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The masses of the earth and sun are \(M_{\mathrm{e}} \approx 6.0 \times 10^{24}\) and \(M_{\mathrm{s}} \approx 2.0 \times 10^{30}\) (both in \(\mathrm{kg}\) ) and their center-to-center distance is \(1.5 \times 10^{8} \mathrm{km}\). Find the position of their \(\mathrm{CM}\) and comment. (The radius of the sun is \(\left.R_{\mathrm{s}} \approx 7.0 \times 10^{5} \mathrm{km} .\right).\)

Short Answer

Expert verified
The center of mass is within the Sun's radius, closer to the Sun due to its larger mass.

Step by step solution

01

Convert Distances to the Same Units

First, make sure all distances are expressed in the same units. The center-to-center distance from the Earth to the Sun is given in kilometers: \(1.5 \times 10^8 \text{ km}\). Convert this distance to meters by multiplying by \(10^3\):\[d = 1.5 \times 10^8 \times 10^3 = 1.5 \times 10^{11} \text{ m}\]
02

Use the Center of Mass Formula

The formula for the center of mass \(x_{\text{cm}}\) between two objects is \[x_{\text{cm}} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2}\] where \(M_1\) and \(M_2\) are the masses and \(x_1\) and \(x_2\) are the positions from a chosen reference point. In this case, take \(x_1 = 0\) for Earth and \(x_2 = 1.5 \times 10^{11} \text{ m}\) for the Sun.
03

Calculate the Center of Mass

Substitute the given values into the center of mass formula:\[x_{\text{cm}} = \frac{(6.0 \times 10^{24} \text{ kg}) \times 0 + (2.0 \times 10^{30} \text{ kg}) \times 1.5 \times 10^{11} \text{ m}}{6.0 \times 10^{24} \text{ kg} + 2.0 \times 10^{30} \text{ kg}}\]Calculating gives:\[x_{\text{cm}} \approx \frac{3.0 \times 10^{41}}{2.0 \times 10^{30}} \approx 1.5 \times 10^{11} \text{ m}\]
04

Compare with the Sun's Radius

The center of mass between the Earth and Sun, found to be \(1.5 \times 10^{11} \text{ m}\), is very close to the position of the Sun because the Sun's mass is much larger than Earth's mass. Compare this to the Sun's radius \(R_s \approx 7.0 \times 10^5 \text{ km} = 7.0 \times 10^8 \text{ m}\), indicating that the center of mass is well within the Sun's volume.

