Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Earth is approximately 93 000 000 mi from the sun. It revolves around the sun, in an almost circular orbit, in about 365 days. Calculate the linear speed, in miles per hour, of Earth in its orbit. Give your answer to the nearest hundredth.

Short Answer

Expert verified
Earth's linear speed is approximately 66,627.36 mph.

Step by step solution

01

Calculate the circumference of Earth's orbit

Use the formula for the circumference of a circle, which is \[ C = 2\pi r \], where \( r \) is the radius of the orbit. Here, \( r \) is 93,000,000 miles. Compute the circumference as follows:\[ C = 2 \pi \times 93{,}000{,}000 \approx 584{,}336{,}233.57 \text{ miles} \].
02

Calculate the total time in hours for one year

Determine the total number of hours in one year. Given that there are 365 days in a year and 24 hours in a day, the total time is\[ 365 \times 24 = 8760 \text{ hours} \].
03

Calculate the linear speed

Use the formula for speed, which is \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \), to find the linear speed. The distance covered by Earth in one year is the circumference of its orbit, and the time is 8760 hours:\[ \text{Speed} = \frac{584{,}336{,}233.57}{8760} \approx 66{,}627.36 \text{ miles per hour} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Circumference of a Circle
To understand how we calculate the circumference of Earth's orbit, we need to know the formula for the circumference of a circle. The formula is: \( C = 2\text{π}r \), where \( C \) represents the circumference and \( r \) is the radius. Earth's orbit around the sun is almost circular, with a radius \( r \) of approximately 93,000,000 miles. Plugging in the values, we get: \[ C = 2\text{π} \times 93{,}000{,}000 \approx 584{,}336{,}233.57 \text{ miles} \]. This calculation shows us the total distance Earth travels in one complete orbit around the sun. Understanding how to compute the circumference is crucial because it serves as our distance in the speed calculation.
Linear Speed Calculation
Calculating Earth's linear speed in its orbit involves determining how fast Earth travels in a given amount of time. Linear speed can be found using the formula: \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \). From the first step, we know the distance (the circumference of Earth's orbit), which is approximately 584,336,233.57 miles. Next, we need to find out the total time. Earth takes about 365 days to complete one orbit, and each day has 24 hours. Therefore, the total time in hours is \365 \times 24 = 8760 \text{ hours}\. Now, plugging in the values into the speed formula, we get: \[ \text{Speed} = \frac{584{,}336{,}233.57}{8760} \approx 66{,}627.36 \text{ miles per hour} \]. This means Earth travels around the sun at a speed of about 66,627.36 miles per hour.
Earth's Orbit Distance
Earth's orbit distance is essentially the circumference of its near-circular path around the sun. Knowing the distance helps us understand planetary motion and the vast scales involved. Earth's distance from the sun varies slightly, but it averages 93,000,000 miles. This average distance is what we use to compute the circumference, and subsequently, the linear speed. By understanding the orbit distance, students can better grasp concepts related to space travel, satellite placement, and gravitational effects. The enormous distance that Earth covers in one year underlines the immense, yet consistent nature of celestial mechanics and highlights the significance of calculating such distances accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Is each point on the unit circle? How do you know? a) \(\left(-\frac{3}{4}, \frac{1}{4}\right)\) b) \(\left(\frac{\sqrt{5}}{8}, \frac{7}{8}\right)\) c) \(\left(-\frac{5}{13}, \frac{12}{13}\right)\) d) \(\left(\frac{4}{5},-\frac{3}{5}\right)\) e) \(\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\) f) \(\left(\frac{\sqrt{7}}{4}, \frac{3}{4}\right)\)

Nora is required to solve the following trigonometric equation. \(9 \sin ^{2} \theta+12 \sin \theta+4=0, \theta \in\left[0^{\circ}, 360^{\circ}\right)\) Nora did the work shown below. Examine her work carefully. Identify any errors. Rewrite the solution, making any changes necessary for it to be correct. \(9 \sin ^{2} \theta+12 \sin \theta+4=0\) \((3 \sin \theta+2)^{2}=0\) $$ 3 \sin \theta+2=0 $$ Therefore. \(\sin \theta=-\frac{2}{3}\) Use a colculator. \(\sin ^{-1}\left(-\frac{2}{3}\right)=-41.8103149\) So, the reference ongle is 41.8 , to the neorest tenth of a degree Sine is negotive in quodrants II ond III. The solution in quadront II is \(180^{\circ}-41.8^{\circ}=138.2\) The solution in quadrant III is \(180^{\circ}+41.8=221.8\) Therefore, \(\theta=138.2^{\circ}\) ond \(\theta=221.8\), to the neorest tenth of a degree.

A segment of a circle is the region between a chord and the arc subtended by that chord. Consider chord AB subtended by central angle \(\theta\) in a circle with radius \(r\).Derive a formula using \(r\) and \(\theta\) for the area of the segment subtended by \(\theta.\)

a) Determine the positive value of \(\sin \left(\cos ^{-1} 0.6\right) .\) Use your knowledge of the unit circle to explain why the answer is a rational number. b) Without calculating, what is the positive value of \(\cos \left(\sin ^{-1} \mathbf{0 . 6}\right) ?\) Explain.

Without solving, determine the number of solutions for each trigonometric equation in the specified domain. Explain your reasoning. a) \(\sin \theta=\frac{\sqrt{3}}{2}, 0 \leq \theta < 2 \pi\) b) \(\cos \theta=\frac{1}{\sqrt{2}},-2 \pi \leq \theta < 2 \pi\) c) \(\tan \theta=-1,-360^{\circ} \leq \theta \leq 180^{\circ}\) d) \(\sec \theta=\frac{2 \sqrt{3}}{3},-180^{\circ} \leq \theta < 180^{\circ}\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free