d by a gravitational force of 9.3 N. What will the force of attraction be if the distance between the two masses e final answer to four decimal places.) ill be 148.8 ON.
d by a gravitational force of 9.3 N. What will the force of attraction be if the distance between the two masses e final answer to four decimal places.) ill be 148.8 ON.
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)...
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![**Question:**
Two masses are attracted by a gravitational force of 9.3 N. What will the force of attraction be if the distance between the two masses is quadrupled? (Round the final answer to four decimal places.)
**Incorrect Response Given:**
The force of attraction will be \(148.8\) N.
**Explanation:**
The gravitational force between two masses is governed by Newton's Law of Universal Gravitation, which is mathematically expressed as:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
Where:
- \( F \) is the gravitational force between the masses,
- \( G \) is the gravitational constant,
- \( m_1 \) and \( m_2 \) are the masses,
- \( r \) is the distance between the centers of the two masses.
When the distance \( r \) is quadrupled, the force \( F \) becomes:
\[ F' = \frac{G \cdot m_1 \cdot m_2}{(4r)^2} = \frac{G \cdot m_1 \cdot m_2}{16r^2} = \frac{1}{16} \cdot F \]
So the new force \( F' \) is \( \frac{1}{16} \) of the original force.
Given the original force \( F = 9.3 \) N, the new force is:
\[ F' = \frac{9.3}{16} \approx 0.5813 \, \text{N} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fefd1da3f-cb32-49f3-8aa7-bea0cbc5b54f%2F33e28e29-8776-4313-9f09-7bf7c47f53d1%2Foej26py_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
Two masses are attracted by a gravitational force of 9.3 N. What will the force of attraction be if the distance between the two masses is quadrupled? (Round the final answer to four decimal places.)
**Incorrect Response Given:**
The force of attraction will be \(148.8\) N.
**Explanation:**
The gravitational force between two masses is governed by Newton's Law of Universal Gravitation, which is mathematically expressed as:
\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]
Where:
- \( F \) is the gravitational force between the masses,
- \( G \) is the gravitational constant,
- \( m_1 \) and \( m_2 \) are the masses,
- \( r \) is the distance between the centers of the two masses.
When the distance \( r \) is quadrupled, the force \( F \) becomes:
\[ F' = \frac{G \cdot m_1 \cdot m_2}{(4r)^2} = \frac{G \cdot m_1 \cdot m_2}{16r^2} = \frac{1}{16} \cdot F \]
So the new force \( F' \) is \( \frac{1}{16} \) of the original force.
Given the original force \( F = 9.3 \) N, the new force is:
\[ F' = \frac{9.3}{16} \approx 0.5813 \, \text{N} \]
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