Answer :
To determine the gravitational force between the two masses, you can use Newton's law of universal gravitation. This law states that every point mass attracts every other point mass in the universe with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between them. Here's the equation:
[tex]\[ \vec{F} = G \frac{m_1 m_2}{r^2} \][/tex]
Where:
- [tex]\(\vec{F}\)[/tex] is the gravitational force.
- [tex]\(G\)[/tex] is the gravitational constant ([tex]\(6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2\)[/tex]).
- [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] are the masses (164 kg and 89.1 kg respectively).
- [tex]\(r\)[/tex] is the distance between the centers of the two masses (68.8 m).
Plugging in the given values, we get:
[tex]\[ \vec{F} = 6.67 \times 10^{-11} \times \frac{164 \times 89.1}{68.8^2} \][/tex]
Evaluating the product of the masses ([tex]\(m_1 m_2\)[/tex]):
[tex]\[ 164 \times 89.1 = 14607.4 \, \text{kg}^2 \][/tex]
Next, evaluate the square of the distance ([tex]\(r^2\)[/tex]):
[tex]\[ 68.8^2 = 4734.24 \, \text{m}^2 \][/tex]
Now, combine these results to complete the fraction:
[tex]\[ \frac{m_1 m_2}{r^2} = \frac{14607.4}{4734.24} \][/tex]
Perform the division:
[tex]\[ \frac{14607.4}{4734.24} \approx 3.084 \, \text{kg} \, \text{m}^{-2} \][/tex]
Finally, multiply this result by the gravitational constant [tex]\(G\)[/tex]:
[tex]\[ \vec{F} = 6.67 \times 10^{-11} \times 3.084 \][/tex]
[tex]\[ \vec{F} \approx 2.06 \times 10^{-10} \, \text{N} \][/tex]
Therefore, the gravitational force between the two masses is:
[tex]\[ \vec{F} \approx 2.06 \times 10^{-10} \, \text{N} \][/tex]
[tex]\[ \vec{F} = G \frac{m_1 m_2}{r^2} \][/tex]
Where:
- [tex]\(\vec{F}\)[/tex] is the gravitational force.
- [tex]\(G\)[/tex] is the gravitational constant ([tex]\(6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2\)[/tex]).
- [tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] are the masses (164 kg and 89.1 kg respectively).
- [tex]\(r\)[/tex] is the distance between the centers of the two masses (68.8 m).
Plugging in the given values, we get:
[tex]\[ \vec{F} = 6.67 \times 10^{-11} \times \frac{164 \times 89.1}{68.8^2} \][/tex]
Evaluating the product of the masses ([tex]\(m_1 m_2\)[/tex]):
[tex]\[ 164 \times 89.1 = 14607.4 \, \text{kg}^2 \][/tex]
Next, evaluate the square of the distance ([tex]\(r^2\)[/tex]):
[tex]\[ 68.8^2 = 4734.24 \, \text{m}^2 \][/tex]
Now, combine these results to complete the fraction:
[tex]\[ \frac{m_1 m_2}{r^2} = \frac{14607.4}{4734.24} \][/tex]
Perform the division:
[tex]\[ \frac{14607.4}{4734.24} \approx 3.084 \, \text{kg} \, \text{m}^{-2} \][/tex]
Finally, multiply this result by the gravitational constant [tex]\(G\)[/tex]:
[tex]\[ \vec{F} = 6.67 \times 10^{-11} \times 3.084 \][/tex]
[tex]\[ \vec{F} \approx 2.06 \times 10^{-10} \, \text{N} \][/tex]
Therefore, the gravitational force between the two masses is:
[tex]\[ \vec{F} \approx 2.06 \times 10^{-10} \, \text{N} \][/tex]