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Key Concepts

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

Earth and Sun System
The Earth and Sun system is a classic example in physics for studying gravitational forces and centers of mass. In this vast system, the Sun and the Earth are considered as two massive bodies held by gravitational attraction. The Sun, being the largest and most massive component, plays a crucial role in determining the dynamic characteristics of this system. It's important to note that although the Earth seems minuscule compared to the Sun, the gravitational interaction is mutual. Both bodies orbit around a common point known as the center of mass (CM).
  • Sun's dominance: The Sun has an immense mass, about 330,000 times that of the Earth, which shapes the gravitational field of the entire solar system.
  • Earth's orbit: The Earth revolves around the Sun in an elliptical orbit, but in a more simplified model, we treat it as a nearly circular path with the center of mass playing a key part.
Understanding how these two bodies interact helps clarify concepts in celestial mechanics, reinforcing the idea that in any two-body system, the center of mass is a crucial location around which both bodies revolve. This understanding is fundamental in fields like astrophysics and cosmology.
Mass Calculation
Calculating the mass of celestial bodies like the Earth and the Sun is an essential part of understanding their gravitational pull and influence. This calculation is crucial in determining the center of mass between them. The masses of the Earth ( \(M_e \approx 6.0 \times 10^{24} \text{ kg}\) ) and the Sun ( \(M_s \approx 2.0 \times 10^{30} \text{ kg}\) ) are already quite staggering numbers, showing how much more massive the Sun is compared to the Earth.
  • Importance of mass difference: The reason the center of mass is closer to the Sun than to the Earth is due to this huge mass difference. In systems where one component's mass is significantly larger, the center of mass shifts towards the more massive object.
  • Simple center of mass formula: The center of mass for two objects can be calculated using the formula \(x_{\text{cm}} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2}\) . Here, the choice of a reference point matters. For simplicity, starting calculations from the Earth's center can simplify computations.
By understanding and calculating the masses accurately, students grasp how it impacts the position and behavior of celestial objects in the sky. This knowledge facilitates a deeper insight into orbits and the balance of forces in space.
Astronomical Distances
The distances in the Earth and Sun system are enormous, which challenges our everyday experience with measuring space. The center-to-center distance or the astronomical unit (AU) between Earth and Sun is approximately \(1.5 \times 10^{8} \text{ km}\) or \(1.5 \times 10^{11} \text{ m}\) when converted. This vast distance is why calculating a center of mass isn't intuitive and requires understanding astronomical scales.
  • Conversion to practical units: It's crucial to convert distances to meters for consistency in calculations, particularly in physics, since the SI unit system is the standard.
  • Comparing astronomical features: When compared to the Sun's radius ( \(7.0 \times 10^{5} \text{ km}\)), the center of mass is found well within the solar structure, reflecting on how large and dense the Sun is compared to the expansive space.
  • Impact on perception: Understanding astronomical distances helps in grasping how far and yet how gravitationally linked celestial bodies are. This comprehension evolves our perception of the solar system's vastness.
Engaging with these large measurements in an accessible way aids students in appreciating the size and scale of our universe, a fundamental aspect of astronomy and cosmology.

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Most popular questions from this chapter

To illustrate the use of a multistage rocket consider the following: (a) A certain rocket carries \(\left.60 \% \text { of its initial mass as fuel. (That is, the mass of fuel is } 0.6 m_{\mathrm{o}} .\right)\) What is the rocket's final speed, accelerating from rest in free space, if it burns all its fuel in a single stage? Express your answer as a multiple of \(v_{\mathrm{ex}} .\) (b) Suppose instead it burns the fuel in two stages as follows: In the first stage it burns a mass \(0.3 m_{\mathrm{o}}\) of fuel. It then jettisons the first-stage fuel tank, which has a mass of \(0.1 m_{\mathrm{o}}\), and then burns the remaining \(0.3 m_{\mathrm{o}}\) of fuel. Find the final speed in this case, assuming the same value of \(v_{\mathrm{ex}}\) throughout, and compare.

Show that the moment of inertia of a uniform solid sphere rotating about a diameter is \(\frac{2}{5} M R^{2}\). The sum (3.31) must be replaced by an integral, which is easiest in spherical polar coordinates, with the axis of rotation taken to be the \(z\) axis. The element of volume is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

Consider a planet orbiting the fixed sun. Take the plane of the planet's orbit to be the \(x y\) plane, with the sun at the origin, and label the planet's position by polar coordinates \((r, \phi)\). (a) Show that the planet's angular momentum has magnitude \(\ell=m r^{2} \omega,\) where \(\omega=\dot{\phi}\) is the planet's angular velocity about the sun. (b) Show that the rate at which the planet "sweeps out area" (as in Kepler's second law) is \(d A / d t=\frac{1}{2} r^{2} \omega,\) and hence that \(d A / d t=\ell / 2 m .\) Deduce Kepler's second law. SECTION 3.5 Angular Momentum for Several Particles

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 \(O\).

Use spherical polar coordinates \(r, \theta, \phi\) to find the CM of a uniform solid hemisphere of radius R, whose flat face lies in the \(x y\) plane with its center at the origin. Before you do this, you will need to convince yourself that the element of volume in spherical polars is \(d V=r^{2} d r \sin \theta d \theta d \phi\). (Spherical polar coordinates are defined in Section 4.8. If you are not already familiar with these coordinates, you should probably not try this problem yet.)

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