Substituting \(d\theta\) with \(蠅dt\), where \(蠅\) is the angular velocity, we get, \(dA = 0.5 * r * r * 蠅 * dt\). On differentiating both sides with respect to time, we obtain the areal velocity, \(dA/dt = 0.5 * r * r * 蠅\).
03

Comparing with given options

Comparing the deduced relation with the options, we see it matches with option (c), ie, \(dA/dt \propto 蠅 r^{2}\). So, the correct relation for the areal velocity is \(dA/dt = 0.5 * r * r * 蠅\), which indicates that the areal velocity is directly proportional to \(蠅 r^{2}\) .

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

The gravitational field due to a mass distribution is \(E=K / x^{3}\) in the \(x\) - direction ( \(K\) is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance \(x\) is (a) \(\mathrm{K} / \underline{x}\) (b) \(K / 2 x\) (c) \(K / x^{2}\) (d) \(K / 2 x^{2}\)

The radii of two planets are respectively \(R_{1}\) and \(R_{2}\) and their densities are respectively \(\rho_{1}\) and \(\rho_{2}\). The ratio of the accelerations due to gravity at their surfaces is (a) \(g_{1}: g_{2}=\frac{\rho_{1}}{R_{1}^{2}}: \frac{\rho_{2}}{R_{2}^{2}}\) (b) \(g_{1}: g_{2}=R_{1} R_{2}: \rho_{1} \rho_{2}\) (c) \(g_{1}: g_{2}=R_{1} \rho_{2}: R_{2} \rho_{1}\) (d) \(g_{1}: g_{2}=R_{1} \rho_{1}: R_{2} \rho_{2}\)

If the radius of the earth were to shrink by \(1 \%\) its mass remaining the same, the acceleration due to gravity on the earth's surface would [Irr-JEE 1981; CPMT 1981; MP PMT 1996, 97; Roorkee 1992; MP PET 1999] (a) Decrease by \(2 \%\) (b) Remain unchanged (c) Increase by \(2 \%\) (d) Increase by \(1 \%\)

Acceleration due to gravity on moon is \(1 / 6\) of the acceleration due to gravity on earth. If the ratio of densities of earth \(\left(\rho_{m}\right)\) and moon \(\left(\rho_{e}\right)\) is \(\left(\frac{\rho_{e}}{\rho_{m}}\right)=\frac{5}{3}\) then radius of moon \(R_{m}\) in terms of \(R_{e}\) will be [MP PMT 2 (a) \(\frac{5}{18} R_{e}\) (b) \(\frac{1}{6} R_{e}\) (c) \(\frac{3}{18} R_{e}\) (d) \(\frac{1}{2 \sqrt{3}} R_{e}\)
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Radiation

Read Explanation

Linear Momentum

Read Explanation

Work Energy and Power

Read Explanation

Fluids

Read Explanation

Circular Motion and Gravitation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.