Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 760

A shell of mass \(\mathrm{M}\) and radius \(\mathrm{R}\) has point mass \(\mathrm{m}\) placed at a distance \(r\) from its center. The gravitational potential energy \(\mathrm{U}(\mathrm{r})-\mathrm{v}\) will be

Short Answer

Expert verified
The gravitational potential energy \(U(r)\) of a point mass \(m\) located at a distance \(r\) from the center of a shell of mass \(M\) and radius \(R\) is given by the formula: \(U(r) = -\dfrac{G m M}{r}\), where G is the gravitational constant.

Step by step solution

01

Write down the formula for potential energy

To calculate the gravitational potential energy, we need to use the formula: \(U(r) = -\dfrac{G m M}{r}\) Where U(r) is the potential energy as a function of r, G is the gravitational constant (approximately \(6.674 \times 10^{-11}\) N(m/kg)²), m is the given point mass, M is the mass of the shell, and r is the distance from the center of the shell.
02

Compute U(r)-v

The exercise asks to compute \(U(r)-v\). But there isn't information about the value of "v" anywhere in the exercise. We'll assume that this is an error and that the desired expression is only U(r). So, we need to find the gravitational potential energy at distance r from the center of the shell. By substituting the given values into the potential energy formula, we get: \(U(r) = -\dfrac{G m M}{r}\) Which is the potential energy as a function of the distance from the center of the shell.

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

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) \(2.45\)

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) \(0.414 \mathrm{R}\)

A satellite is moving around the earth with speed \(\mathrm{v}\) in a circular orbit of radius \(\mathrm{r}\). If the orbit radius is decreased by \(1 \%\) its speed will (A) increase by \(1 \%\) (B) increase by \(0.5 \%\) (C) decreased by \(1 \%\) (C) Decreased by \(0.5 \%\)

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If the radius of the earth is \(\mathrm{R}\), the radius of planet would be (A) \(2 \mathrm{R}\) (B) \(4 \mathrm{R}\) (C) \(1 / 4 \mathrm{R}\) (D) \(\mathrm{R} / 2\)

The escape velocity for a rocket from earth is \(11.2 \mathrm{kms}^{-1}\) value on a planet where acceleration due to gravity is double that on earth and diameter of the planet is twice that of earth will be $=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}$ (A) \(11.2\) (B) \(22.4\) (C) \(5.6\) (C) \(53.6\)
See all solutions

Recommended explanations on English Textbooks

Blog

Read Explanation

Linguistic Terms

Read Explanation

Global English

Read Explanation

Argumentative Essay

Read Explanation

Synthesis Essay

Read Explanation

Summary Text

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