At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

Use the following information to determine the distance between Earth and its moon.

| Earth's mass = [tex]\(6.0 \times 10^{24} \text{ kg}\)[/tex] | Moon's tangential speed = [tex]\(1,025 \text{ m/s}\)[/tex] |
| Earth's radius = [tex]\(6.4 \times 10^6 \text{ m}\)[/tex] | Moon's mass = [tex]\(7.3 \times 10^{22} \text{ kg}\)[/tex] |

Possible distances:
A. [tex]\(4.6 \times 10^6 \text{ m}\)[/tex]
B. [tex]\(3.8 \times 10^8 \text{ m}\)[/tex]
C. [tex]\(3.9 \times 10^{11} \text{ m}\)[/tex]

Sagot :

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]