WILLIAMS W. answered • 11/08/23
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a) To find the gravitational force acting on the moon from the earth, you can use Newton's law of universal gravitation:
\[F = \frac{G \cdot (m_1 \cdot m_2)}{r^2}\]
Where:
- \(F\) is the gravitational force,
- \(G\) is the gravitational constant (\(6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2\)),
- \(m_1\) is the mass of the Earth (\(5.98 \times 10^{24} \, \text{kg}\)),
- \(m_2\) is the mass of the Moon (\(7.35 \times 10^{22} \, \text{kg}\)), and
- \(r\) is the distance between the centers of the Earth and the Moon (\(3.8 \times 10^8 \, \text{m}\)).
Once you find the gravitational force (\(F\)), you can calculate the centripetal acceleration (\(a\)) using the formula for centripetal force:
\[F = m \cdot a\]
Where \(m\) is the mass of the Moon. The gravitational force (\(F\)) acts as the centripetal force.
b) To find the velocity of the Moon in its orbit, you can use the centripetal acceleration formula:
\[a = \frac{v^2}{r}\]
Where:
- \(a\) is the centripetal acceleration,
- \(v\) is the velocity you want to find, and
- \(r\) is the distance between the Earth and the Moon.
You already calculated the centripetal acceleration in part (a), and you have the value of \(r\). Solve for \(v\).
c) To find the time it takes for the Moon to complete one orbit around the Earth (its orbital period), you can use the formula:
\[T = \frac{2\pi r}{v}\]
Where:
- \(T\) is the time (in seconds) for one orbit,
- \(r\) is the distance between the Earth and the Moon (as given),
- \(v\) is the velocity you found in part (b), and
- \(\pi\) is a mathematical constant (approximately 3.14159).
Since you want the time in days, you'll need to convert the result to days. There are 86,400 seconds in a day.
