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Chapter 9: Problem 36

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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Most popular questions from this chapter

A car starts from rest and accelerates around a flat curve of radius \(R=36 \mathrm{~m}\). The tangential component of the car's acceleration remains constant at \(a_{\mathrm{t}}=3.3 \mathrm{~m} / \mathrm{s}^{2},\) while the centripetal acceleration increases to keep the car on the curve as long as possible. The coefficient of friction between the tires and the road is \(\mu=0.95 .\) What distance does the car travel around the curve before it begins to skid? (Be sure to include both the tangential and centripetal components of the acceleration.)

A race car is making a U-turn at constant speed. The coefficient of friction between the tires and the track is \(\mu_{\mathrm{s}}=1.20 .\) If the radius of the curve is \(10.0 \mathrm{~m},\) what is the maximum speed at which the car can turn without sliding? Assume that the car is performing uniform circular motion.

Determine the linear and angular speeds and accelerations of a speck of dirt located \(2.0 \mathrm{~cm}\) from the center of a CD rotating inside a CD player at 250 rpm.

Consider a \(53-\mathrm{cm}\) -long lawn mower blade rotating about its center at 3400 rpm. a) Calculate the linear speed of the tip of the blade. b) If safety regulations require that the blade be stoppable within \(3.0 \mathrm{~s}\), what minimum angular acceleration will accomplish this? Assume that the angular acceleration is constant.

Unlike a ship, an airplane does not use its rudder to turn. It turns by banking its wings: The lift force, perpendicular to the wings, has a horizontal component, which provides the centripetal acceleration for the turn, and a vertical component, which supports the plane's weight. (The rudder counteracts yaw and thus it keeps the plane pointed in the direction it is moving.) The famous spy plane, the SR-71 Blackbird, flying at \(4800 \mathrm{~km} / \mathrm{h}\), has a turning radius of \(290 . \mathrm{km} .\) Find its banking angle.
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