Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 813

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 (A) decrease by \(2 \%\) (B) remain Unchanged (C) increase by \(2 \%\) (D) increases by \(1 \%\)

Short Answer

Expert verified
If the radius of the Earth were to shrink by 1% while its mass remains the same, the acceleration due to gravity on the Earth's surface would increase by approximately 2.03%. The answer is closest to (C) increase by 2%.

Step by step solution

01

Relevant Formulas

In order to solve the problem, we will need the formula for gravitational acceleration, which is: \( g = \frac{GM}{r^2} \) Where: - \(g\) is the gravitational acceleration - \(G\) is the gravitational constant (approximately \(6.674 \times 10^{-11} m^3 kg^{-1} s^{-2}\)) - \(M\) is the mass of the body (Earth in our case) - \(r\) is the radius of the body
02

Initial Gravitational Acceleration

First, let's determine the initial gravitational acceleration before the radius shrinks. We will define the initial values as: - Radius: \(r\) - Gravitational Acceleration: \(g\) Now we can express the initial gravitational acceleration as: \( g = \frac{GM}{r^2} \)
03

Shrunk Radius and New Gravitational Acceleration

The exercise tells us that the radius of the Earth shrinks by 1%. Therefore, the new radius will be: \( r' = r - 0.01r = 0.99r \) Now let's find the new gravitational acceleration (\(g'\)) with the new radius: \(g' = \frac{GM}{(0.99r)^2} = \frac{GM}{0.9801r^2}\)
04

Comparing New and Old Gravitational Acceleration

Now we will compare the new gravitational acceleration (\(g'\)) to the old one (\(g\)), by taking the ratio of \(g'\) to \(g\): \(\frac{g'}{g} = \frac{GM}{0.9801r^2} \times \frac{r^2}{GM} = \frac{1}{0.9801} \approx 1.0203\) Which indicates that the gravitational acceleration increased. To find the percentage increase, we can subtract 1 and multiply by 100: \((1.0203 - 1) \times 100 = 2.03 \% \)
05

Conclusion

Therefore, if the radius of the Earth were to shrink by 1% while the mass remains the same, the acceleration due to gravity on the Earth's surface would increase by approximately 2.03%. The answer is closest to (C) increase by 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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Which one of following statements regarding artificial satellite of earth is incorrect (A) The orbital velocity depends on the mass of the satellite (B) A minimum velocity of \(8 \mathrm{kms}^{-1}\) is required by a satellite to orbit quite close to the earth. (C) The period of revolution is large if the radius of its orbit is large (D) The height of geostationary satellite is about \(36000 \mathrm{~km}\) from earth

The Gravitational P.E. of a body of mass \(\mathrm{m}\) at the earth's surface is \(-\mathrm{mgRe}\). Its gravitational potential energy at a height \(\operatorname{Re}\) from the earth's surface will be \(=\ldots \ldots \ldots\) here (Re is the radius of the earth) (A) \(-2 \mathrm{mgRe}\) (B) \(2 \mathrm{mgRe}\) (C) \((1 / 2) \mathrm{mg} \mathrm{Re}\) (D) \(-(1 / 2) \mathrm{mg} \operatorname{Re}\)

Direction (Read the following questions and choose) (A) If both Assertion and Reason are true and the Reason is correct explanation of assertion (B) If both Assertion and Reason are true, but reason is not correct explanation of the Assertion (C) If Assertion is true, but the Reason is false (D) If Assertion is false, but the Reason is true Assertion: The value of acc. due to gravity \((\mathrm{g})\) does not depend upon mass of the body Reason: This follows from \(\mathrm{g}=\left[(\mathrm{GM}) / \mathrm{R}^{2}\right]\), where \(\mathrm{M}\) is mass of planet (earth) and \(\mathrm{R}\) is radius of planet (earth) (a) \(\mathrm{A}\) (b) \(\mathrm{B}\) (c) \(\mathrm{C}\) (d) D

The height at which the weight of a body becomes \(1 / 16\) th its weight on the surface of (radius \(\mathrm{R}\) ) is (A) \(3 \mathrm{R}\) (B) \(4 \mathrm{R}\) (C) \(5 \mathrm{R}\) (D) \(15 \mathrm{R}\)

If \(\mathrm{g}\) is the acceleration due to gravity at the earth's surface and \(\mathrm{r}\) is the radius of the earth, the escape velocity for the body to escape out of earth's gravitational field is \(\ldots \ldots \ldots\) (A) \(\mathrm{gr}\) (B) \(\sqrt{(2 \mathrm{gr})}\) (C) \(\mathrm{g} / \mathrm{r}\) (D) \(\mathrm{r} / \mathrm{g}\)
See all solutions

Recommended explanations on English Textbooks

Phonetics

Read Explanation

Argumentative Essay

Read Explanation

English Language Study

Read Explanation

Semiotics

Read Explanation

Pragmatics

Read Explanation

Blog

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.

Sign-up for free