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Chapter 2: Problem 43

The apparent size of an object in the sky is proportional to its actual diameter divided by its distance. The Moon has a radius of \(1,737 \mathrm{km},\) with an average distance of \(3.780 \times 10^{5} \mathrm{km}\) from Earth. The Sun has a radius of \(696,000 \mathrm{km},\) with an average distance of \(1.496 \times 10^{8} \mathrm{km}\) from Earth. Show that the apparent sizes of the Moon and Sun in our sky are approximately the same.

Short Answer

Expert verified
The apparent sizes of the Moon (\(9.19 \times 10^{-3}\)) and the Sun (\(9.30 \times 10^{-3}\)) are approximately the same.

Step by step solution

01

- Understanding Apparent Size Formula

The apparent size of an object in the sky can be described by the formula: \[ \text{Apparent Size} \text{ (AS)} \text{ is proportional to } \frac{\text{Diameter}}{\text{Distance}}.\] In this case, we will compare the apparent sizes of the Moon and the Sun.
02

- Calculate Diameter of the Moon

The radius of the Moon is given as 1,737 km. Therefore, the diameter of the Moon is: \[ \text{Diameter of Moon} = 2 \times \text{Radius} = 2 \times 1,737 \text{ km} = 3,474 \text{ km} \]
03

- Calculate Apparent Size of the Moon

With the diameter of the Moon as 3,474 km and the average distance to Earth as \( 3.780 \times 10^5 \) km, the apparent size of the Moon is: \[ \text{Apparent Size of Moon} \text{ (ASM)} = \frac{3,474 \text{ km}}{3.780 \times 10^5 \text{ km}} \]
04

- Simplify the Apparent Size of the Moon

Perform the division: \[ \text{ASM} = \frac{3,474}{3.780 \times 10^5} \ \text{ASM} = \frac{3,474}{378,000} \ \text{ASM} \thickapprox 9.19 \times 10^{-3} \]
05

- Calculate Diameter of the Sun

The radius of the Sun is given as 696,000 km. Therefore, the diameter of the Sun is: \[ \text{Diameter of Sun} = 2 \times \text{Radius} = 2 \times 696,000 \text{ km} = 1,392,000 \text{ km} \]
06

- Calculate Apparent Size of the Sun

With the diameter of the Sun as 1,392,000 km and the average distance to Earth as \( 1.496 \times 10^8 \) km, the apparent size of the Sun is: \[ \text{Apparent Size of Sun} \text{ (ASS)} = \frac{1,392,000 \text{ km}}{1.496 \times 10^8 \text{ km}} \]
07

- Simplify the Apparent Size of the Sun

Perform the division: \[ \text{ASS} = \frac{1,392,000}{1.496 \times 10^8} \ \text{ASS} = \frac{1,392,000}{149,600,000} \ \text{ASS} \thickapprox 9.30 \times 10^{-3} \]
08

- Compare Apparent Sizes

The apparent size of the Moon is approximately \(9.19 \times 10^{-3}\), and the apparent size of the Sun is approximately \(9.30 \times 10^{-3}\). Since these values are very close to each other, we can conclude that the apparent sizes of the Moon and the Sun in our sky are approximately the same.

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Key Concepts

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

apparent size formula
The apparent size of celestial objects like the Moon and the Sun in the sky can be determined using a simple formula. This formula states that the apparent size (AS) is proportional to the object's actual diameter divided by its distance from the observer. Mathematically, this can be represented as:
\[\text{AS} \text{ is proportional to } \frac{\text{Diameter}}{\text{Distance}}\] In this exercise, we use this formula to compare the apparent sizes of the Moon and the Sun. The proportionality allows us to see why objects with vastly different actual sizes can appear similar in size when viewed from Earth.
Moon
The Moon, Earth's natural satellite, has some interesting measurements. First, let's understand its dimensions before applying the apparent size formula:
  • The Moon's radius: 1,737 km
  • The distance from Earth: approximately 378,000 km
Using the radius, we can calculate the Moon's diameter as follows:
\[\text{Diameter of Moon} = 2 \times 1,737 \text{ km} = 3,474 \text{ km}\] Now, we apply the apparent size formula:
\[\text{Apparent Size of Moon} = \frac{3,474 \text{ km}}{378,000 \text{ km}}\] Simplifying this, we get:
\[\text{ASM} = \frac{3,474}{378,000} \thickapprox 9.19 \times 10^{-3}\] This result shows that the Moon has a small apparent size value when seen from Earth, despite its significant actual size.
Sun
The Sun, being a massive star, has vastly different dimensions compared to the Moon:
  • The Sun's radius: 696,000 km
  • The distance from Earth: approximately 149,600,000 km
We start by calculating the Sun's diameter:
\[\text{Diameter of Sun} = 2 \times 696,000 \text{ km} = 1,392,000 \text{ km}\] Next, we apply the apparent size formula to the Sun:
\[\text{Apparent Size of Sun} = \frac{1,392,000 \text{ km}}{149,600,000 \text{ km}}\] Simplifying this, we obtain:
\[\text{ASS} = \frac{1,392,000}{149,600,000} \thickapprox 9.30 \times 10^{-3}\] Surprisingly, this value is very close to that of the Moon, allowing us to understand why the Sun and Moon appear nearly the same size in our sky, even though the Sun is much larger.

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Most popular questions from this chapter

Using a protractor, you estimate an angle of \(40^{\circ}\) between your zenith and Polaris. What is the altitude of Polaris? What is your latitude? Are you in the continental United States or Canada?

Earth is spinning along at \(1,674 \mathrm{km} / \mathrm{h}\) at the equator. Use this fact, along with the length of the day, to calculate Earth's equatorial diameter.

How would the length of the eclipse season change if the plane of the Moon's orbit were inclined less than its current \(5.2^{\circ}\) to the plane of Earth's orbit? Explain your answer.

Go to the "NASA Eclipse" website (http://eclipse.gsfc.nasa. gov/eclipse.html). When is the next lunar eclipse? Will it be visible at your location if the skies are clear? Is it a total or partial eclipse? When is the next solar eclipse? Will it be visible at your location? Compare the fraction of Earth that the solar eclipse will affect with the fraction for the lunar eclipse. Why are lunar eclipses visible in so many more locations?

A lunar eclipse occurs when _______ shadow falls on _______. a. Earth's; the Moon b. the Moon's; Earth c. the Sun's; the Moon d. the Sun's; Earth
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