Answer :
### 10. Calculating the Gravitational Force Between the Large Ball and the Small Ball
To calculate the gravitational force between two objects, we use Newton's Law of Universal Gravitation, which states that:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the gravitational force between the two objects.
- [tex]\( G \)[/tex] is the gravitational constant, approximately [tex]\( 6.67430 \times 10^{-11} \, \text{N}(m^2/kg^2) \)[/tex].
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects.
- [tex]\( r \)[/tex] is the distance between the centers of the two objects.
Given:
- Mass of the Large Ball, [tex]\( m_{\text{large ball}} = 9.0 \times 10^{-2} \, \text{kg} \)[/tex]
- Mass of the Small Ball, [tex]\( m_{\text{small ball}} = 5.0 \times 10^{-3} \, \text{kg} \)[/tex]
- Distance between the balls, [tex]\( r = 0.5 \, \text{m} \)[/tex]
Plug these values into the formula:
[tex]\[ F = 6.67430 \times 10^{-11} \, \frac{(9.0 \times 10^{-2} \, \text{kg}) \times (5.0 \times 10^{-3} \, \text{kg})}{(0.5 \, \text{m})^2} \][/tex]
First, calculate the product of the masses:
[tex]\[ m_1 \times m_2 = 9.0 \times 10^{-2} \, \text{kg} \times 5.0 \times 10^{-3} \, \text{kg} = 4.5 \times 10^{-4} \, \text{kg}^2 \][/tex]
Then, calculate the square of the distance:
[tex]\[ r^2 = (0.5 \, \text{m})^2 = 0.25 \, \text{m}^2 \][/tex]
Now, substitute these values back into the force equation:
[tex]\[ F = 6.67430 \times 10^{-11} \, \frac{4.5 \times 10^{-4} \, \text{kg}^2}{0.25 \, \text{m}^2} \][/tex]
Simplify the fraction:
[tex]\[ F = 6.67430 \times 10^{-11} \, \times 1.8 \times 10^{-3} \][/tex]
Finally, multiply the constants:
[tex]\[ F \approx 1.201374 \times 10^{-13} \, \text{N} \][/tex]
Thus, the gravitational force between the large ball and the small ball is approximately:
[tex]\[ \boxed{1.201374 \times 10^{-13} \, \text{N}} \][/tex]
### 11. Explanation of Why Objects Fall to Earth and Not Towards Each Other
In the second part of the question, we are asked to explain why the objects (large ball, small ball, and feather) fall to Earth rather than moving towards each other. To understand this, let's compare the gravitational forces involved:
- Earth and Large Ball: [tex]\( 0.882 \, \text{N} \)[/tex]
- Earth and Small Ball: [tex]\( 0.049 \, \text{N} \)[/tex]
- Large Ball and Feather: [tex]\( 2.41 \times 10^{-14} \, \text{N} \)[/tex]
- Large Ball and Small Ball: [tex]\( 1.201374 \times 10^{-13} \, \text{N} \)[/tex]
These gravitational forces indicate the relative strength of the attraction:
- The gravitational force between the Earth and the large ball ([tex]\( 0.882 \, \text{N} \)[/tex]) is significantly larger than the force between the large ball and the small ball ([tex]\( 1.201374 \times 10^{-13} \, \text{N} \)[/tex]).
- Similarly, the gravitational force between the Earth and the small ball ([tex]\( 0.049 \, \text{N} \)[/tex]) is much stronger compared to the force between the large ball and the small ball.
- The force between the large ball and the feather is even smaller, at [tex]\( 2.41 \times 10^{-14} \, \text{N} \)[/tex].
The vastly greater gravitational force of Earth on these objects means that when they are dropped, the force pulling them towards Earth is overwhelmingly stronger compared to the mutually attractive forces between the small objects themselves. Each object's fall to Earth is dominated by the Earth's massive gravitational pull, making it so they do not move towards each other but rather accelerate downwards towards the planet's surface.
In summary, the gravitational attraction of Earth on each object is vastly stronger than the gravitational attractions between the objects themselves. Therefore, when these objects are dropped, they fall towards Earth rather than moving towards each other.
