Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 81

Prove that \(U=-2 K\) for any gravitational circular orbit. [Hint: Use Newton's second law to relate the gravitational force to the acceleration required to maintain uniform circular motion.]

Short Answer

Expert verified
Question: Prove that the relationship between gravitational potential energy (U) and kinetic energy (K) for any gravitational circular orbit is given by \(U=-2 K\). Solution: To prove that \(U=-2 K\) for any gravitational circular orbit, we need to: 1. Define the expressions for potential energy (U) and kinetic energy (K) in a gravitational system 2. Relate the gravitational force and centripetal force using Newton's second law 3. Substitute the expression for orbital velocity into the kinetic energy formula 4. Show the relationship between potential and kinetic energies by comparing their expressions Following these steps, we find that \(U = - 2K\) for any gravitational circular orbit.

Step by step solution

01

Define potential and kinetic energies

For a gravitational system, the potential energy (U) is given by the formula: \(U = -\frac{GMm}{r}\) Where G is the gravitational constant, M is the mass of the central object (e.x. the Earth), m is the mass of the object in orbit (e.x. a satellite), and r is the distance between their centers of mass. The kinetic energy (K) is given by the formula: \(K = \frac{1}{2}mv^2\) Where v is the orbital velocity of the object in orbit (i.e., the satellite).
02

Relate the gravitational force and centripetal force

According to Newton's second law, in a circular orbit, the gravitational force acts as the centripetal force that keeps the object in orbit. Therefore, we have: \(F_g = F_c\) \(\frac{GMm}{r^2} = m\frac{v^2}{r}\) Now, let's rearrange the formula for v: \(v^2 =\frac{GM}{r}\)
03

Substitute the expression for orbital velocity into the kinetic energy formula

Now, let's substitute the expression for \(v^2\) found in step 2 into the formula for the kinetic energy: \(K = \frac{1}{2}m\frac{GM}{r}\)
04

Show the relationship between potential and kinetic energies

Next, let's compare the expressions for potential energy (U) and kinetic energy (K) to find the relationship between them: \(U = -\frac{GMm}{r}\) \(K = \frac{1}{2}m\frac{GM}{r}\) Notice that if you multiply the equation for K by 2, the right-hand side of the equation will match the right-hand side of the equation for U but with opposite signs: \(2 K = -U\) This proves that for any gravitational circular orbit: \(U = - 2K\)

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

A 4.0 -kg block is released from rest at the top of a frictionless plane of length \(8.0 \mathrm{m}\) that is inclined at an angle of \(15^{\circ}\) to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point \(5.0 \mathrm{m}\) along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer? A 4.0 -kg block is released from rest at the top of a frictionless plane of length \(8.0 \mathrm{m}\) that is inclined at an angle of \(15^{\circ}\) to the horizontal. A cord is attached to the block and trails along behind it. When the block reaches a point \(5.0 \mathrm{m}\) along the incline from the top, someone grasps the cord and exerts a constant tension parallel to the incline. The tension is such that the block just comes to rest when it reaches the bottom of the incline. (The person's force is a nonconservative force.) What is this constant tension? Solve the problem twice, once using work and energy and again using Newton's laws and the equations for constant acceleration. Which method do you prefer?
Show that 1 kilowatt-hour (kW-h) is equal to 3.6 MJ.
The maximum speed of a child on a swing is \(4.9 \mathrm{m} / \mathrm{s}\) The child's height above the ground is \(0.70 \mathrm{m}\) at the lowest point in his motion. How high above the ground is he at his highest point?
When the spring on a toy gun is compressed by a distance \(x\), it will shoot a rubber ball straight up to a height of \(h .\) Ignoring air resistance, how high will the gun shoot the same rubber ball if the spring is compressed by an amount \(2 x ?\) Assume \(x<<h\)
The escape speed from the surface of the Earth is $11.2 \mathrm{km} / \mathrm{s} .$ What would be the escape speed from another planet of the same density (mass per unit volume) as Earth but with a radius twice that of Earth?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Thermodynamics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Force

Read Explanation

Measurements

Read Explanation

Modern Physics

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