Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 9

The first cosmic velocity is the velocity of motion on a circular orbit of radius close to the radius of the earth. Find the magnitude of the first cosmic velocity \(v_{1}\) and show that \(v_{2}=\sqrt{2} v_{1}\) (cf. Section 3B).

Short Answer

Expert verified
Answer: The second cosmic velocity (escape velocity) is equal to the square root of 2 times the first cosmic velocity (orbital velocity), represented as \(v_{2} = \sqrt{2} v_{1}\).

Step by step solution

01

Write down the Newton's Law of Gravitation under a circular orbit

Recall that Newton's Law of Gravitation states that the gravitational force between two masses (in this case, Earth and the object) is given by: \(F = G\frac{m_{1}m_{2}}{r^2}\), where \(F\) is the gravitational force, \(G\) is the gravitational constant, \(m_{1}\) and \(m_{2}\) are the masses of the two objects, and \(r\) is the distance between the center of those objects. For an object in a circular orbit, the gravitational force is equal to the centripetal force, which is given by: \(F = m_{2}\frac{v_{1}^2}{r}\).
02

Set gravitational force equal to centripetal force and solve for \(v_{1}\)

To find the first cosmic velocity, we need to set the gravitational force and centripetal force equal to each other. This gives us the equation: \(G\frac{m_{1}m_{2}}{r^2} = m_{2}\frac{v_{1}^2}{r}\). Now, we can solve for \(v_{1}\): \(v_{1}^2 = \frac{G m_{1}}{r}\) \(v_{1} = \sqrt{\frac{G m_{1}}{r}}\)
03

Write down the equation for the second cosmic velocity

The equation for the second cosmic velocity, also known as escape velocity, is given by: \(v_{2} = \sqrt{\frac{2G m_{1}}{r}}\), which represents the minimum velocity required for an object to break free from Earth's gravitational pull.
04

Show that \(v_{2} = \sqrt{2} v_{1}\)

Now we can substitute the equation for the first cosmic velocity from Step 2 into the equation for the second cosmic velocity in Step 3: \(v_{2} = \sqrt{\frac{2G m_{1}}{r}} = \sqrt{2} \sqrt{\frac{G m_{1}}{r}} = \sqrt{2}v_{1}\) Therefore, the second cosmic velocity \(v_{2}\) is equal to \(\sqrt{2}\) times the first cosmic velocity \(v_{1}\).

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 the radius of a planet is \(\alpha\) times the radius of the earth and its mass \(\beta\) times that of the earth, find the ratio of the acceleration of the force of gravity and the first and second cosmic velocities to the corresponding quantities for the earth.

Show that the angle \(\Phi\) between the pericenter and apocenter is equal to the semiperiod of an oscillation in the one-dimensional system with potential energy \(W(x)=U(M / x)+\left(x^{2} / 2\right)\). $$ \Phi=\int_{x_{\min }}^{x_{\max }} \frac{d x}{\sqrt{2(E-W)}} $$

Let \(U(r)=-k r^{-\beta}, 0<\beta<2\). Find \(\Phi_{0}=\lim _{E \rightarrow-0} \Phi\).

How does the running velocity of an animal on level ground and uphill depend on the size \(L\) of the animal?

How does the height of an animal's jump depend on its size?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Space Physics

Read Explanation

Wave Optics

Read Explanation

Thermodynamics

Read Explanation

Fluids

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