Balls A and B attract each other gravitationally with a force of magnitude F at distance R. If we triple the mass of ball B and triple the separation of the balls to 3R, what is the magnitude of their attractive force now? A. F B. F/3 C. 3F D. F/9 OD OB O C O A

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**Question:**

Balls A and B attract each other gravitationally with a force of magnitude F at distance R. If we triple the mass of ball B and triple the separation of the balls to 3R, what is the magnitude of their attractive force now?

**Options:**

A. F  
B. F/3  
C. 3F  
D. F/9  

**Answer Choices:**

- ○ D
- ○ B
- ○ C
- ○ A

**Explanation:**

Using Newton’s law of universal gravitation, the gravitational force between two masses is given by:

\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

Where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between their centers. 

If the mass of ball B is tripled, it becomes \( 3m_2 \). If the separation is tripled to \( 3R \), the new force \( F' \) is:

\[ F' = \frac{G \cdot m_1 \cdot (3m_2)}{(3R)^2} = \frac{G \cdot m_1 \cdot 3m_2}{9R^2} \]

Thus,

\[ F' = \frac{1}{3} \left(\frac{G \cdot m_1 \cdot m_2}{R^2}\right) = \frac{1}{3}F \]

Therefore, the correct answer is B. F/3.
Transcribed Image Text:**Question:** Balls A and B attract each other gravitationally with a force of magnitude F at distance R. If we triple the mass of ball B and triple the separation of the balls to 3R, what is the magnitude of their attractive force now? **Options:** A. F B. F/3 C. 3F D. F/9 **Answer Choices:** - ○ D - ○ B - ○ C - ○ A **Explanation:** Using Newton’s law of universal gravitation, the gravitational force between two masses is given by: \[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \] Where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between their centers. If the mass of ball B is tripled, it becomes \( 3m_2 \). If the separation is tripled to \( 3R \), the new force \( F' \) is: \[ F' = \frac{G \cdot m_1 \cdot (3m_2)}{(3R)^2} = \frac{G \cdot m_1 \cdot 3m_2}{9R^2} \] Thus, \[ F' = \frac{1}{3} \left(\frac{G \cdot m_1 \cdot m_2}{R^2}\right) = \frac{1}{3}F \] Therefore, the correct answer is B. F/3.
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