Answer :
To calculate the gravitational force between the large ball and the small ball, we use Newton's law of universal gravitation. The formula for the gravitational force between two masses is given by:
[tex]\[ F = G \frac{m_{1} m_{2}}{r^{2}} \][/tex]
where
- [tex]\( F \)[/tex] is the gravitational force,
- [tex]\( G \)[/tex] is the gravitational constant [tex]\((6.67430 \times 10^{-11} \ \text{N} \cdot \text{m}^{2} \cdot \text{kg}^{-2})\)[/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 masses.
Given:
- [tex]\( m_{\text{small ball}} = 5.0 \times 10^{-3} \ \text{kg} \)[/tex]
- [tex]\( m_{\text{large ball}} = 9.0 \times 10^{2} \ \text{kg} \)[/tex]
- [tex]\( r = 0.5 \ \text{m} \)[/tex]
Now, substituting these values into the formula:
[tex]\[ F = (6.67430 \times 10^{-11}) \frac{(5.0 \times 10^{-3})(9.0 \times 10^{2})}{(0.5)^{2}} \][/tex]
First, calculate the product of the masses:
[tex]\[ m_{1} m_{2} = (5.0 \times 10^{-3}) \times (9.0 \times 10^{2}) = 4.5 \times 10^{0} = 4.5 \][/tex]
Next, calculate the square of the distance:
[tex]\[ r^{2} = (0.5)^{2} = 0.25 \][/tex]
Then, substitute these values into the formula:
[tex]\[ F = (6.67430 \times 10^{-11}) \frac{4.5}{0.25} \][/tex]
[tex]\[ F = (6.67430 \times 10^{-11}) \times 18.0 \][/tex]
[tex]\[ F = 1.201374 \times 10^{-9}\ \text{N} \][/tex]
Therefore, the gravitational force between the large ball and the small ball is:
[tex]\[ F = 1.201374 \times 10^{-9}\ \text{N} \][/tex]
11. When the large ball, small ball, and feather are dropped side by side, the gravitational forces between the objects themselves are extremely small compared to the gravitational force exerted by the Earth on these objects.
For instance, the force between the Earth and each object (large ball, small ball, and feather) is much greater:
- Earth and Large Ball: [tex]\( 0.882 \ \text{N} \)[/tex]
- Earth and Small Ball: [tex]\( 0.049 \ \text{N} \)[/tex]
- Earth and Feather: very negligible but still greater than inter-object forces.
These forces vastly overshadow the gravitational force between the large ball and the small ball ([tex]\( 1.201374 \times 10^{-9}\ \text{N} \)[/tex]), and between other small combinations such as the large ball and feather ([tex]\( 2.41 \times 10^{-44}\ \text{N} \)[/tex]), causing all objects to fall to Earth rather than move toward each other.
Therefore, the objects fall to Earth due to the Earth's significantly stronger gravitational pull on each of them compared to the negligible gravitational forces they exert on each other.
[tex]\[ F = G \frac{m_{1} m_{2}}{r^{2}} \][/tex]
where
- [tex]\( F \)[/tex] is the gravitational force,
- [tex]\( G \)[/tex] is the gravitational constant [tex]\((6.67430 \times 10^{-11} \ \text{N} \cdot \text{m}^{2} \cdot \text{kg}^{-2})\)[/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 masses.
Given:
- [tex]\( m_{\text{small ball}} = 5.0 \times 10^{-3} \ \text{kg} \)[/tex]
- [tex]\( m_{\text{large ball}} = 9.0 \times 10^{2} \ \text{kg} \)[/tex]
- [tex]\( r = 0.5 \ \text{m} \)[/tex]
Now, substituting these values into the formula:
[tex]\[ F = (6.67430 \times 10^{-11}) \frac{(5.0 \times 10^{-3})(9.0 \times 10^{2})}{(0.5)^{2}} \][/tex]
First, calculate the product of the masses:
[tex]\[ m_{1} m_{2} = (5.0 \times 10^{-3}) \times (9.0 \times 10^{2}) = 4.5 \times 10^{0} = 4.5 \][/tex]
Next, calculate the square of the distance:
[tex]\[ r^{2} = (0.5)^{2} = 0.25 \][/tex]
Then, substitute these values into the formula:
[tex]\[ F = (6.67430 \times 10^{-11}) \frac{4.5}{0.25} \][/tex]
[tex]\[ F = (6.67430 \times 10^{-11}) \times 18.0 \][/tex]
[tex]\[ F = 1.201374 \times 10^{-9}\ \text{N} \][/tex]
Therefore, the gravitational force between the large ball and the small ball is:
[tex]\[ F = 1.201374 \times 10^{-9}\ \text{N} \][/tex]
11. When the large ball, small ball, and feather are dropped side by side, the gravitational forces between the objects themselves are extremely small compared to the gravitational force exerted by the Earth on these objects.
For instance, the force between the Earth and each object (large ball, small ball, and feather) is much greater:
- Earth and Large Ball: [tex]\( 0.882 \ \text{N} \)[/tex]
- Earth and Small Ball: [tex]\( 0.049 \ \text{N} \)[/tex]
- Earth and Feather: very negligible but still greater than inter-object forces.
These forces vastly overshadow the gravitational force between the large ball and the small ball ([tex]\( 1.201374 \times 10^{-9}\ \text{N} \)[/tex]), and between other small combinations such as the large ball and feather ([tex]\( 2.41 \times 10^{-44}\ \text{N} \)[/tex]), causing all objects to fall to Earth rather than move toward each other.
Therefore, the objects fall to Earth due to the Earth's significantly stronger gravitational pull on each of them compared to the negligible gravitational forces they exert on each other.
