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Chapter 8: Problem 53

Pluto and Charon. Pluto's diameter is approximately \(2370 \mathrm{km},\) and the diameter of its satellite Charon is 1250 \(\mathrm{km}\) . Although the distance varies, they are often about \(19,700 \mathrm{km}\) apart, center to center. Assuming that both Pluto and Charon have the same composition and hence the same average density, find the location of the center of mass of this system relative to the center of Pluto.

Short Answer

Expert verified
The center of mass is approximately 2229 km from Pluto's center.

Step by step solution

01

Understand the Concept of Center of Mass

The center of mass of a system can be found using the formula:\[x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2}\]where \(m_1\) and \(m_2\) are the masses of Pluto and Charon, and \(x_1\) and \(x_2\) are their positions. Since we are asked to find the center of mass relative to Pluto's center, \(x_1 = 0\). Hence:\[x_{cm} = \frac{m_2 \cdot x_2}{m_1 + m_2}\]
02

Calculate Volumes of Pluto and Charon

Assuming the bodies are spherical, their volumes are given by:\[ V = \frac{4}{3} \pi r^3 \]For Pluto, with a diameter of 2370 km:\[ V_{Pluto} = \frac{4}{3} \pi \left( \frac{2370}{2} \right)^3 \]For Charon, with a diameter of 1250 km:\[ V_{Charon} = \frac{4}{3} \pi \left( \frac{1250}{2} \right)^3 \]
03

Use Volume to Find Masses

Since density is the same for both, the masses are directly proportional to the volumes.This implies:\[ m_{Pluto} = \rho \cdot V_{Pluto}, \quad m_{Charon} = \rho \cdot V_{Charon}\]Cancel \( \rho \) when computing their ratio:\[ \text{Mass ratio} = \frac{m_{Charon}}{m_{Pluto}} = \frac{V_{Charon}}{V_{Pluto}}\]
04

Calculate the Mass Ratio

Calculate the volumes and find:\[ \frac{V_{Charon}}{V_{Pluto}} = \frac{\left( \frac{1250}{2} \right)^3}{\left( \frac{2370}{2} \right)^3} \approx \frac{1}{8.853}\]
05

Find the Center of Mass Position

Let \(x_2 = 19700\, \mathrm{km}\) (distance between centers). Calculate the center of mass:\[ x_{cm} = \frac{1 \, \cdot 19700}{1 + 8.853} \approx 2229.35 \, \mathrm{km}\]
06

State Final Result in Relation to Pluto

Since \(x_{cm}\) is measured from the center of Pluto, the center of mass is approximately 2229 km away from Pluto's center.

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Key Concepts

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

Mass Ratio
Understanding the concept of mass ratio is essential in problems dealing with celestial bodies, like Pluto and Charon. When two objects have the same density, their mass is directly proportional to their volume. This implies that the mass ratio can be expressed as the ratio of their volumes:
  • If the volume of Pluto is significantly larger than Charon, Pluto will also have a much larger mass.
  • The mass ratio is valuable for finding the center of mass because it helps us understand which object contributes more to the system's total mass.
In our case, the mass ratio of Charon to Pluto can be calculated by comparing their respective volumes. This ratio provides critical insight into the gravitational interplay and helps determine where the center of mass is located relative to Pluto's center.
The calculated ratio of the two masses was approximately 1:8.853, with Charon being the lighter body.
Volume Calculation
To calculate the center of mass for Pluto and Charon, determining their volumes is crucial. Volume calculation for spherical objects requires the formula:\[ V = \frac{4}{3} \pi r^3 \]where \( V \) is volume and \( r \) is the radius. Both Pluto and Charon are essentially assumed to be perfect spheres for simplicity.
First, convert the diameter to radius by dividing by 2. For Pluto, with a 2370 km diameter, the radius is 1185 km. For Charon, with a 1250 km diameter, the radius is 625 km.The next step:
  • Plug these values into the sphere volume formula to compute their respective volumes.
This provides the values needed to explore how their mass, and subsequently mass ratio, compare when density is the same. The accurate computation is essential to ensuring precision in further calculations involving the mass ratio and center of mass.
Sphere Volume Formula
The sphere volume formula is central to calculating the properties of celestial bodies such as Pluto and Charon, since they are considered spherical in this exercise. The formula is given by:\[ V = \frac{4}{3} \pi r^3 \]This equation is used to calculate the volume of both spheres.
Using the formula:
  • Insert the radius of each celestial body. For Pluto, it is 1185 km and for Charon, it is 625 km.
  • Apply the cubic calculation to find the volume specific to each sphere.
This formula helps us leverage the given diameters to derive the correct mass relationships between Pluto and Charon. It points out how much space each one occupies, ultimately informing us about their respective pulls in a gravitational context.
Distance Between Celestial Bodies
The distance between celestial bodies is critical when evaluating gravitational forces and calculating the center of mass of a dual-body system like Pluto-Charon. The key measurement here is the distance from one center to the other, a constant that affects how the system balances:
  • In this problem, the distance between Pluto and Charon is typically around 19,700 km.
  • This measure is used as \( x_2 \) in the center of mass formula, locating Charon's contribution.
The calculated center of mass, when combined with the mass ratio, highlights the system's stability and dynamics at this distance.
Even though the point is typically not right between the two bodies due to their size and mass variance ratio, this concept helps students visualize the gravitational relationship. Remember, this distance is pivotal for determining how external forces might influence their mutual orbits.

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Most popular questions from this chapter

Force of a Baseball Swing. A baseball has mass 0.145 \(\mathrm{kg}\) . (a) If the velocity of a pitched ball has a magnitude of 45.0 \(\mathrm{m} / \mathrm{s}\) and the batted ball's velocity is 55.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. (b) If the ball remains in contact with the bat for 2.00 \(\mathrm{ms}\) , find the magnitude of the average force applied by the bat.

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