/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 The moon completes one (circular... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The moon completes one (circular) orbit of the earth in 27.3 days. The distance from the earth to the moon is \(3.84 \times 10^{8} \mathrm{m} .\) What is the moon's centripetal acceleration?

Short Answer

Expert verified
The moon's centripetal acceleration is approximately \(0.00272 \mathrm{m/s^2}\)

Step by step solution

01

Convert Period to SI Units

The period is given in days, but we need it in seconds for our calculations as we're dealing with SI units. We can do this by multiplying by the number of seconds in a day. Doing this, we get \((27.3 \times 24 \times 60 \times 60) = 2358720 \mathrm{s}\)
02

Compute the Orbital Speed

The speed in which the moon makes one complete orbit, or the distance in one period, is given by the circumference of the circular path, which is \(2\pi r\) = \(2\pi \times (3.84 \times 10^{8}) = 2.41 \times 10^{9} \mathrm{m}\). The speed can be found by dividing this distance by the period, i.e., \(\frac{2.41 \times 10^{9}}{2358720}\), which gives roughly \(1022 \mathrm{m/s}\).
03

Compute the Centripetal Acceleration

Finally, we can calculate the centripetal acceleration using the formula \(a = \frac{v^2}{r}\). Substituting for \(v\) and \(r\) from previous steps, the computation becomes \(a = \frac{(1022)^2}{3.84 \times 10^{8}}\), which gives approximately \(0.00272 \mathrm{m/s^2}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Circular Motion
Exploring the nature of circular motion is quite fascinating. It occurs when an object moves along a circular path, which is the case for countless celestial bodies, including our own moon as it orbits Earth. A pivotal aspect of circular motion is centripetal acceleration, which is not an outward force but rather the acceleration that keeps an object moving in a circle, directed towards the center of the rotation.

The key formula we use here is:
\[ a_{\text{centripetal}} = \frac{v^2}{r} \]
where \( a_{centripetal} \) is the centripetal acceleration, \( v \) is the orbital speed of the moon, and \( r \) is the radius of the moon's orbit. This expression allows us to understand how the velocity of the orbiting body and the radius of the path fundamentally determine the required centripetal acceleration to maintain its circular path.
Orbital Period Conversion
To work with most physical equations effectively, it's essential to understand the importance of unit conversion, particularly when dealing with time and the orbital period of celestial bodies. The orbital period of an object, like the moon around Earth, is usually given in days. However, for calculations, especially those following the International System of Units (SI units), it is crucial to convert this period into seconds.

To convert days into seconds, we simply utilize the fact that:
\[ 1 \text{ day} = 24 \text{ hours} = 24 \times 60 \text{ minutes} = 24 \times 60 \times 60 \text{ seconds} \]

By performing this conversion, we ensure that we can plug accurate values into formulas and derive correct computations for parameters such as orbital speed and centripetal acceleration.
Gravitational Forces
The wonder of circular motion in orbits is predominantly governed by gravitational forces, a fundamental interaction that clarifies not only the celestial dance of moons and planets but also informs a diverse array of phenomena across the cosmos. Gravitational forces provide the necessary centripetal force for celestial bodies to maintain their orbits.

Newton's law of universal gravitation, expressed as:
\[ F = G \frac{m_1 m_2}{r^2} \]

illustrates the relationship between two masses (\(m_1\) and \(m_2\)) and the distance (\r) between their centers. The gravitational constant (\(G\)) normalizes this equation. This force is what propels objects into orbital motion and keeps them there, with the centripetal acceleration we calculate being a direct consequence of these gravitational interactions.
SI Units
Utilizing SI units, the modern form of the metric system, is a uniform method for scientists worldwide to communicate their measurements and calculations clearly. SI units include the meter for distance, kilogram for mass, and second for time. These standardized units form a coherent system that allows us to compare and reproduce experiments and calculations.

When addressing problems related to physics and astronomy, converting to and employing SI units is indispensable. It ensures precision and universality in scientific discourse.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two children who are bored while waiting for their flight at the airport decide to race from one end of the \(20-\mathrm{m}\) -long moving sidewalk to the other and back. Phillippe runs on the sidewalk at \(2.0 \mathrm{m} / \mathrm{s}\) (relative to the sidewalk). Renee runs on the floor at \(2.0 \mathrm{m} / \mathrm{s} .\) The sidewalk moves at \(1.5 \mathrm{m} / \mathrm{s}\) relative to the floor. Both make the turn instantly with no loss of speed. a. Who wins the race? b. By how much time does the winner win?

A wildlife researcher is tracking a flock of geese. The geese fly \(4.0 \mathrm{km}\) due west, then turn toward the north by \(40^{\circ}\) and \(\mathrm{fly}\). another \(4.0 \mathrm{km} .\) How far west are they of their initial position? What is the magnitude of their displacement?

A pilot in a small plane encounters shifting winds. He flies \(26.0 \mathrm{km}\) northeast, then \(45.0 \mathrm{km}\) due north. From this point, he flies an additional distance in an unknown direction, only to find himself at a small airstrip that his map shows to be \(70.0 \mathrm{km}\) directly north of his starting point. What were the length and direction of the third leg of his trip?

The archerfish uses a remarkable method for catching insects sitting on branches or leaves above the waterline. The fish rises to the surface and then shoots out a stream of water precisely aimed to knock the insect off its perch into the water, where the archerfish gobbles it up. Scientists have measured the speed of the water stream exiting the fish's mouth to be \(3.7 \mathrm{m} / \mathrm{s}\). An archerfish spots an insect sitting \(19 \mathrm{cm}\) above the waterline and a horizontal distance of \(30 \mathrm{cm}\) away. The fish aims its stream at an angle of \(39^{\circ}\) from the waterline. Does the stream hit its mark?

Josh is climbing up a steep \(34^{\circ}\) slope, moving at a steady 0.75 \(\mathrm{m} / \mathrm{s}\) along the ground. How many meters of elevation does he gain in one minute of this climb?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.