Question

The distance from Earth to the Moon is approximately \(240,000 \mathrm{mi.}\) (a) What is this distance in meters? (b) The peregrine falcon has been measured as traveling up to 350 \(\mathrm{km} /\) hr in a dive. If this falcon could fly to the Moon at this speed, how many seconds would it take? (c) The speed of light is \(3.00 \times 10^{8} \mathrm{m} / \mathrm{s} .\) How long does it take for light to travel from Earth to the Moon and back again? (\boldsymbol{d} ) ~ E a r t h ~ t r a v e l s ~ a r o u n d ~the Sun at an average speed of 29.783 \(\mathrm{km} / \mathrm{s} .\) Convert this speed to miles per hour.

Short answer

The distance from Earth to the Moon in meters is approximately \(386,241,600 \mathrm{m}\). It would take the falcon about 3,971,060 seconds to travel to the Moon at its dive speed. It takes light approximately 2.575 seconds to travel from Earth to the Moon and back. The Earth's average speed around the Sun is approximately 66,611.6 miles per hour.

Step-by-step solution

Part (a): Converting miles to meters

To convert miles to meters, we can use the conversion factor: \(1 \mathrm{mile} = 1,609.344 \mathrm{m}\). Multiply the distance in miles by the conversion factor: \(240,000 \mathrm{mi} \times 1,609.344 \frac{\mathrm{m}}{\mathrm{mi}} = 386,241,600 \mathrm{m}\) The distance from Earth to the Moon in meters is approximately \(386,241,600 \mathrm{m}\).

Part (b): Time for falcon to travel to the Moon

First, convert the falcon's speed from km/hr to m/s: \(350 \frac{\mathrm{km}}{\mathrm{hr}} \times \frac{1,000 \mathrm{m}}{1 \mathrm{km}} \times \frac{1 \mathrm{hr}}{3,600 \mathrm{s}} = \frac{350,000}{3,600} \mathrm{m/s} = 97.222\bar{2} \frac{\mathrm{m}}{\mathrm{s}}\) Now we can compute the time it would take the falcon to travel to the Moon by dividing the distance by the speed: \(t = \frac{386,241,600 \mathrm{m}}{97.222\bar{2} \frac{\mathrm{m}}{\mathrm{s}}} = 3,971,060\mathrm{s}\) It would take the falcon about 3,971,060 seconds to travel to the Moon at its dive speed.

Part (c): Time for light to travel to Moon and back

The speed of light is given as \(3.00 \times 10^8 \frac{\mathrm{m}}{\mathrm{s}}\). First, calculate how long it takes for light to travel to the Moon: \(t_{1} = \frac{386,241,600 \mathrm{m}}{3.00 \times 10^8 \frac{\mathrm{m}}{\mathrm{s}}} = 1.2875 \mathrm{s}\) Since light needs to travel back to Earth, this time must be doubled: \(t_{\text{roundtrip}} = 2t_{1} = 2 \times 1.2875 \mathrm{s} = 2.575 \mathrm{s}\) It takes light approximately 2.575 seconds to travel from Earth to the Moon and back.

Part (d): Converting Earth's speed to mph

To convert Earth's average speed from km/s to mph, use these conversion factors: \(1\mathrm{mi} = 1.609344\mathrm{km}\) and \(1\mathrm{hr} = 3600\mathrm{s}\): \(29.783 \frac{\mathrm{ km}}{\mathrm{s}} \times \frac{1 \mathrm{mi}}{1.609344 \mathrm{km}} \times \frac{3600 \mathrm{s}}{1\mathrm{hr}} = 66,611.6\frac{\mathrm{mi}}{\mathrm{hr}}\) The Earth's average speed around the Sun is approximately 66,611.6 miles per hour.

Key concepts

Distance Conversion

When we talk about distance conversion, it's essential to understand the relationship between different units of measurement. In the original exercise, the task was to convert the distance from Earth to the Moon from miles to meters. The conversion factor used is one mile equals approximately 1,609.344 meters. By multiplying the distance in miles (240,000) by this conversion factor, we find that the distance is approximately 386,241,600 meters.

Understanding distance conversion is essential when working with different systems of measurement. Here are some key aspects to consider:

• Know the conversion factors between common units such as miles to kilometers or meters.
• Always use the correct conversion factor to ensure accuracy in calculations.
• Multiplication or division is used depending on whether you are converting to a larger or smaller unit of measurement.

Speed Calculation

Speed calculation involves determining how fast an object is moving. For part (b) of the exercise, the peregrine falcon's speed needs to be converted from kilometers per hour (km/hr) to meters per second (m/s). To do this, we multiply by 1,000 to convert kilometers to meters and divide by 3,600 to convert hours to seconds. The resulting speed is approximately 97.222 m/s.

Once we have the speed in the desired unit, we can calculate travel time using the formula: \[ \text{Time} (t) = \frac{\text{Distance} (d)}{\text{Speed} (s)} \] This simple formula allows us to find that it would take the falcon about 3,971,060 seconds to reach the Moon if it could sustain its top speed. Here are a few tips to remember when doing speed calculations:

• Make sure that the distance and speed are in compatible units before calculating the time.
• Use the formula consistently to ensure accurate results.
• Check your conversions carefully to avoid errors.

Light Travel Time

The speed of light is an essential constant in physics, approximately \(3 \times 10^8 \text{ m/s} \). In the context of this exercise, it describes how quickly light can travel from Earth to the Moon and back. For this, we only need to calculate the time for a one-way trip and then double it to account for the round trip.

Using the formula: \[ t = \frac{d}{s} \] where \( d \) is the distance and \( s \) is the speed, we find that the one-way trip takes about 1.2875 seconds. Thus, the roundtrip takes about 2.575 seconds. The concept of light travel time provides insights into how fast light moves and supports various scientific endeavors.

• Remember that light speed is constant in a vacuum and is denoted as \(c\).
• Light travel time is crucial in astronomy and helps astronomers measure distances in space.
• Always apply the roundtrip concept to measure the total time for travel scenarios that involve a return journey.

Scientific Notation

Scientific notation is a way to express very large or very small numbers conveniently. In this exercise, the speed of light is given as \(3.00 \times 10^8 \text{ m/s} \). This notation is beneficial because it simplifies calculations and makes reading and writing large numbers easier.

Scientific notation follows the form:\[ a \times 10^n \] where \( a \) is a number between 1 and 10, and \( n \) is an integer. Here are some reasons why it’s useful:

• It clearly shows the order of magnitude of a number, making it easier to compare very large or very small quantities.
• It simplifies arithmetic operations such as multiplication and division.
• It’s widely used in science and engineering to handle data efficiently.

Understanding scientific notation will help you better grasp the numerical sizes of figures commonly used in physical sciences.