Calculate earths average speed relative to the sun

The planet Earth orbits the sun with a period of one year. In this problem, you may assume the radius of the Earths orbit is 1.50x10^8 km.

(A) Calculate the Earths average speed relative to the sun in meters per second.
(B) what is the earths average velocity relative to the sun over a period of one year in meters per second?

Solution (A): The earth's orbit around the sun insignificantly differs from a circular one. This means that its path is equal to the circumference of the circle "S=2\\pi R=2\\cdot 3.14\\cdot 1.50\\cdot 10^8km=9.42\\cdot 10^{11}m". This is the path the Earth travels in the year in which it is contained "t=1year=365 [day]\\cdot 24 [hour]\\cdot 60 [min]\\cdot 60s=3.15\\cdot 10^7 s"

By determining the average speed we write "v=\\frac{S}{t}=\\frac{9.42\\cdot 10^{11}m}{3.15\\cdot 10^7 s}=3.0\\cdot 10^4m\/s" .

Solution (B): Relative to the sun the earth each year returns to the same point from which it started moving a year ago. Therefore relative to the sun it passes zero distance "\\vec D=\\vec 0 m" per year. By determining the average velocity we write "\\vec V=\\frac {\\vec D}{t}=\\vec 0" m/s.

Answers: (A)  the Earths average speed relative to the sun in meters per second is "3.0\\cdot 10^4m\/s"

(B) The earths average velocity relative to the sun over a period of one year is "0 m\/s"

Question:

(a) Calculate Earth's average speed relative to the Sun.

(b) What is its average velocity over a period of one year?

Speed and velocity are two different quantities that are often confused. The former is a scalar quantity described only by magnitude. On the other hand, the latter is characterized by both magnitude and direction. With this, we can distinguish average velocity over average speed. The average speed of a particle is just the total distance over the total time required to travel that distance; average velocity is the total displacement over the total time. Note here that displacement is a position vector.

Answer and Explanation: 1

(a) To determine the average speed of the Earth as it moves around the Sun, we have to determine the radius of the Earth's orbit, circumference and the time it takes to complete the orbit. This are known to be,

• Radius of the Earth's orbit: {eq}\begin{align*} R=150,000,000~km \end{align*} {/eq}

• Circumference of the Earth's Orbit: {eq}\begin{align*} C=2\pi R=2\pi (150,000,000)=942,477,796.1~km \end{align*} {/eq}

• Time to complete one revolution: {eq}\begin{align*} t_{total}=365.2422~days = 8765.8128~hrs \end{align*} {/eq}

With this known parameters, the speed of the Earth around the sun is calculated to be

$$\begin{align*} V&=\frac{C}{t_{total}}&\text{[Plug in the given]}\\\\ V&=\frac{942,477,796.1}{8765.8128}&\text{[Divide]}\\\\ V&=107,517.4451~\frac{km}{hr}&\text{[Convert]}\\\\ V&=\left( 107,517.4451~\frac{km}{hr}\right)\cdot \left( \frac{1,000~m}{1~km}\right)\cdot \left( \frac{1~hr}{3600~sec}\right)&\text{[Simplify]}\\\\ V&=29,865.95696\frac{m}{s}\\\\ V&\approx \boxed{ 30,000~\frac{m}{s}} \end{align*} $$

(b) The Earth completes one whole revolution around the sun over a period of one year. With this, the displacement of our planet is zero because it goes back to its original position. Since displacement is zero, then the average velocity is also zero.

Learn more about this topic:

Speed and Velocity: Difference and Examples

from

Chapter 3 / Lesson 5

This lesson presents the differences between speed and velocity based on the differences between scalars and vectors. It also introduces the concepts of constant, average, and instantaneous speed and velocity.

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Question:

1. (a) Calculate Earth's average speed relative to the Sun.

(b) What is its average velocity over a period of one year?

2. A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation.

(a) Calculate the average speed of the blade tip in the helicopter's frame of reference.

(b) What is its average velocity over one revolution?

The average speed of an object is a scalar quantity that corresponds to the average rate at which it covers a certain distance over a set period of time. Note that this is different from the average velocity, which is a vector quantity.

Answer and Explanation: 1

1.a. In order to determine the Earth's average speed relative to the Sun, we just have to divide the distance traveled from Earth to the Sun, {eq}2\pi r {/eq} (where {eq}r = 1.5 \times 10^{11} \, m {/eq}), over the average length of a year ({eq}t = 365.25 \, days {/eq}). Thus,

{eq}\begin{align} \bar v_{Earth} & = \frac{\text{Distance traveled}}{\text{Time}} \\[1ex] &= \frac{2\pi r}{t} \\[1ex] &= \frac{2\pi (1.5 \times 10^{11} \, m)}{365.25 \, days} \times \frac{1 \, day}{24 \, hr} \times \frac{1 \, hr}{60 \, min} \times \frac{1 \, min}{60 \, s} \\[1ex] &= 2.99 \times 10^4 \, \frac {m}{s} \\[1ex] & \approx \boxed {3 \times 10^4 \, \frac {m}{s}} \\[1ex] \end{align} {/eq}

1.b. We know that the Earth will return to its starting point/original position with respect to the sun after one year. Thus, the displacement is zero--making the average velocity as:

{eq}\boxed {v = 0 \, \frac {m}{s}} \\ \\ {/eq}

2.a. In order to determine the average speed of the blade tip, we just have to divide the distance traveled, {eq}2\pi r {/eq} ({eq}r = 5 \, m {/eq}), over the time elapsed ({eq}t = 60 \, s / 100 \, rev {/eq}) such that:

{eq}\begin{align} \bar v_{blade \, tip} & = \frac{\text{ Distance traveled}}{\text{Time elapsed}} \\[1ex] &= \frac{2\pi r}{t} \\[1ex] &= \frac{2\pi \times 5.00 \, m}{60 \, s / 100 \, rev} \\[1ex] &= \boxed {52.36 \, \frac {m}{s}} \end{align} {/eq}

2.b.So after one revolution, the blade tip returns to its starting point/original position. Thus, the displacement is zero--making the average velocity as:

{eq}\boxed {v = 0 \, \frac {m}{s}} \\ \\ {/eq}

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