Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 815

An artificial satellite moving in a circular orbit around earth has a total (kinetic + potential energy) \(E_{0}\), its potential energy is (A) \(-\mathrm{E}_{0}\) (B) \(1.5 \mathrm{E}_{0}\) (C) \(2 \mathrm{E}_{0}\) (D) \(\mathrm{E}_{0}\)

Short Answer

Expert verified
The potential energy (U) of an artificial satellite moving in a circular orbit around Earth is twice the total energy (E): \(U = -2E_{0}\). The correct option is (C) \(2 \mathrm{E}_{0}\).

Step by step solution

01

Recall the formulas for gravitational potential energy and kinetic energy

Gravitational potential energy (U) is given by the formula: \(U = -\dfrac{GMm}{r}\), where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the centers of the Earth and the satellite. Kinetic energy (K) is given by the formula: \(K = \dfrac{1}{2}mv^2\), where m is the mass of the satellite, and v is its orbital velocity.
02

Express the orbital velocity in terms of the gravitational constant, Earth's mass, and distance

For a satellite in a circular orbit, the centripetal force is balanced by the gravitational force. That is, \(m\dfrac{v^2}{r} = \dfrac{GMm}{r^2}\). We can rearrange this equation for the orbital velocity: \(v^2 = \dfrac{GM}{r}\).
03

Substitute the orbital velocity into the kinetic energy formula

Now, use the expression for orbital velocity found in Step 2 to substitute into the kinetic energy formula: \(K = \dfrac{1}{2}m\left(\dfrac{GM}{r}\right)\).
04

Find the total energy (kinetic and potential energy) of the satellite

Add the kinetic and potential energy formulas to find the total energy of the satellite (E): \(E_{0} = U + K = -\dfrac{GMm}{r} + \dfrac{1}{2}m\left(\dfrac{GM}{r}\right)\).
05

Solve for potential energy in terms of total energy

We have \(E_{0} = -\dfrac{GMm}{r} + \dfrac{1}{2}m\left(\dfrac{GM}{r}\right)\). Simplifying this equation, we get: \(E_{0} = - \dfrac{1}{2}\dfrac{GMm}{r}\). Thus, potential energy (U) in terms of total energy (E) is twice of the total energy: \(U = -2E_{0}\).
06

Choose the correct answer from the given options

From the result obtained in Step 5, the potential energy (U) is -2 times the total energy (E). Hence, the correct option is (C) \(2 \mathrm{E}_{0}\).

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 the following graphs represents correctly the variation of the gravitational field with the distance (r) from the center of spherical shell of mass \(\mathrm{M}\) and radius a

Weight of a body is maximum at (A) moon (B) poles of earth (C) Equator of earth (D) Center of earth

Energy required to move a body of mass \(\mathrm{m}\) from from an orbit of radius \(2 \mathrm{R}\) to \(3 \mathrm{R}\) is \(\ldots \ldots \ldots \ldots\) (A) \(\left[(\mathrm{GMm}) /\left(12 \mathrm{R}^{2}\right)\right]\) (B) \(\left[(\mathrm{GMm}) /\left(3 \mathrm{R}^{2}\right)\right]\) (C) \([(\mathrm{GMm}) /(8 \mathrm{R})]\) (D) \([(\mathrm{GMm}) /(6 \mathrm{R})]\)

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) \(\mathrm{R} / 2\)

A particle of mass \(10 \mathrm{~g}\) is kept on the surface of a uniform sphere of mass \(100 \mathrm{~kg}\) and radius \(10 \mathrm{~cm}\). Find the work to be done against the gravitational force between them to take the particle is away from the sphere \(\left(\mathrm{G}=6.67 \times 10^{-11} \mathrm{SI}\right.\) unit \()\) (A) \(6.67 \times 10^{-9} \mathrm{~J}\) (B) \(6.67 \times 10^{-10} \mathrm{~J}\) (C) \(13.34 \times 10^{-10} \mathrm{~J}\) (D) \(3.33 \times 10^{-10} \mathrm{~J}\)
See all solutions

Recommended explanations on English Textbooks

Free Response Essay

Read Explanation

English Grammar Summary

Read Explanation

Syntax

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

Textual Analysis

Read Explanation

Language Acquisition

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