Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 71

If the upper atmosphere of Jupiter has a temperature of \(160 \mathrm{K}\) and the escape speed is \(60 \mathrm{km} / \mathrm{s}\), would an astronaut expect to find much hydrogen there?

Short Answer

Expert verified
Based on the calculations, the root mean square speed of hydrogen molecules in Jupiter's upper atmosphere is 1.77 x 10^3 m/s, which is significantly lower than the escape speed of 60,000 m/s. Therefore, hydrogen can be found in the upper atmosphere of Jupiter.

Step by step solution

01

Identify the given information

We are given the following information: - Temperature of Jupiter's upper atmosphere: \(T = 160 \mathrm{K}\) - Escape speed: \(v_\text{escape} = 60 \mathrm{km} / \mathrm{s} = 60,000 \mathrm{m} / \mathrm{s}\) - We need to determine if the hydrogen (H\(_2\)) can be found in the upper atmosphere.
02

Convert temperature to energy

First, we need to find the average kinetic energy of hydrogen molecules in the atmosphere. We will use the following equation: $$\frac{3}{2} k_\mathrm{B} T = \frac{1}{2} m_\mathrm{H_2} v_\mathrm{rms}^{2},$$ where \(k_\mathrm{B}\) is the Boltzmann constant (\(1.38 \times 10^{-23} \mathrm{J} / \mathrm{K}\)), \(T\) is the temperature of the atmosphere, \(m_\mathrm{H_2}\) is the mass of a hydrogen molecule, and \(v_\text{rms}\) is the root mean square speed of the hydrogen molecules.
03

Calculate the mass of a hydrogen molecule

For H\(_2\), we need to find the molar mass using the atomic mass of hydrogen (approximately \(1 \mathrm{amu}\)): $$m_\mathrm{H_2} = 2 \times \frac{1 \mathrm{g}}{\text{mol}} \times \frac{1 \text{kg}}{1000 \mathrm{g}} \times \frac{1 \text{mol}}{6.022 \times 10^{23} \text{particles}},$$ which simplifies to: $$m_\mathrm{H_2} = 3.32 \times 10^{-27} \text{kg}$$.
04

Calculate the root mean square speed

Now, we use the temperature-energy equation to solve for \(v_\text{rms}\): $$v_\mathrm{rms} = \sqrt{\frac{3 k_\mathrm{B} T}{m_\mathrm{H_2}}}.$$ Plug in the values: $$v_\mathrm{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \mathrm{J / K} \times 160 \mathrm{K}}{3.32 \times 10^{-27} \text{kg}}} = 1.77\times 10^3 \mathrm{m} / \mathrm{s}.$$
05

Compare root mean square speed with escape speed

Now we need to compare the calculated root mean square speed with the escape speed of Jupiter. If the root mean square speed is significantly lower than the escape speed, hydrogen will not be able to escape Jupiter's atmosphere: $$v_\mathrm{rms} = 1.77 \times 10^3 \mathrm{m} / \mathrm{s} \quad \text{ vs.} \quad v_\text{escape} = 60,000 \mathrm{m} / \mathrm{s}.$$ Given that the root mean square speed of hydrogen is far lower than the escape speed, we can expect to find hydrogen molecules in the upper atmosphere of Jupiter.

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 2.4 -m length of copper pipe extends directly from a hot-water heater in a basement to a faucet on the first floor of a house. If the faucet isn't fixed in place, how much will it rise when the pipe is heated from $20.0^{\circ} \mathrm{C}\( to \)90.0^{\circ} \mathrm{C} .$ Ignore any increase in the size of the faucet itself or of the water heater.
An iron cannonball of radius \(0.08 \mathrm{m}\) has a cavity of radius $0.05 \mathrm{m}$ that is to be filled with gunpowder. If the measurements were made at a temperature of \(22^{\circ} \mathrm{C}\) how much extra volume of gunpowder, if any, will be required to fill 500 cannonballs when the temperature is \(30^{\circ} \mathrm{C} ?\)
The coefficient of linear expansion of brass is \(1.9 \times 10^{-5}\) $^{\circ} \mathrm{C}^{-1} .\( At \)20.0^{\circ} \mathrm{C},$ a hole in a sheet of brass has an area of \(1.00 \mathrm{mm}^{2} .\) How much larger is the area of the hole at $30.0^{\circ} \mathrm{C} ?(\text { Wile tutorial: loop around the equator })$
A steel sphere with radius \(1.0010 \mathrm{cm}\) at \(22.0^{\circ} \mathrm{C}\) must slip through a brass ring that has an internal radius of $1.0000 \mathrm{cm}$ at the same temperature. To what temperature must the brass ring be heated so that the sphere, still at \(22.0^{\circ} \mathrm{C},\) can just slip through?
A scuba diver has an air tank with a volume of \(0.010 \mathrm{m}^{3}\) The air in the tank is initially at a pressure of \(1.0 \times 10^{7} \mathrm{Pa}\) Assuming that the diver breathes \(0.500 \mathrm{L} / \mathrm{s}\) of air, find how long the tank will last at depths of (a) \(2.0 \mathrm{m}\) and (b) $20.0 \mathrm{m} .\( (Make the same assumptions as in Example \)13.6 .)$
See all solutions

Recommended explanations on Physics Textbooks

Force

Read Explanation

Quantum Physics

Read Explanation

Nuclear Physics

Read Explanation

Oscillations

Read Explanation

Classical Mechanics

Read Explanation

Magnetism

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