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Chapter 3: Problem 62

Astro We know that the Moon revolves around Earth during a period of \(27.3\) days. The average distance from center of Earth to the center of the Moon is \(3.84 \times\) \(10^{8} \mathrm{~m}\). What is the acceleration of the Moon due to its motion around Earth?

Short Answer

Expert verified
The acceleration of the Moon due to its motion around the Earth is approximately \(0.00272 \, \text{m/s}^2\).

Step by step solution

01

Convert the Time Period into Seconds

To work with SI units, the time period of the Moon's revolution must be converted into seconds. Given that one day contains 24 hours, one hour contains 60 minutes, and one minute contains 60 seconds, the conversion is performed as follows: \(27.3 \, \text{days} \times 24 \, \text{hours/day} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 2358720 \, \text{seconds}\)
02

Compute the Speed of the Moon

Knowing that the speed is the distance over time, and the Moon's orbit is a circle, the distance traveled is the circumference of the orbit, given by \(2 \pi r\), where \(r\) is the radius of the orbit. The speed of the Moon is then computed as follows:\(\text{Speed} = \frac{2 \pi r}{T} = \frac{2 \pi (3.84 \times 10^{8} \, \text{m})}{2358720 \, \text{seconds}} = 1022.44 \, \text{m/s}\)
03

Calculate the Acceleration of the Moon

Using the Centripetal Acceleration formula \(a = \frac{v^2}{r}\), where \(v\) is the speed and \(r\) is the radius, the acceleration of the moon can be calculated as follows:\(\text{Acceleration} = \frac{(1022.44 \, \text{m/s})^2}{3.84 \times 10^{8} \, \text{m}} = 0.00272 \, \text{m/s}^2\)

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Key Concepts

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

Orbital Mechanics
Orbital mechanics is a fascinating area of physics that studies the motion of objects in space governed by gravity. When we talk about the Moon revolving around the Earth, we are dealing with a real-world application of these principles.

The Moon's orbit is a near-perfect circle, allowing us to simplify many calculations. This orbit is similar to pathways of satellites which follow predictable paths around a central mass, due to gravitational forces. Each path is determined by the balance between the forward motion of the object and the gravitational pull of the body it is orbiting around.

In the case of the Moon, its orbital period (the time it takes to complete one full orbit around the Earth) is crucial. This period, along with the distance between the Earth and the Moon, helps us understand the Moon's speed and, as a result, its centripetal acceleration. An understanding of these parameters is key to calculating important metrics like acceleration and velocity that are essential in both theoretical and practical applications of space exploration.
SI Units Conversion
Whenever you're working with physics, understanding and converting to SI units is a necessity. In our problem, we're dealing with time and all measurements have to be in consistent units to ensure accuracy in further calculations.

The SI unit for time is seconds, so when the time period of the Moon's orbit is given in days, converting it to seconds is crucial. This requires multiplying the number of days by the number of hours in a day (24), then by the number of minutes in an hour (60), and finally by the number of seconds in a minute (60). The formula is as follows:
  • Time in seconds = Days 脳 24 脳 60 脳 60
  • For our calculation: 27.3 days 脳 24 脳 60 脳 60 = 2,358,720 seconds.
This conversion ensures our calculations for speed and acceleration are precise and compatible with all other SI based calculations.
Circular Motion
When the Moon revolves around Earth, it undergoes circular motion which is a type of motion in which an object moves along the circumference of a circle with a fixed radius. The concepts of speed, distance, and acceleration are key.

The Moon's speed is determined by the distance it travels (the circumference of its orbit) divided by its orbital period in seconds. The Moon's orbit can be treated as a circle, and its circumference is given by the formula:
  • Circumference = 2蟺r, where r is the radius of the orbit.
Once we know the speed, we can find the centripetal acceleration which keeps the Moon in its orbit. This acceleration is calculated using the net inward force on the Moon that keeps it moving in a circle. The formula used is:
  • Centripetal Acceleration = \( \frac{v^2}{r} \)
where \( v \) is the speed, and \( r \) is the radius of the circular path. This acceleration is what pulls the Moon towards Earth, maintaining its orbit.

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Most popular questions from this chapter

Sports Steve Young stands on the 20-yard line, poised to throw long. He throws the ball at initial velocity \(v_{0}\) equal to \(15.0 \mathrm{~m} / \mathrm{s}\) and releases it at an angle \(\theta\) equal to \(45.0^{\circ}\). (a) Having faked an end around, Jerry Rice comes racing past Steve at a constant velocity \(V_{J}\) equal to \(8.00 \mathrm{~m} / \mathrm{s}\), heading straight down the field. Assuming that Jerry catches the ball at the same height above the ground that Steve throws it, how long must Steve wait to throw, after Jerry goes past, so that the ball falls directly into Jerry's hands? (b) As in part (a), Jerry is coming straight past Steve at \(V_{\mathrm{J}}\) equal to \(8.00 \mathrm{~m} / \mathrm{s}\). But just as Jerry goes past, Steve starts to run in the same direction as Jerry with \(V_{\mathrm{s}}\) equal to \(1.50 \mathrm{~m} / \mathrm{s}\). How long must Steve wait to release the ball so that it falls directly into Jerry's hands? SSM

\(-\vec{A}\) is \(66.0 \mathrm{~m}\) long at a \(28^{\circ}\) angle with respect to the \(+x\) axis. \(\vec{B}\) is \(40.0 \mathrm{~m}\) long at a \(56^{\circ}\) angle with respect to the \(-x\) axis. What is the sum of vectors \(\vec{A}\) and \(\vec{B}\) (magnitude and angle with the \(+x\) axis)?

You drop a rock from rest from the top of a tall building. (a) How far has the rock fallen in \(2.50\) s? (b) What is the velocity of the rock after it has fallen \(11.0 \mathrm{~m}\) ? (c) It takes \(0.117 \mathrm{~s}\) for the rock to pass a \(2.00-\mathrm{m}\) high window. How far from the top of the building is the top of the window?

A zookeeper is trying to shoot a monkey sitting at the top of a tree with a tranquilizer gun. If the monkey drops from the tree at the same instant that the zookeeper fires, where should the zookeeper aim if he wants to hit the monkey? (Neglect any effects due to air resistance.) A. Aim straight at the monkey. B. Aim lower than the monkey. C. Aim higher than the monkey. D. Aim to the right of the monkey. E. It's impossible to determine.

In 1892 George W. G. Ferris designed a carnival ride in the shape of a large wheel. This Ferris wheel had a diameter of \(76 \mathrm{~m}\) and rotated one revolution every \(20 \mathrm{~min}\). What was the magnitude of the acceleration that riders experienced? SSM
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