/*! 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

Spring scale \(A\) is attached to the ceiling. A 10.0 -kg mass is suspended from the scale. A second spring scale, \(\mathrm{B}\), is hanging from a hook at the bottom of the 10.0 -kg mass and a 4.0 -kg mass hangs from the second spring scale. (a) What are the readings of the two scales if the masses of the scales are negligible? (b) What are the readings if each scale has a mass of $1.0 \mathrm{kg} ?$
A skydiver, who weighs \(650 \mathrm{N},\) is falling at a constant speed with his parachute open. Consider the apparatus that connects the parachute to the skydiver to be part of the parachute. The parachute pulls upward with a force of \(620 \mathrm{N} .\) (a) What is the force of the air resistance acting on the skydiver? (b) Identify the forces and the interaction partners of each force exerted on the skydiver. (c) Identify the forces and interaction partners of each force exerted on the parachute.
Two of Robin Hood's men are pulling a sledge loaded with some gold along a path that runs due north to their hideout. One man pulls his rope with a force of \(62 \mathrm{N}\) at an angle of \(12^{\circ}\) east of north and the other pulls with the same force at an angle of \(12^{\circ}\) west of north. Assume the ropes are parallel to the ground. What is the sum of these two forces on the sledge?
A book rests on the surface of the table. Consider the following four forces that arise in this situation: (a) the force of the Earth pulling on the book, (b) the force of the table pushing on the book, (c) the force of the book pushing on the table, and (d) the force of the book pulling on the Earth. The book is not moving. Which pair of forces must be equal in magnitude and opposite in direction even though they are not an interaction pair?
The coefficient of static friction between a block and a horizontal floor is \(0.40,\) while the coefficient of kinetic friction is \(0.15 .\) The mass of the block is \(5.0 \mathrm{kg} .\) A horizontal force is applied to the block and slowly increased. (a) What is the value of the applied horizontal force at the instant that the block starts to slide? (b) What is the net force on the block after it starts to slide?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Mechanics and Materials

Read Explanation

Physical Quantities And Units

Read Explanation

Thermodynamics

Read Explanation

Engineering Physics

Read Explanation

Fundamentals of 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.