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The masses of the earth and moon are \(M_{\mathrm{e}} \approx 6.0 \times 10^{24}\) and \(M_{\mathrm{m}} \approx 7.4 \times 10^{22}\) (both in kg) and their center to center distance is \(3.8 \times 10^{5} \mathrm{km}\). Find the position of their \(\mathrm{CM}\) and comment. (The radius of the earth is \(\left.R_{e} \approx 6.4 \times 10^{3} \mathrm{km} .\right).\)

Short Answer

Expert verified
The center of mass is inside the Earth, about 4630 km from its center.

Step by step solution

01

Understand the Formula for Center of Mass

The center of mass (CM) of two objects can be found using the formula \( x = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2} \), where \(m_1\) and \(m_2\) are the masses of the two objects, and \(x_1\) and \(x_2\) are their positions. Here, we take the position of the Earth as 0, and the distance \(x_2\) to be the distance between the Earth and the Moon.
02

Assign Known Values to Variables

Assign the known values: \(m_1 = M_{\mathrm{e}} = 6.0 \times 10^{24} \text{ kg} \), \(m_2 = M_{\mathrm{m}} = 7.4 \times 10^{22} \text{ kg} \), and the distance \(x_2 = 3.8 \times 10^{5} \text{ km} = 3.8 \times 10^{8} \text{ m} \). We set \(x_1 = 0\) because we measure distances from the Earth's center.
03

Apply the Formula

Substitute the values into the center of mass formula: \[ x = \frac{6.0 \times 10^{24} \cdot 0 + 7.4 \times 10^{22} \cdot 3.8 \times 10^{8}}{6.0 \times 10^{24} + 7.4 \times 10^{22}} \] Calculate the numerator and the denominator separately.
04

Calculate Numerator and Denominator

Calculate the numerator: \( 7.4 \times 10^{22} \cdot 3.8 \times 10^{8} = 2.812 \times 10^{31} \). Calculate the denominator: \( 6.0 \times 10^{24} + 7.4 \times 10^{22} = 6.074 \times 10^{24} \).
05

Solve for the Center of Mass

Now divide the calculated numerator by the denominator: \[ x = \frac{2.812 \times 10^{31}}{6.074 \times 10^{24}} \approx 4.63 \times 10^{6} \text{ m} \]
06

Interpret the Result

The center of mass between the Earth and the Moon is approximately \(4.63 \times 10^{6}\) meters from Earth's center. Since this distance is less than the Earth's radius of \(6.4 \times 10^6\) m, the CM is inside the Earth.

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Key Concepts

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

Earth and Moon System
The Earth and Moon system is a fascinating example of gravitational interaction in celestial mechanics. These two massive celestial bodies are bound together by gravity, leading to the Moon orbiting around the Earth. This system behaves as though all the mass were concentrated at a single point, known as the center of mass (CM).
This center is crucial in understanding the dynamics and stability of their orbit. Although the Earth is vastly more massive than the Moon, the Moon's gravitational influence is significant despite its smaller size.
  • The Earth has a mass of approximately \(6.0 \times 10^{24}\) kg.
  • The Moon's mass is roughly \(7.4 \times 10^{22}\) kg.
  • The distance from Earth to Moon is \(3.8 \times 10^5\) km.
These numbers illustrate the scales at play in the Earth and Moon system, highlighting the delicate gravitational dance between the planet and its natural satellite.
Center of Mass Calculation
Calculating the center of mass (CM) in any two-body system helps us understand where the balance point is, given their mass distribution. In the Earth and Moon calculation, this point is the average position of all the mass in the system.
For our certified space neighbors, we use the formula:\[x = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2}\]where:
  • \(m_1\) is Earth's mass \(6.0 \times 10^{24}\) kg.
  • \(m_2\) is Moon's mass \(7.4 \times 10^{22}\) kg.
  • \(x_1\) is Earth's position we take as 0.
  • \(x_2\) is the Earth-Moon distance \(3.8 \times 10^8\) m.
By substituting these values into the formula, the CM is found to be \(4.63 \times 10^6\) m from Earth's center. This calculation results in a point lying within the Earth's radius. Understanding this helps explain why the Earth wobbles slightly due to the Moon despite the vastness of space.
Celestial Mechanics
Celestial mechanics is the branch of astronomy that deals with the motions and gravitational interactions of astronomical objects. It uses concepts such as center of mass to predict and explain the movement of celestial bodies like planets, moons, and stars.
The knowledge of center of mass in systems like Earth and Moon is pivotal. It helps scientists predict not just their orbits but also any perturbations influencing them. These calculations allow us to foresee eclipse paths, tides, and the overall gravitational pull that affects our planet.
  • Accurate measurements of mass and distance are crucial in this field.
  • Use of Newton's laws and gravitational theories help predict celestial motions.
  • Understanding celestial mechanics is essential for space exploration and understanding natural phenomena.
Through advancements in celestial mechanics, our comprehension of cosmic dynamics continues to deepen, aiding not just physicists but also creating practical applications in navigation and satellite technology.

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Most popular questions from this chapter

Consider a gun of mass \(M\) (when unloaded) that fires a shell of mass \(m\) with muzzle speed \(v\). (That is, the shell's speed relative to the gun is \(v\).) Assuming that the gun is completely free to recoil (no external forces on gun or shell), use conservation of momentum to show that the shell's speed relative to the ground is \(v /(1+m / M).\)

A particle moves under the influence of a central force directed toward a fixed origin \(O\). (a) Explain why the particle's angular momentum about \(O\) is constant. (b) Give in detail the argument that the particle's orbit must lie in a single plane containing \(O\).

A shell traveling with speed \(v_{\mathrm{o}}\) exactly horizontally and due north explodes into two equal-mass fragments. It is observed that just after the explosion one fragment is traveling vertically up with speed \(v_{\mathrm{o}} .\) What is the velocity of the other fragment?

Consider a uniform solid disk of mass \(M\) and radius \(R\), rolling without slipping down an incline which is at angle \(\gamma\) to the horizontal. The instantaneous point of contact between the disk and the incline is called \(P\). (a) Draw a free-body diagram, showing all forces on the disk. (b) Find the linear acceleration \(\dot{v}\) of the disk by applying the result \(\dot{\mathbf{L}}=\Gamma^{\text {ext }}\) for rotation about \(P .\) (Remember that \(L=I \omega\) and the moment of inertia for rotation about a point on the circumference is \(\frac{3}{2} M R^{2} .\) The condition that the disk not slip is that \(v=R \omega \text { and hence } \dot{v}=R \dot{\omega} .)\) (c) Derive the same result by applying \(\dot{\mathbf{L}}=\mathbf{\Gamma}^{\mathrm{ext}}\) to the rotation about the CM. (In this case you will find there is an extra unknown, the force of friction. You can eliminate this by applying Newton's second law to the motion of the CM. The moment of inertia for rotation about the \(\mathrm{CM}\) is \(\frac{1}{2} M R^{2}\).)

Consider a system comprising two extended bodies, which have masses \(M_{1}\) and \(M_{2}\) and centers of mass at \(\mathbf{R}_{1}\) and \(\mathbf{R}_{2}\). Prove that the \(\mathrm{CM}\) of the whole system is at $$ \mathbf{R}=\frac{M_{1} \mathbf{R}_{1}+M_{2} \mathbf{R}_{2}}{M_{1}+M_{2}} $$ This beautiful result means that in finding the CM of a complicated system, you can treat its component parts just like point masses positioned at their separate centers of mass - even when the component parts are themselves extended bodies.

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