/*! 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 33 In the solar system, the inter-p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 33

In the solar system, the inter-planetary region has chunks of matter (much smaller in size compared to planets) called asteroids. They (A) will not move around the sun, since they have very small masses compared to the sun. (B) will move in an irregular way because of their small masses and will drift away into outer space. (C) will move around the sun in closed orbits but not obey Kepler's laws. (D) will move in orbits like planets and obey Kepler's laws.

Short Answer

Expert verified
The correct answer is choice (D): "Asteroids will move in orbits like planets and obey Kepler's laws."

Step by step solution

01

Statement (A) Analysis

Statement (A) says that asteroids will not move around the sun due to their small masses compared to the sun. However, this statement is inaccurate. Even though asteroids have smaller masses compared to the sun, they are still influenced by the sun's gravitational force and will move in its gravitational field.
02

Statement (B) Analysis

Statement (B) suggests that asteroids will move in an irregular way and drift away into outer space due to their small masses. While their small mass might cause a more significant influence of other forces (e.g., radiative forces or disturbance from other celestial bodies), the dominant force acting on them in the solar system is still the sun's gravity. This gravitational force keeps them bound to the solar system, and hence, statement (B) is not accurate.
03

Statement (C) Analysis

Statement (C) claims that asteroids will move around the sun in closed orbits but not obey Kepler's laws. However, this statement is incorrect. As long as a celestial body is influenced predominantly by the gravitational pull of another body (in this case, the sun), it will still follow Kepler's laws regardless of its size.
04

Statement (D) Analysis

Lastly, statement (D) asserts that asteroids will move in orbits like planets and obey Kepler's laws. This statement is accurate. Despite their smaller masses, asteroids are still influenced primarily by the sun's gravity, which causes them to move in orbits and follow Kepler's laws, just like the planets in our solar system.
05

Conclusion

The correct answer is choice (D): "Asteroids will move in orbits like planets and obey Kepler's laws."

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

At a height above the surface of the earth equal to the radius of the earth the value of \(g\) (acceleration due to gravity on the surface of the earth) will be nearly (A) Zero (B) \(\sqrt{g}\) (C) \(\frac{g}{4}\) (D) \(\frac{g}{2}\)

The ratio of the radii of the planets \(P_{1}\) and \(P_{2}\) is \(k_{1}\). The ratio of the acceleration due to the gravity on them is \(k_{2}\). The ratio of the escape velocities from them will be (A) \(k_{1} k_{2}\) (B) \(\sqrt{k_{1} k_{2}}\) (C) \(\sqrt{\left(k_{1} / k_{2}\right)}\) (D) \(\sqrt{\left(k_{2} / k_{1}\right)}\)

A solid sphere of uniform density and radius 4 units is located with its centre at the origin \(O\) of co-ordinates. Two spheres of equal radii 1 units, with their centres at \(A(-2,0,0)\) and \(B(2,0,0)\) respectively, are taken out of the solid leaving behind spherical cavities as shown in Fig. 7.12. Then (A) the gravitational field due to this object at the origin is zero. (B) the gravitational field at the point \(B(2,0,0)\) is zero. (C) the gravitational potential is same at all points on the circle \(y^{2}+z^{2}=36\) (D) the gravitational potential is same at all points on the circle \(y^{2}+z^{2}=4\)

If three uniform spheres, each having mass \(M\) and radius \(R\), are kept in such a way that each touches the other two, the magnitude of the gravitational force on any sphere due to the other two is (A) \(\frac{G M^{2}}{4 R^{2}}\) (B) \(\frac{2 G M^{2}}{R^{2}}\) (C) \(\frac{2 G M^{2}}{4 R^{2}}\) (D) \(\frac{\sqrt{3} G M^{2}}{4 R^{2}}\)

If the length of a simple pendulum is equal to the radius \(R\) of the earth, its time period will be (A) \(2 \pi \sqrt{R / g}\) (B) \(2 \pi \sqrt{R / 2 g}\) (C) \(2 \pi \sqrt{2 R / g}\) (D) \(\pi \sqrt{R / 2 g}\)
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Fundamentals of Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Quantum Physics

Read Explanation

Scientific Method Physics

Read Explanation

Wave Optics

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.