/*! 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 46 The Position of the Moon Relativ... [FREE SOLUTION] | 91Ó°ÊÓ

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The Position of the Moon Relative to the center of the Earth, the position of the Moon can be approximated by $$\begin{aligned} \overrightarrow{\mathbf{r}}=&\left(3.84 \times 10^{8} \mathrm{m}\right)\left\\{\cos \left[\left(2.46 \times 10^{-6} \mathrm{radians} / \mathrm{s}\right) \mathrm{f}\right] \hat{\mathrm{x}}\right.\\\ &\left.+\sin \left[\left(2.46 \times 10^{-6} \mathrm{radians} / \mathrm{s}\right) t\right] \hat{\mathrm{y}}\right\\} \end{aligned}$$ where \(t\) is measured in seconds. (a) Find the magnitude and direction of the Moon's average velocity between \(t=0\) and \(t=7.38\) days. (This time is one- quarter of the 29.5 days it takes the Moon to complete one orbit.) (b) Is the instantaneous speed of the Moon greater than, less than, or the same as the average speed found in part (a)? Explain.

Short Answer

Expert verified
Average speed is less than instantaneous speed.

Step by step solution

01

Expression for Average Velocity

The average velocity of an object is defined as its displacement divided by the time interval. Given the position vector \( \overrightarrow{\mathbf{r}}(t) \), the displacement \( \Delta \overrightarrow{\mathbf{r}} \) is \( \overrightarrow{\mathbf{r}}(t_f) - \overrightarrow{\mathbf{r}}(t_i) \). The time interval is \( \Delta t = t_f - t_i \). Here, \( t_i = 0 \) and \( t_f = 7.38 \times 24 \times 3600 \) seconds. We need to calculate \( \overrightarrow{\mathbf{r}}(0) \) and \( \overrightarrow{\mathbf{r}}(t_f) \).
02

Calculate Position Vectors at t=0 and t=7.38 days

When \( t = 0 \), we find:\[ \overrightarrow{\mathbf{r}}(0) = (3.84 \times 10^{8} \mathrm{m})\cos(0)\hat{\mathrm{x}} + (3.84 \times 10^{8} \mathrm{m})\sin(0)\hat{\mathrm{y}} = (3.84 \times 10^{8} \mathrm{m})\hat{\mathrm{x}} \]Calculate \( t = 7.38 \times 24 \times 3600 \), then evaluate:\[ \overrightarrow{\mathbf{r}}(t) = (3.84 \times 10^{8} \mathrm{m})\cos\left((2.46 \times 10^{-6} \mathrm{rad/s}) \cdot t\right)\hat{\mathrm{x}} + (3.84 \times 10^{8} \mathrm{m})\sin\left((2.46 \times 10^{-6} \mathrm{rad/s}) \cdot t\right)\hat{\mathrm{y}} \]
03

Calculate Displacement \( \Delta \overrightarrow{\mathbf{r}} \)

The displacement vector is:\[ \Delta \overrightarrow{\mathbf{r}} = \overrightarrow{\mathbf{r}}(t_f) - \overrightarrow{\mathbf{r}}(0) \]Plug in the results from Step 2 to find the coordinates of the displacement vector. Simplify to calculate the components of \( \Delta \overrightarrow{\mathbf{r}} \).
04

Compute the Average Velocity

The average velocity \( \overrightarrow{\mathbf{v_{avg}}} \) can now be determined:\[ \overrightarrow{\mathbf{v_{avg}}} = \frac{\Delta \overrightarrow{\mathbf{r}}}{\Delta t} \]Use the results from Steps 2 and 3 for \( \Delta \overrightarrow{\mathbf{r}} \) and divide by \( \Delta t = 7.38 \times 24 \times 3600 \) seconds to find the magnitude and direction of the average velocity.
05

Compare Instantaneous Speed and Average Speed

The instantaneous speed of the Moon is given by the magnitude of the velocity vector, which is constant due to circular motion. Hence, the instantaneous speed is equal to the orbital speed:\[ v = \omega r = 2.46 \times 10^{-6} \times 3.84 \times 10^{8} \]The magnitude of the average velocity is generally less due to the circular path.

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Key Concepts

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

Displacement
Displacement is a fundamental concept in physics that refers to the change in position of an object. It is a vector quantity, meaning it has both magnitude and direction. In the context of the Moon's position relative to the Earth, displacement can be thought of as the straight line distance and direction from the Moon's starting point to its final point in its orbit during a given time interval.
For the Moon's motion from time 0 to 7.38 days (which covers a quarter of its orbit), the displacement is found using its initial and final position vectors. The position vectors describe the Moon’s location in terms of its x and y components, which are calculated using trigonometric functions. To compute the displacement, we simply subtract the initial position vector from the final position vector, resulting in a new vector that shows the net change in position.
This displacement is then divided by the time interval to obtain the average velocity. Unlike distance traveled, the displacement vector points directly across the orbital path, often resulting in a shorter path than the actual path traveled by the Moon.
Instantaneous Speed
Instantaneous speed refers to the magnitude of an object's velocity at any given moment. Unlike average speed, which is calculated over a time period, instantaneous speed tells us how fast an object is moving at a specific instant. For the Moon, this speed remains relatively constant due to its uniform circular motion around the Earth.
Even though the Moon's velocity direction constantly changes as it orbit in a circular path, the magnitude of its instantaneous speed remains steady. To find this speed, we multiply the angular velocity, denoted as \( \omega \), with the radius of the orbit \( r \). The formula is given by \( v = \omega r \) which gives us the speed in meters per second (m/s).
This calculation reveals that the instantaneous speed is constant and does not vary with time, unlike average speed, which takes the circular displacement into consideration possibly leading to a different magnitude. This constancy is a special characteristic of objects in circular motion.
Circular Motion
Circular motion occurs when an object moves in a path along the circumference of a circle. It is characterized by two main properties: the centripetal force, which acts towards the center of the circle, and a constant speed along the circular path, though the velocity direction constantly changes.
For circular motion like the Moon's orbit, while the instantaneous speed remains constant, the direction of velocity changes continuously. It's this changing direction which ensures that while the speed (magnitude of velocity) is constant, the object remains on the circular path. This path is completed in cycles, such as the 29.5-day lunar orbit.
Additionally, the concept of angular velocity, \( \omega \), becomes significant and is used to relate linear speed and the radius of the circle as mentioned in the instantaneous speed section. In this case, \( \omega \) is used to calculate both the speed and the necessary centripetal acceleration that keeps the Moon moving in its orbit.
Understanding circular motion is crucial, as it applies to various physical and astronomical phenomena, offering insights into the natural, constant rotation and revolution of celestial bodies.

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