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Calculate the solid angles subtended by the moon and by the sun, both as seen from the earth. Comment on your answers. (The radii of the moon and sun are \(R_{\mathrm{m}}=1.74 \times 10^{6} \mathrm{m}\) and \(R_{\mathrm{s}}=6.96 \times 10^{8}\) m. Their distances from earth are \(\left.d_{\mathrm{m}}=3.84 \times 10^{8} \mathrm{m} \text { and } d_{\mathrm{s}}=1.50 \times 10^{11} \mathrm{m} .\right)\)

Short Answer

Expert verified
The solid angles are \( 6.67 \times 10^{-5} \text{ sr} \) for the moon, and \( 6.80 \times 10^{-5} \text{ sr} \) for the sun. They are similar, explaining solar eclipses.

Step by step solution

01

Understand Solid Angle Concept

The solid angle is a measure of how large an object appears to an observer looking from a particular point. It is measured in steradians (sr) and calculated using the formula \( \Omega = \frac{A}{r^2} \), where \( A \) is the apparent surface area of the object and \( r \) is the distance from the object.
02

Formulate Solid Angle for Spherical Objects

For spherical objects like the moon and sun, the apparent surface area \( A \) observer from point at distance \( r \) is the surface area of a spherical cap. The formula for the solid angle is \( \Omega = 2\pi (1 - \cos \theta) \), where \( \theta \) is the angular radius subtended by the object. This can be simplified as \( \Omega = \frac{\pi R^2}{d^2} \), where \( R \) is the radius of the spherical body and \( d \) is the distance from the observer.
03

Calculate Solid Angle of the Moon

Using the formula \( \Omega_m = \frac{\pi R_m^2}{d_m^2} \). Substitute the given values: \( R_m = 1.74 \times 10^6 \) m and \( d_m = 3.84 \times 10^8 \) m. Now calculate: \[ \Omega_m = \frac{\pi (1.74 \times 10^6)^2}{(3.84 \times 10^8)^2} \approx 6.67 \times 10^{-5} \text{ sr} \].
04

Calculate Solid Angle of the Sun

Similarly, using the formula \( \Omega_s = \frac{\pi R_s^2}{d_s^2} \). Substitute the given values: \( R_s = 6.96 \times 10^8 \) m and \( d_s = 1.50 \times 10^{11} \) m. Calculate: \[ \Omega_s = \frac{\pi (6.96 \times 10^8)^2}{(1.50 \times 10^{11})^2} \approx 6.80 \times 10^{-5} \text{ sr} \].
05

Comment on the Results

Both solid angles are very similar: the moon's solid angle is \( 6.67 \times 10^{-5} \text{ sr} \) and the sun's solid angle is \( 6.80 \times 10^{-5} \text{ sr} \). This explains why the moon can cover the sun almost perfectly during a solar eclipse from Earth's perspective.

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Key Concepts

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

Understanding Steradians
A steradian is a unit of measure for solid angles, similar to how radians measure angles in a plane. Imagine you are standing at the center of a sphere and looking at a patch on its surface. The solid angle tells you how much of your view is taken up by that patch. This is measured in steradians (sr).
Think of it like a cone emanating from your point of view and covering a part of the sphere. The key formula to calculate a solid angle is \( \Omega = \frac{A}{r^2} \), where \( A \) is the surface area of the object’s projection and \( r \) is the distance to the object.
  • A sphere has a total solid angle of \( 4\pi \) steradians.
  • This unit helps us understand objects in 3D space, like celestial bodies or antennas.
Basics of Spherical Geometry
Spherical geometry deals with shapes on the surface of a sphere, unlike traditional geometry that focuses on flat planes. This is crucial for understanding how large spherical objects like the sun and the moon appear from Earth. The essential aspects include:
  • **Spherical Caps:** Just like a circle is a 2D projection of a 1D line, a spherical cap on the surface of a sphere is a 3D projection of a 2D circle.
  • **Angles on Spheres:** Given the radius \( R \) and the distance \( d \), the apparent size is computed using \( \Omega = \frac{\pi R^2}{d^2} \).
This formula simplifies the complexity of spherical projections and helps calculate how celestial bodies appear in the sky.
The Mysteries of Solar Eclipses
A solar eclipse is a fascinating celestial event where the moon passes between the Earth and the sun, blocking sunlight partially or completely. The reason the moon can cover the sun almost perfectly lies in their solid angles being nearly identical:
  • The moon's solid angle is approximately \( 6.67 \times 10^{-5} \) sr.
  • The sun's solid angle is approximately \( 6.80 \times 10^{-5} \) sr.
  • This closeness means that, despite the vast difference in actual sizes, they appear similarly sized from Earth’s perspective.
These nearly equal solid angles allow the moon to obscure the sun almost entirely during a total eclipse, leading to a breathtaking alignment of cosmic proportions.