To calculate the gravitational force between two objects, we use Newton's Law of Universal Gravitation, which states that:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the gravitational force between the two objects.
- [tex]\( G \)[/tex] is the gravitational constant, approximately [tex]\( 6.67430 \times 10^{-11} \, \text{N}(m^2/kg^2) \)[/tex].
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects.
- [tex]\( r \)[/tex] is the distance between the centers of the two objects.
Given:
- Mass of the Large Ball, [tex]\( m_{\text{large ball}} = 9.0 \times 10^{-2} \, \text{kg} \)[/tex]
- Mass of the Small Ball, [tex]\( m_{\text{small ball}} = 5.0 \times 10^{-3} \, \text{kg} \)[/tex]
- Distance between the balls, [tex]\( r = 0.5 \, \text{m} \)[/tex]
Plug these values into the formula:
[tex]\[ F = 6.67430 \times 10^{-11} \, \frac{(9.0 \times 10^{-2} \, \text{kg}) \times (5.0 \times 10^{-3} \, \text{kg})}{(0.5 \, \text{m})^2} \][/tex]
First, calculate the product of the masses:
[tex]\[ m_1 \times m_2 = 9.0 \times 10^{-2} \, \text{kg} \times 5.0 \times 10^{-3} \, \text{kg} = 4.5 \times 10^{-4} \, \text{kg}^2 \][/tex]
Then, calculate the square of the distance:
[tex]\[ r^2 = (0.5 \, \text{m})^2 = 0.25 \, \text{m}^2 \][/tex]
Now, substitute these values back into the force equation:
[tex]\[ F = 6.67430 \times 10^{-11} \, \frac{4.5 \times 10^{-4} \, \text{kg}^2}{0.25 \, \text{m}^2} \][/tex]
Simplify the fraction:
[tex]\[ F = 6.67430 \times 10^{-11} \, \times 1.8 \times 10^{-3} \][/tex]
Finally, multiply the constants:
[tex]\[ F \approx 1.201374 \times 10^{-13} \, \text{N} \][/tex]
Thus, the gravitational force between the large ball and the small ball is approximately:
[tex]\[ \boxed{1.201374 \times 10^{-13} \, \text{N}} \][/tex]
### 11. Explanation of Why Objects Fall to Earth and Not Towards Each Other
In the second part of the question, we are asked to explain why the objects (large ball, small ball, and feather) fall to Earth rather than moving towards each other. To understand this, let's compare the gravitational forces involved:
- Earth and Large Ball: [tex]\( 0.882 \, \text{N} \)[/tex]
- Earth and Small Ball: [tex]\( 0.049 \, \text{N} \)[/tex]
- Large Ball and Feather: [tex]\( 2.41 \times 10^{-14} \, \text{N} \)[/tex]
- Large Ball and Small Ball: [tex]\( 1.201374 \times 10^{-13} \, \text{N} \)[/tex]
These gravitational forces indicate the relative strength of the attraction:
- The gravitational force between the Earth and the large ball ([tex]\( 0.882 \, \text{N} \)[/tex]) is significantly larger than the force between the large ball and the small ball ([tex]\( 1.201374 \times 10^{-13} \, \text{N} \)[/tex]).
- Similarly, the gravitational force between the Earth and the small ball ([tex]\( 0.049 \, \text{N} \)[/tex]) is much stronger compared to the force between the large ball and the small ball.
- The force between the large ball and the feather is even smaller, at [tex]\( 2.41 \times 10^{-14} \, \text{N} \)[/tex].
The vastly greater gravitational force of Earth on these objects means that when they are dropped, the force pulling them towards Earth is overwhelmingly stronger compared to the mutually attractive forces between the small objects themselves. Each object's fall to Earth is dominated by the Earth's massive gravitational pull, making it so they do not move towards each other but rather accelerate downwards towards the planet's surface.
In summary, the gravitational attraction of Earth on each object is vastly stronger than the gravitational attractions between the objects themselves. Therefore, when these objects are dropped, they fall towards Earth rather than moving towards each other.
