Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 708

Density of the earth is doubled keeping its radius constant then acceleration, due to gravity will be \(-m s^{-2}\) \(\left(\mathrm{g}=9.8 \mathrm{~ms}^{2}\right)\) (A) \(19.6\) (B) \(9.8\) (C) \(4.9\) (D) \(2.45\)

Short Answer

Expert verified
The new acceleration due to gravity is \(19.6 \text{ m/s²}\), which corresponds to the choice (A).

Step by step solution

01

Write the formula for acceleration due to gravity

The formula for acceleration due to gravity is: \[g = \frac{GM}{r^2}\] Where G is the gravitational constant, M is the mass of the Earth, and r is the radius of the Earth.
02

Write the mass-density relationship formula

The mass-density relationship is given by: \[M = \rho V\] Where M is the mass, ρ is the density, and V is the volume.
03

Substitute the mass-density relationship in the gravity formula

Replace M in the gravity formula using the mass-density relationship: \[g = \frac{G \cdot (\rho V)}{r^2}\]
04

Rewrite the formula in terms of radius

Since the radius of the Earth is constant, we can rewrite the volume in terms of radius (using the formula for the volume of a sphere, \(V = \frac{4}{3} \pi r^3\)): \[g = \frac{G \cdot (\rho \cdot \frac{4}{3} \pi r^3)}{r^2} =\frac{4}{3}G\pi r \rho\]
05

Calculate acceleration due to gravity when Earth's density is doubled

Now, we double the density (2ρ) and find the new acceleration due to gravity: \[g' = \frac{4}{3}G\pi r (2 \rho) \]
06

Calculate the ratio of the new gravity to the initial gravity

To find the new gravity, observe the ratio between the new gravity and the initial gravity: \[\frac{g'}{g} = \frac{\frac{4}{3}G\pi r (2 \rho)}{\frac{4}{3}G\pi r \rho} = 2\]
07

Calculate the new acceleration due to gravity

Since the ratio of the new gravity to the initial gravity is 2, the new acceleration due to gravity (g') is twice the initial gravity (9.8 m/s²): \[g' = 2 \cdot 9.8 = 19.6 \text{ m/s²}\] Answer: The new acceleration due to gravity is 19.6 m/s², which corresponds to the choice (A).

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

If \(\mathrm{V}_{\mathrm{e}}\) and \(\mathrm{V}_{\mathrm{o}}\) are represent the escape velocity and orbital velocity of satellite corresponding to a circular orbit of -adius \(\mathrm{r}\), then A) \(\mathrm{V}_{\mathrm{e}}=\mathrm{V}_{\mathrm{o}}\) (B) \(\sqrt{2} \mathrm{~V}_{\mathrm{o}}=\mathrm{V}_{\mathrm{e}}\) C) \(\mathrm{V}_{\mathrm{e}}=\left(\mathrm{V}_{\mathrm{O}} / \sqrt{2}\right)\) (D) \(\mathrm{V}_{\mathrm{e}}\) and \(\mathrm{V}_{\mathrm{o}}\) are not related

The mass of a space ship is \(1000 \mathrm{~kg} .\) It is to be launched from earth's surface out into free space the value of \(\mathrm{g}\) and \(\mathrm{R}\) (radius of earth) are \(10 \mathrm{~ms}^{-2}\) and \(6400 \mathrm{~km}\) respectively the required energy for this work will be $=\ldots \ldots \ldots .$ J (A) \(6.4 \times 10^{11}\) (B) \(6.4 \times 10^{8}\) (C) \(6.4 \times 10^{9}\) (D) \(6.4 \times 10^{10}\)

Given mass of the moon is \((1718)\) of the mass of the earth and corresponding radius is \((1 / 4)\) of the earth, If escape velocity on the earth surface is \(11.2 \mathrm{kms}^{-1}\) the value of same on the surface of moon is $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$. (A) \(0.14\) (B) \(0.5\) (C) \(2.5\) (D) 5

At what distance from the center of earth, the value of acceleration due to gravity \(g\) will be half that of the surfaces \((\mathrm{R}=\) Radius of earth \()\) (A) \(2 \mathrm{R}\) (B) \(\mathrm{R}\) (C) \(1.414 \mathrm{R}\) (D) \(0.414 \mathrm{R}\)

If the height of a satellite from the earth is negligible in comparison of the radius of the earth \(\mathrm{R}\), the orbital velocity of the satellite is (A) \(\mathrm{gR}\) (B) \((\mathrm{gR} / 2)\) (C) \(\sqrt{(g} / \mathrm{R})\) (D) \(\sqrt{(g R)}\)
See all solutions

Recommended explanations on English Textbooks

Sociolinguistics

Read Explanation

Text Comparison

Read Explanation

Phonology

Read Explanation

History of English Language

Read Explanation

Phonetics

Read Explanation

Semiotics

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