### Physics Problem: Gravitational Attraction of Compressed Matter Balls**Problem Statement:**Two identical tiny balls of highly compressed matter are 1.50 meters apart. When released in an orbiting space station, they accelerate toward each other at 2.00 cm/s². Given:- Separation distance (\( d \)): 1.50 m- Acceleration (\( a \)): 2.00 cm/s² \( = 0.02 \) m/s²- Gravitational constant (\( G \)): \( 6.67 \times 10^{-11} \) N \( \cdot \) m²/kg²**Question:**What is the mass of each of them?**Equation for gravitational force:**\[ F = G \frac{m_1 m_2}{d^2} \]Where:- \( F \) is the gravitational force,- \( G \) is the gravitational constant,- \( m_1 \) and \( m_2 \) are the masses of the balls,- \( d \) is the separation distance.Since the balls are identical, \( m_1 = m_2 = m \). Additionally, using Newton's second law of motion (\( F = ma \)), and substituting into the gravitational force equation, we get:\[ m \cdot a = G \frac{m^2}{d^2} \]\[ a = G \frac{m}{d^2} \]\[ m = \frac{a d^2}{G} \]**Calculation:**Plugging in the values:\[ m = \frac{0.02 \cdot (1.50)^2}{6.67 \times 10^{-11}} \]\[ m = \frac{0.02 \cdot 2.25}{6.67 \times 10^{-11}} \]\[ m = \frac{0.045}{6.67 \times 10^{-11}} \]\[ m \approx 6.75 \times 10^{8} \text{ kg} \]**Answer:**The mass of each ball is:\[ 6.75 \times 10^{8} \text{ kg} \]

Gravity is a natural force that attracts objects with mass towards each other. It is one of the…

Two identical tiny balls of highly compressed matter are 1.50 apart. When released in an orbiting space station, they accelerate toward each other at 2.00 cm/s². (G= 6.67 x 10-11 N-m²/kg²) What is the mass of each of them? x 10 kg

Two identical tiny balls of highly compressed matter are 1.50 apart. When released in an orbiting space station, they accelerate toward each other at 2.00 cm/s². (G= 6.67 x 10-11 N-m²/kg²) What is the mass of each of them? x 10 kg

College Physics

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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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According to classical physics, time flow is the same for every observer. But according to einstein's theory, time flow is different for different observers. The idea that time flow is not the same for all may sound strange because we may not experience this effect in everyday life. This effect is only observed when an object moves with some velocity comparable to the velocity of light. This velocity is termed relativistic velocity.

Question

### Physics Problem: Gravitational Attraction of Compressed Matter Balls

**Problem Statement:**
Two identical tiny balls of highly compressed matter are 1.50 meters apart. When released in an orbiting space station, they accelerate toward each other at 2.00 cm/s².

Given:
- Separation distance (\( d \)): 1.50 m
- Acceleration (\( a \)): 2.00 cm/s² \( = 0.02 \) m/s²
- Gravitational constant (\( G \)): \( 6.67 \times 10^{-11} \) N \( \cdot \) m²/kg²

**Question:**
What is the mass of each of them?

**Equation for gravitational force:**
\[ F = G \frac{m_1 m_2}{d^2} \]
Where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant,
- \( m_1 \) and \( m_2 \) are the masses of the balls,
- \( d \) is the separation distance.

Since the balls are identical, \( m_1 = m_2 = m \). Additionally, using Newton's second law of motion (\( F = ma \)), and substituting into the gravitational force equation, we get:
\[ m \cdot a = G \frac{m^2}{d^2} \]
\[ a = G \frac{m}{d^2} \]
\[ m = \frac{a d^2}{G} \]

**Calculation:**
Plugging in the values:
\[ m = \frac{0.02 \cdot (1.50)^2}{6.67 \times 10^{-11}} \]
\[ m = \frac{0.02 \cdot 2.25}{6.67 \times 10^{-11}} \]
\[ m = \frac{0.045}{6.67 \times 10^{-11}} \]
\[ m \approx 6.75 \times 10^{8} \text{ kg} \]

**Answer:**
The mass of each ball is:
\[ 6.75 \times 10^{8} \text{ kg} \]

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