Answer :
To determine the distance between the Earth and its moon using the given information, follow these steps:
1. Identify the relevant formulas:
The force of gravity between two masses is given by Newton's law of gravitation:
[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]\(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)[/tex]),
- [tex]\(m_1\)[/tex] is the mass of the Earth ([tex]\(6.0 \times 10^{24} \, \text{kg}\)[/tex]),
- [tex]\(m_2\)[/tex] is the mass of the Moon ([tex]\(7.3 \times 10^{22} \, \text{kg}\)[/tex]),
- [tex]\(r\)[/tex] is the distance between the centers of the Earth and the Moon.
2. Centripetal force consideration:
The Moon is in a circular orbit around the Earth, and the centripetal force required to keep the Moon in orbit is provided by the gravitational force. This centripetal force can be expressed as:
[tex]\[ F = \frac{m_2 v^2}{r} \][/tex]
where:
- [tex]\(v\)[/tex] is the tangential speed of the Moon ([tex]\(1025 \, \text{m/s}\)[/tex]).
3. Equate the gravitational force to the centripetal force:
[tex]\[ G \frac{m_1 m_2}{r^2} = \frac{m_2 v^2}{r} \][/tex]
Since [tex]\(m_2\)[/tex] appears on both sides of the equation, it can be canceled out:
[tex]\[ G \frac{m_1}{r^2} = \frac{v^2}{r} \][/tex]
4. Solve for [tex]\(r\)[/tex]:
Rearrange the equation to solve for [tex]\(r\)[/tex]:
[tex]\[ G \frac{m_1}{r} = v^2 \][/tex]
[tex]\[ r = \frac{G m_1}{v^2} \][/tex]
5. Substitute the values:
[tex]\[ r = \frac{6.67430 \times 10^{-11} \times 6.0 \times 10^{24}}{1025^2} \][/tex]
6. Calculate [tex]\(r\)[/tex]:
Performing the calculations:
[tex]\[ r \approx 381161689.47055316 \, \text{m} \][/tex]
Thus, the distance between the Earth and its moon is approximately [tex]\(381161689.47055316 \, \text{m}\)[/tex], which can be rounded to [tex]\(3.8 \times 10^8 \, \text{m}\)[/tex] for simplicity.
Therefore, the closest answer from the provided options is:
[tex]\[ 3.8 \times 10^8 \, \text{m} \][/tex]
1. Identify the relevant formulas:
The force of gravity between two masses is given by Newton's law of gravitation:
[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]\(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)[/tex]),
- [tex]\(m_1\)[/tex] is the mass of the Earth ([tex]\(6.0 \times 10^{24} \, \text{kg}\)[/tex]),
- [tex]\(m_2\)[/tex] is the mass of the Moon ([tex]\(7.3 \times 10^{22} \, \text{kg}\)[/tex]),
- [tex]\(r\)[/tex] is the distance between the centers of the Earth and the Moon.
2. Centripetal force consideration:
The Moon is in a circular orbit around the Earth, and the centripetal force required to keep the Moon in orbit is provided by the gravitational force. This centripetal force can be expressed as:
[tex]\[ F = \frac{m_2 v^2}{r} \][/tex]
where:
- [tex]\(v\)[/tex] is the tangential speed of the Moon ([tex]\(1025 \, \text{m/s}\)[/tex]).
3. Equate the gravitational force to the centripetal force:
[tex]\[ G \frac{m_1 m_2}{r^2} = \frac{m_2 v^2}{r} \][/tex]
Since [tex]\(m_2\)[/tex] appears on both sides of the equation, it can be canceled out:
[tex]\[ G \frac{m_1}{r^2} = \frac{v^2}{r} \][/tex]
4. Solve for [tex]\(r\)[/tex]:
Rearrange the equation to solve for [tex]\(r\)[/tex]:
[tex]\[ G \frac{m_1}{r} = v^2 \][/tex]
[tex]\[ r = \frac{G m_1}{v^2} \][/tex]
5. Substitute the values:
[tex]\[ r = \frac{6.67430 \times 10^{-11} \times 6.0 \times 10^{24}}{1025^2} \][/tex]
6. Calculate [tex]\(r\)[/tex]:
Performing the calculations:
[tex]\[ r \approx 381161689.47055316 \, \text{m} \][/tex]
Thus, the distance between the Earth and its moon is approximately [tex]\(381161689.47055316 \, \text{m}\)[/tex], which can be rounded to [tex]\(3.8 \times 10^8 \, \text{m}\)[/tex] for simplicity.
Therefore, the closest answer from the provided options is:
[tex]\[ 3.8 \times 10^8 \, \text{m} \][/tex]