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Chapter 1: Problem 54

The circumference of a circle is given by \(C=2 \pi r,\) where \(r\) is the radius of the circle. a. Calculate the approximate circumference of Earth's orbit around the Sun, assuming that the orbit is a circle with a radius of \(1.5 \times 10^{8} \mathrm{km}\). b. Noting that there are 8,766 hours in a year, how fast, in kilometers per hour, does Earth move in its orbit? c. How far along in its orbit does Earth move in one day?

Short Answer

Expert verified
a. \(9.42 \times 10^{8} \text{ km}\). b. \(1.07 \times 10^{5} \text{ km/hour}\). c. \(2.57 \times 10^{6} \text{ km} \).

Step by step solution

01

Identify and Understand the Formula

The circumference of a circle is given by the formula: \[ C = 2 \pi r \] where \( r \) is the radius of the circle. In this exercise, the radius is the distance from Earth to the Sun, which is \( r = 1.5 \times 10^{8} \text{ km} \).
02

Calculate the Circumference of Earth's Orbit (Part a)

Substitute \( r = 1.5 \times 10^{8} \text{ km} \) into the circumference formula: \[ C = 2 \pi (1.5 \times 10^{8} \text{ km}) \] This simplifies to: \[ C \approx 2 \times 3.14 \times 1.5 \times 10^{8} \] \[ C \approx 9.42 \times 10^{8} \text{ km} \] So, the approximate circumference of Earth's orbit is \( 9.42 \times 10^{8} \text{ km} \).
03

Convert Time to Hours (Part b)

There are 8,766 hours in a year. This information is necessary to calculate the speed of Earth in its orbit.
04

Calculate the Speed of Earth in its Orbit (Part b)

The speed of Earth in its orbit can be found using the formula: \[ \text{Speed} = \frac{C}{\text{Time}} \] Using \( C = 9.42 \times 10^{8} \text{ km} \) and \( \text{Time} = 8,766 \text{ hours} \), we have: \[ \text{Speed} = \frac{9.42 \times 10^{8} \text{ km}}{8,766 \text{ hours}} \] \[ \text{Speed} \approx 1.07 \times 10^{5} \text{ km/hour} \] So, the Earth moves at approximately \( 1.07 \times 10^{5} \text{ km/hour} \) in its orbit.
05

Calculate Distance Traveled in One Day (Part c)

To find the distance Earth travels in one day, first convert one day into hours: \[ 1 \text{ day} = 24 \text{ hours} \] Then use the speed calculated in Step 4: \[ \text{Distance} = \text{Speed} \times \text{Time} \] \[ \text{Distance} = 1.07 \times 10^{5} \text{ km/hour} \times 24 \text{ hours} \] \[ \text{Distance} \approx 2.57 \times 10^{6} \text{ km} \] So, Earth travels approximately \( 2.57 \times 10^{6} \text{ km} \) in one day.

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Key Concepts

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

Circumference of a Circle
The circumference of a circle is the distance around the circle, which can be found with the formula:

\(C = 2 \pi r\).

In this formula, \(C\) represents the circumference, \(r\) is the radius of the circle, and \(\pi\) (approximately 3.14) is a constant.

For example, to find the circumference of Earth's orbit around the Sun, we assume the orbit is a perfect circle with a radius of \(1.5 \times 10^{8} \text{ km}\).
Plugging in the values, we get:

\[C = 2 \pi (1.5 \times 10^{8} \text{ km}) = 9.42 \times 10^{8} \text{ km}\].

So, the circumference of Earth's orbit is approximately \( 9.42 \times 10^{8} \text{ km}\).
Earth's Orbital Speed
Orbital speed is the speed at which Earth travels around the Sun. To calculate this speed, we need to know the circumference of Earth's orbit (\(C\)) and the time it takes for Earth to complete one orbit (the period), which is one year or 8,766 hours.

The formula to find speed (\(v\)) is:

\[v = \frac{C}{t}\].

Using the circumference \(C\) we calculated earlier (\(9.42 \times 10^{8} \text{ km}\)), and the time \(t\) = 8,766 hours, we get:

\[v = \frac{9.42 \times 10^{8} \text{ km}}{8,766 \text{ hours}} \approx 1.07 \times 10^{5} \text{ km/hour}\].

Thus, Earth's orbital speed is approximately \( 1.07 \times 10^{5} \text{ km/hour}\).
Distance Traveled in Orbit
To determine how far Earth travels in its orbit in a certain period, like a day, we use the orbital speed we calculated previously.

First, convert one day into hours, since speed is given in kilometers per hour.
\(1 \text{ day} = 24 \text{ hours}\).

Now, use the following formula to calculate distance (\(d\)):

\[d = v \times t\].

With \(v = 1.07 \times 10^{5} \text{ km/hour}\) and \(t = 24 \text{ hours}\), we get:

\[d = 1.07 \times 10^{5} \times 24 = 2.57 \times 10^{6} \text{ km}\].

So, in one day, Earth travels approximately \( 2.57 \times 10^{6} \text{ km}\).
Astronomical Calculations
Astronomical calculations involve the application of mathematical equations to understand celestial phenomena.

These calculations can include:
  • The distance between celestial objects.
  • The speed of objects in space.
  • The time it takes for objects to orbit around others.
In our exercise, we use these calculations to understand Earth's movement around the Sun by:
  • Calculating the circumference of Earth's orbit
  • Determining Earth's orbital speed.
  • Finding out how far Earth travels in one day.
Such calculations help us grasp the vastness of space and the dynamics of celestial movements.
Unit Conversions
Unit conversions are fundamental in solving astronomical problems, as measurements are often given in different units.

Here are some essential conversions:
  • There are 1,000 meters in a kilometer.
  • There are 24 hours in a day.
  • There are 8,766 hours in a year.
In our example, we had to convert:
  • Radius from kilometers to the circle's circumference.
  • Time in years to hours.
  • Orbital speed per hour to distance traveled per day.
By mastering these conversions, you can seamlessly navigate between different units, making calculations more coherent and easier to comprehend.

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