/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 149 (a) If a spacecraft moves in a s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 149

(a) If a spacecraft moves in a straight line between the Earth and the Sun, at what point would the force of gravity on the spacecraft due to the Sun be as large as that due to the Earth? (b) If the spacecraft is close to, but not at, this equilibrium point, does the net force on the spacecraft tend to push it toward or away from the equilibrium point? [Hint: Imagine the spacecraft a small distance \(d\) closer to the Earth and find out which gravitational force is stronger.]

Short Answer

Expert verified
Answer: The equilibrium distance is given by the formula \(r_e = \frac{m_ER^2}{2m_ER + (m_E + m_S)r_e}\), where \(r_e\) is the distance from the Earth to the spacecraft, \(m_E\) is Earth's mass, \(m_S\) is the Sun's mass, and \(R\) is the distance between Earth and the Sun. When the spacecraft is slightly closer to the Earth, the net force pushes it away from the Earth and towards the equilibrium point.

Step by step solution

01

Write the equations for gravitational force

Let \(F_e\) be the gravitational force due to the Earth on the spacecraft and \(F_s\) be the gravitational force due to the Sun on the spacecraft. Using the gravitational force formula, we have: \(F_e = G\frac{m_Em}{r_e^2}\) (Earth's gravitational force) \(F_s = G\frac{m_Sm}{r_s^2}\) (Sun's gravitational force) Where \(m_E\) is Earth's mass, \(m_S\) is the Sun's mass, \(m\) is the spacecraft's mass, \(r_e\) is the distance from the Earth to the spacecraft, and \(r_s\) is the distance from the Sun to the spacecraft.
02

Set up the equilibrium condition

The equilibrium condition is that at a certain point, the gravitational force due to the Earth equals the gravitational force due to the Sun. Therefore, we have: \(F_e = F_s\) Plugging in the equations from Step 1, we get: \(G\frac{m_Em}{r_e^2} = G\frac{m_Sm}{r_s^2}\)
03

Solve for the distance from the Earth

To find the equilibrium point, we need to find the distance \(r_e\) from the Earth when the forces are equal. We can simplify the equation from Step 2 by canceling out like terms: \(\frac{m_E}{r_e^2} = \frac{m_S}{r_s^2}\) We know that \(r_e + r_s\) is the distance between Earth and the Sun, which we can denote as \(R\). So, \(r_s = R - r_e\). Now, we substitute \(r_s\) in the equation: \(\frac{m_E}{r_e^2} = \frac{m_S}{(R - r_e)^2}\) Now, we can solve for \(r_e\): \(m_ER^2 - 2m_ERr_e + m_Er_e^2 = m_Sr_e^2\)
04

Find the equilibrium distance

Solve for \(r_e\) to find the equilibrium distance: \(r_e = \frac{m_ER^2}{2m_ER + (m_E + m_S)r_e}\)
05

Analyze the net force when the spacecraft is close to the equilibrium point

Let the spacecraft be a small distance \(d\) closer to the Earth. The gravitational force due to the Earth becomes stronger, while the gravitational force due to the Sun becomes weaker. This means that \(F_e > F_s\). Since the Earth's gravitational force is stronger, the net force on the spacecraft pushes it away from the Earth and towards the equilibrium point.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 box sits on a horizontal wooden ramp. The coefficient of static friction between the box and the ramp is 0.30 You grab one end of the ramp and lift it up, keeping the other end of the ramp on the ground. What is the angle between the ramp and the horizontal direction when the box begins to slide down the ramp? (tutorial: crate on ramp)
You grab a book and give it a quick push across the top of a horizontal table. After a short push, the book slides across the table, and because of friction, comes to a stop. (a) Draw an FBD of the book while you are pushing it. (b) Draw an FBD of the book after you have stopped pushing it, while it is sliding across the table. (c) Draw an FBD of the book after it has stopped sliding. (d) In which of the preceding cases is the net force on the book not equal to zero? (e) If the book has a mass of \(0.50 \mathrm{kg}\) and the coefficient of friction between the book and the table is \(0.40,\) what is the net force acting on the book in part (b)? (f) If there were no friction between the table and the book, what would the free-body diagram for part (b) look like? Would the book slow down in this case? Why or why not?
(a) What is the magnitude of the gravitational force that the Earth exerts on the Moon? (b) What is the magnitude of the gravitational force that the Moon exerts on the Earth? See the inside front and back covers for necessary information.
A crow perches on a clothesline midway between two poles. Each end of the rope makes an angle of \(\theta\) below the horizontal where it connects to the pole. If the weight of the crow is \(W,\) what is the tension in the rope? Ignore the weight of the rope.
Tamar wants to cut down a large, dead poplar tree with her chain saw, but she does not want it to fall onto the nearby gazebo. Yoojin comes to help with a long rope. Yoojin, a physicist, suggests they tie the rope taut from the poplar to the oak tree and then pull sideways on the rope as shown in the figure. If the rope is 40.0 m long and Yoojin pulls sideways at the midpoint of the rope with a force of \(360.0 \mathrm{N},\) causing a \(2.00-\mathrm{m}\) sideways displacement of the rope at its midpoint, what force will the rope exert on the poplar tree? Compare this with pulling the rope directly away from the poplar with a force of \(360.0 \mathrm{N}\) and explain why the values are different. [Hint: Until the poplar is cut through enough to start falling, the rope is in equilibrium.] (IMAGE CAN'T COPY)
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Work Energy and Power

Read Explanation

Physical Quantities And Units

Read Explanation

Famous Physicists

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Radiation

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.