Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 797

The orbital speed of jupiter is (A) greater than the orbital speed of earth (B) less than the orbital speed of earth (C) equal to the orbital speed of earth (D) zero

Short Answer

Expert verified
The orbital speed of Jupiter (\(13,100 \, m/s\)) is less than the orbital speed of Earth (\(29,500 \, m/s\)). Therefore, the correct answer is (B) less than the orbital speed of Earth.

Step by step solution

01

Understand orbital speed

Orbital speed is the speed at which an object, in this case, a planet, moves in its orbit around another object, usually the Sun. The orbital speed depends on the mass of the object being orbited and the distance between the two objects.
02

Calculate Earth's orbital speed

The Earth's orbital speed can be found using the formula for the orbital speed of an object around another object, given by: \[v_{Earth} = \sqrt{\frac{G * M_{Sun}}{r_{Earth}}}\] Where: - \(v_{Earth}\) is the orbital speed of Earth - \(G\) is the gravitational constant (\(6.674 \times 10^{-11} \, N \cdot m^2/kg^2\)) - \(M_{Sun}\) is the mass of the Sun (\(1.989 \times 10^{30} \, kg\)) - \(r_{Earth}\) is the average distance from Earth to the Sun (about \(1.496 \times 10^{11}\, m\)) Plugging in the values and calculating Earth's orbital speed: \[v_{Earth} = \sqrt{\frac{6.674 \times 10^{-11} * 1.989 \times 10^{30}}{1.496 \times 10^{11}}} \approx 29,500 \, m/s\]
03

Calculate Jupiter's orbital speed

Similarly, we can find Jupiter's orbital speed using the same formula: \[v_{Jupiter} = \sqrt{\frac{G * M_{Sun}}{r_{Jupiter}}}\] Where: - \(v_{Jupiter}\) is the orbital speed of Jupiter - \(r_{Jupiter}\) is the average distance from Jupiter to the Sun (about \(7.785 \times 10^{11}\, m\)) Plugging in the values and calculating Jupiter's orbital speed: \[v_{Jupiter} = \sqrt{\frac{6.674 \times 10^{-11} * 1.989 \times 10^{30}}{7.785 \times 10^{11}}} \approx 13,100 \, m/s\]
04

Compare the orbital speeds of Earth and Jupiter

Now that we have calculated the orbital speeds of both Earth and Jupiter, we can compare them: - Earth's orbital speed: \(29,500 \, m/s\) - Jupiter's orbital speed: \(13,100 \, m/s\) As we can see, Jupiter's orbital speed is lower than Earth's. So, our answer is: (B) less than the orbital speed of earth

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

The acceleration due to gravity near the surface of a planet of radius \(\mathrm{R}\) and density \(\mathrm{d}\) is proportional to (A) \(\mathrm{d} / \mathrm{R}^{2}\) (B) \(\mathrm{d} \mathrm{R}^{2}\) (C) \(\mathrm{dR}\) (D) \(\mathrm{d} / \mathrm{R}\)

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) \(\sqrt{(g R)}\)

Rockets are launched in eastward direction to take advantage of (A) the clear sky on eastern side (B) the thiner atmosphere on this side (C) earth's rotation (D) earth's tilt

The acceleration due to gravity on a planet is same as that on earth and its radius is four times that of earth. What will be the value of escape velocity on that planet if it is \(\mathrm{V}_{\mathrm{e}}\) on the earth (A) \(\mathrm{V}_{\mathrm{e}}\) (B) \(2 \mathrm{~V}_{\mathrm{e}}\) (C) \(4 \mathrm{~V}_{\mathrm{e}}\) (D) \(\mathrm{V}_{\mathrm{e}} / 2\)

4 A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the center of earth in new orbit is two times of the earlier orbit. The time period in second orbit is $\ldots \ldots \ldots \ldots$ hours. (A) \(4.8\) (B) \(48 \sqrt{2}\) (C) 24 (D) \(24 \sqrt{2}\)
See all solutions

Recommended explanations on English Textbooks

Creative Story

Read Explanation

Cues and Conventions

Read Explanation

Phonetics

Read Explanation

Phonology

Read Explanation

Multiple Choice Questions

Read Explanation

Blog

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