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What is the centripetal acceleration of the Moon? The period of the Moon's orbit about the Earth is 27.3 days, measured with respect to the fixed stars. The radius of the Moon's orbit is \(R_{M}=3.85 \cdot 10^{8} \mathrm{~m}\).

Short Answer

Expert verified
Question: Determine the centripetal acceleration of the Moon given that its orbit has a radius of \(3.85 \cdot 10^8\) meters, and it takes 27.3 days to complete one orbit. Answer: To determine the centripetal acceleration of the Moon, follow these steps: 1. Calculate the circumference of the Moon's orbit using its radius. 2. Find the velocity of the Moon using the circumference and period of the orbit. 3. Compute the centripetal acceleration using the formula \(a_c = \dfrac{v^2}{r}\). After completing these steps, you will find that the centripetal acceleration of the Moon is approximately \(2.72 \times 10^{-3}\, m/s^2\).

Step by step solution

01

Find the Circumference of the Moon's orbit

We can find the circumference of the Moon's orbit using the formula \(C = 2\pi R_M\). Let's calculate the circumference: \(C = 2 \pi (3.85 \cdot 10^{8})\)
02

Find the Velocity of the Moon

To find the velocity of the Moon, we can use the formula \(v = \dfrac{C}{T}\), where \(T\) is the period of the Moon's orbit. We are given that \(T\) is 27.3 days, but we need to convert it to seconds. \(T = 27.3 \cdot 24 \cdot 60 \cdot 60\) seconds Now, we can find the velocity: \(v = \dfrac{2 \pi (3.85 \cdot 10^{8})}{27.3 \cdot 24 \cdot 60 \cdot 60}\)
03

Calculate the Centripetal Acceleration

Now that we have calculated the velocity \(v\), we can find the centripetal acceleration using the formula \(a_c = \dfrac{v^2}{R_M}\). Substitute the values and compute the result: \(a_c = \dfrac{(\dfrac{2 \pi (3.85 \cdot 10^{8})}{27.3 \cdot 24 \cdot 60 \cdot 60})^2}{3.85 \cdot 10^{8}}\) Simplify and compute the result to get the centripetal acceleration of the Moon.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moon's Orbit
The Moon orbits the Earth in a near-circular path. This path is quite large, having a radius of about \(3.85 \cdot 10^{8} \text{ meters}\). This distance is what we call the Moon's orbital radius. Understanding this orbit is essential. It helps us calculate other important features, like velocity and acceleration later on.

As the Moon travels around Earth, it maintains a consistent path. This means it experiences what is called
  • centripetal force, pulling it toward Earth's center,
  • keeping it from flying off into space.
This force is responsible for the Moon's constant motion. It is an important concept when studying not only lunary movements but also planetary systems and satellites.
Velocity Calculation
To calculate the velocity of the Moon along its orbit, we first find the orbit's circumference. The circumference is the total distance the Moon travels in one complete orbit.

Here's how you calculate it: start with the Moon's radius, \(R_M\), and use the formula for circumference, \(C = 2 \pi R_M\). So, plugging in the Moon's orbital radius, the circumference is calculated as:

\[C = 2 \pi (3.85 \cdot 10^{8})\]

This circumference value gives the full path length. Knowing the path length is half the work. Now, divide this by the period of orbit (converted to seconds) to find the velocity.

Using the relation: \( v = \dfrac{C}{T} \), where \(T\) is the Moon's period:

\[v = \dfrac{2 \pi (3.85 \cdot 10^{8})}{27.3 \cdot 24 \cdot 60 \cdot 60}\]
This calculation will yield the Moon's average speed as it orbits Earth.
Period of Orbit
The period of the Moon's orbit refers to the time it takes to make one complete circle around Earth. For the Moon, this period is precisely 27.3 days, when measured relative to fixed stars.

This is significant because it determines how long it takes for the Moon to return to a similar position in space. But to apply this period in calculations, it needs conversion into seconds. Here’s how it’s done:

The conversion involves multiplying:
  • 27.3 days by
  • 24 hours per day,
  • 60 minutes per hour, and
  • 60 seconds per minute.

That gives you the total seconds, so:

\[T = 27.3 \cdot 24 \cdot 60 \cdot 60\]

Calculating this period in seconds allows us to use it in formulas for velocity and centripetal acceleration, providing meaningful insights into Moon’s orbital dynamics.
Circumference Formula
The circumference formula is essential in many orbital calculations. It's expressed as \(C = 2\pi R\), where \(R\) represents the radius of the circular path.

The constant \( \pi \) (Pi) is central here, approximately equal to 3.14159. It relates the diameter of a circle to its circumference and is a crucial element in trigonometry and geometry.

The formula helps determine distances traveled by orbiting bodies, like the Moon. For our calculation of the Moon's orbit, this formula simplified the process by allowing direct application of the given radius.

Using the formula correctly is vital. It ensures accurate distance calculations and hence accurate predictions about orbital mechanics. This knowledge applies to anything with a circular orbit, from planets to satellites, forming a foundational concept in not just physics but also engineering and astronomy.

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Most popular questions from this chapter

A vinyl record that is initially turning at \(33 \frac{1}{3}\) rpm slows uniformly to a stop in a time of \(15 \mathrm{~s}\). How many rotations are made by the record while stopping?

An object is moving in a circular path. If the centripetal force is suddenly removed, how will the object move? a) It will move radially outward. b) It will move radially inward. c) It will move vertically downward. d) It will move in the direction in which its velocity vector points at the instant the centripetal force vanishes.

A ball attached to the end of a string is swung around in a circular path of radius \(r\). If the radius is doubled and the linear speed is kept constant, the centripetal acceleration a) remains the same. b) increases by a factor of 2 . c) decreases by a factor of 2 . d) increases by a factor of 4 e) decreases by a factor of 4 .

Suppose you are riding on a roller coaster, which moves through a vertical circular loop. Show that your apparent weight at the bottom of the loop is six times your weight when you experience weightlessness at the top, independent of the size of the loop. Assume that friction is negligible.

The motor of a fan turns a small wheel of radius \(r_{\mathrm{m}}=\) \(2.00 \mathrm{~cm} .\) This wheel turns a belt, which is attached to a wheel of radius \(r_{f}=3.00 \mathrm{~cm}\) that is mounted to the axle of the fan blades. Measured from the center of this axle, the tip of the fan blades are at a distance \(r_{\mathrm{b}}=15.0 \mathrm{~cm} .\) When the fan is in operation, the motor spins at an angular speed of \(\omega=1200\). rpm. What is the tangential speed of the tips of the fan blades?

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