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Most popular questions from this chapter

By integrating the element of solid angle \((14.15), d \Omega=\sin \theta d \theta d \phi\), over all directions, verify that the solid angle corresponding to all directions is 4 \(\pi\) steradians.

Our definition of the scattering cross section, \(N_{\mathrm{sc}}=N_{\text {inc }} n_{\text {tar }} \sigma,\) applies to an experiment using a narrow beam of projectiles all of which pass through a wide target assembly. Experimenters sometimes use a wide incident beam, which completely engulfs a small target assembly (the beam of photons from a car's headlamp, directed at a small piece of plastic, for example). Show that in this case \(N_{\mathrm{sc}}=n_{\text {inc }} N_{\mathrm{tar}} \sigma\) where \(n_{\text {inc }}\) is the density (number/area) of the incident beam, viewed head-on, and \(N_{\text {tar }}\) is the total number of targets in the target assembly.

In deriving the cross section for scattering by a hard sphere, we used the "law of reflection," that the angles of incidence and reflection of a particle bouncing off a hard sphere are equal, as in Figure 14.10. Use conservation of energy and angular momentum to prove this law. (The definition of "hard-sphere scattering" is that a projectile bounces with its kinetic energy unchanged. That the force is spherically symmetric implies, as usual, that angular momentum about the sphere's center is conserved.)

(a) A certain nucleus has radius \(5 \mathrm{fm}\). ( \(1 \mathrm{fm}=10^{-15} \mathrm{m}\).) Find its cross section \(\sigma\) in barns. (1 barn \(=10^{-28} \mathrm{m}^{2} .\) ) (b) Do the same for an atom of radius \(0.1 \mathrm{nm}\). \(\left(1 \mathrm{nm}=10^{-9} \mathrm{m} .\right)\)

[Computer] Consider a point projectile moving in a fixed, spherical force whose potential energy is $$U(r)=\left\\{\begin{array}{ll} -U_{0} & (0 \leq r \leq R) \\ 0 & (R < r) \end{array}\right.\quad\quad\quad\quad\quad\quad(14.60)$$ where \(U_{\mathrm{o}}\) is a positive constant. This so-called spherical well represents a projectile which moves freely in either of the regions \(r < R\) and \(R < r_{1}\) but, when it crosses the boundary \(r=R_{1}\) receives a radially inward impulse that changes its kinetic energy by \(\pm U_{\mathrm{o}}\left(+U_{\mathrm{o}}\right.\) going inward, \(-U_{\circ}\) going outward). (a) Sketch the orbit of a projectile that approaches the well with momentum \(p_{\mathrm{o}}\) and impact parameter \(b < R .\) (b) Use conservation of energy to find the momentum \(p\) of the projectile inside the well \((r < R)\). Let \(\zeta\) denote the momentum ratio \(\zeta=p_{\mathrm{o}} / p\) and let \(d\) denote the projectile's distance of closest approach to the origin. Use conservation of angular momentum to show that \(d=\zeta b\). (c) Use your sketch to prove that the scattering angle \(\theta\) is $$\theta=2\left(\arcsin \frac{b}{R}-\arcsin \frac{\zeta b}{R}\right).\quad\quad\quad\quad\quad\quad(14.60)$$ This gives \(\theta\) as a function of \(b\), which is what you need to get the cross section. The relation depends on the momentum ratio \(\zeta,\) which in turn depends on the incoming momentum \(p_{\mathrm{o}}\) and the well depth \(U_{\mathrm{o}} .\) Plot \(\theta\) as a function of \(b\) for the case that \(\zeta=0.5 .\) ( \(\mathbf{d}\) ) By differentiating \(\theta\) with respect to \(b\), find an expression for the differential cross section as a function of \(b\), and make a plot of \(d \sigma / d \Omega\) against \(\theta\) for the case that \(\zeta=0.5 .\) Comment. [Hint: To plot as a function of \(\theta\) you don't need to solve for \(b\) in terms of \(\theta ;\) instead, you can make a parametric plot of the point \((\theta, d \sigma / d \Omega)\) as a function of the parameter b running from 0 to \(R\).] (e) By integrating \(d \sigma / d \Omega\) over all directions, find the total cross section.

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