The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force FSM that the sun exerts moon is perpendicular to the force FEM that the earth exerts on the moon. The masses are: mass of sun = 1.99 × 1030 kg, ma = 5.98 x 1024 kg, mass of moon = 7.35 x 1022 kg. The distances shown in the drawing are rsM = 1.50 × 1011 m and EM = 3.85 Determine the magnitude of the net gravitational force on the moon. Number : Sun Inits TSM FSM FEM Earth Moon TEM
The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force FSM that the sun exerts moon is perpendicular to the force FEM that the earth exerts on the moon. The masses are: mass of sun = 1.99 × 1030 kg, ma = 5.98 x 1024 kg, mass of moon = 7.35 x 1022 kg. The distances shown in the drawing are rsM = 1.50 × 1011 m and EM = 3.85 Determine the magnitude of the net gravitational force on the moon. Number : Sun Inits TSM FSM FEM Earth Moon TEM
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
Related questions
Concept explainers
Gravitational force
In nature, every object is attracted by every other object. This phenomenon is called gravity. The force associated with gravity is called gravitational force. The gravitational force is the weakest force that exists in nature. The gravitational force is always attractive.
Acceleration Due to Gravity
In fundamental physics, gravity or gravitational force is the universal attractive force acting between all the matters that exist or exhibit. It is the weakest known force. Therefore no internal changes in an object occurs due to this force. On the other hand, it has control over the trajectories of bodies in the solar system and in the universe due to its vast scope and universal action. The free fall of objects on Earth and the motions of celestial bodies, according to Newton, are both determined by the same force. It was Newton who put forward that the moon is held by a strong attractive force exerted by the Earth which makes it revolve in a straight line. He was sure that this force is similar to the downward force which Earth exerts on all the objects on it.
Question
![### Example Problem: Gravitational Forces in the Sun-Earth-Moon System
#### Problem Description:
The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force \(\vec{F}_{SM}\) that the sun exerts on the moon is perpendicular to the force \(\vec{F}_{EM}\) that the earth exerts on the moon. The masses are:
- Mass of the sun (\(m_{S}\)) = \(1.99 \times 10^{30}\) kg
- Mass of the earth (\(m_{E}\)) = \(5.98 \times 10^{24}\) kg
- Mass of the moon (\(m_{M}\)) = \(7.35 \times 10^{22}\) kg
The distances shown in the drawing are:
- Distance between the sun and the moon (\(r_{SM}\)) = \(1.50 \times 10^{11}\) m
- Distance between the earth and the moon (\(r_{EM}\)) = \(3.85 \times 10^{8}\) m
Determine the magnitude of the net gravitational force on the moon.
#### Diagram Explanation:
The diagram illustrates the following key points:
- The Sun is represented as a yellow star on the left side of the diagram.
- The Earth is depicted as a blue sphere with green continents at a lower vertical position.
- The Moon is shown as a smaller gray sphere in the upper region of the drawing, located such that it forms a right angle with lines extending from the Earth and the Sun.
- Two forces are acting on the Moon:
- \(\vec{F}_{SM}\) (blue arrow pointing from the Sun to the Moon) represents the gravitational force exerted by the Sun on the Moon.
- \(\vec{F}_{EM}\) (blue arrow pointing from the Earth to the Moon) represents the gravitational force exerted by the Earth on the Moon.
- The distance \(r_{SM}\) is the horizontal distance between the Sun and the Moon.
- The distance \(r_{EM}\) is the vertical distance between the Earth and the Moon.
### Calculation:
Fill in the calculated values to determine the net gravitational force on the Moon:
\[
\text{Number: } \_\_\_\_\_\_
\]
\[
\text{Units: } \_\_\_\_\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f0a09bd-ef71-43b6-abcd-653755da1dfb%2Fb15aed62-59bb-4a4b-98ac-11187c06f2a9%2Fedxp7j_processed.png&w=3840&q=75)
Transcribed Image Text:### Example Problem: Gravitational Forces in the Sun-Earth-Moon System
#### Problem Description:
The drawing (not to scale) shows one alignment of the sun, earth, and moon. The gravitational force \(\vec{F}_{SM}\) that the sun exerts on the moon is perpendicular to the force \(\vec{F}_{EM}\) that the earth exerts on the moon. The masses are:
- Mass of the sun (\(m_{S}\)) = \(1.99 \times 10^{30}\) kg
- Mass of the earth (\(m_{E}\)) = \(5.98 \times 10^{24}\) kg
- Mass of the moon (\(m_{M}\)) = \(7.35 \times 10^{22}\) kg
The distances shown in the drawing are:
- Distance between the sun and the moon (\(r_{SM}\)) = \(1.50 \times 10^{11}\) m
- Distance between the earth and the moon (\(r_{EM}\)) = \(3.85 \times 10^{8}\) m
Determine the magnitude of the net gravitational force on the moon.
#### Diagram Explanation:
The diagram illustrates the following key points:
- The Sun is represented as a yellow star on the left side of the diagram.
- The Earth is depicted as a blue sphere with green continents at a lower vertical position.
- The Moon is shown as a smaller gray sphere in the upper region of the drawing, located such that it forms a right angle with lines extending from the Earth and the Sun.
- Two forces are acting on the Moon:
- \(\vec{F}_{SM}\) (blue arrow pointing from the Sun to the Moon) represents the gravitational force exerted by the Sun on the Moon.
- \(\vec{F}_{EM}\) (blue arrow pointing from the Earth to the Moon) represents the gravitational force exerted by the Earth on the Moon.
- The distance \(r_{SM}\) is the horizontal distance between the Sun and the Moon.
- The distance \(r_{EM}\) is the vertical distance between the Earth and the Moon.
### Calculation:
Fill in the calculated values to determine the net gravitational force on the Moon:
\[
\text{Number: } \_\_\_\_\_\_
\]
\[
\text{Units: } \_\_\_\_\
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Recommended textbooks for you
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
College Physics
Physics
ISBN:
9781305952300
Author:
Raymond A. Serway, Chris Vuille
Publisher:
Cengage Learning
University Physics (14th Edition)
Physics
ISBN:
9780133969290
Author:
Hugh D. Young, Roger A. Freedman
Publisher:
PEARSON
Introduction To Quantum Mechanics
Physics
ISBN:
9781107189638
Author:
Griffiths, David J., Schroeter, Darrell F.
Publisher:
Cambridge University Press
Physics for Scientists and Engineers
Physics
ISBN:
9781337553278
Author:
Raymond A. Serway, John W. Jewett
Publisher:
Cengage Learning
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:
9780321820464
Author:
Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:
Addison-Wesley
College Physics: A Strategic Approach (4th Editio…
Physics
ISBN:
9780134609034
Author:
Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:
PEARSON