Tag Archives: earth speed

Problem 6-4: Period, angular velocity, and linear velocity of the Earth

(a) What is the period of rotation of Earth in seconds? (b) What is the angular velocity of Earth? (c) Given that Earth has a radius of 6.4×106 m at its equator, what is the linear velocity at Earth’s surface?

Solution:

Part A

The period of a rotating body is the time it takes for 1 full revolution. The Earth rotates about its axis, and complete 1 full revolution in 24 hours. Therefore, the period is

Period=24 hoursPeriod=24 hours×3600 seconds1 hourPeriod=86400 seconds  (Answer)\begin{align*} \text{Period} & = 24 \ \text{hours} \\ \\ \text{Period} & = 24 \ \text{hours} \times \frac{3600 \ \text{seconds}}{1 \ \text{hour}} \\ \\ \text{Period} & = 86400 \ \text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The angular velocity ω\omega is the rate of change of an angle,

ω=ΔθΔt,\omega = \frac{\Delta \theta}{\Delta t},

where a rotation Δθ\Delta \theta takes place in a time Δt\Delta t.

From the given problem, we are given the following: Δθ=2πradian=1 revolution\Delta \theta = 2\pi \text{radian} = 1 \ \text{revolution}, and Δt=24 hours=1440 minutes=86400 seconds\Delta t =24\ \text{hours} = 1440 \ \text{minutes}= 86400 \ \text{seconds}. Therefore, the angular velocity is

ω=ΔθΔtω=1 revolution1440 minutesω=6.94×10−4 rpm  (Answer)\begin{align*} \omega & = \frac{\Delta\theta}{\Delta t} \\ \\ \omega & = \frac{1 \ \text{revolution}}{1440 \ \text{minutes}}\\ \\ \omega & = 6.94 \times 10^{-4}\ \text{rpm}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

We can also express the angular velocity in units of radians per second. That is

ω=ΔθΔtω=2π radian86400 secondsω=7.27×10−5 radians/second  (Answer)\begin{align*} \omega & = \frac{\Delta\theta}{\Delta t} \\ \\ \omega & = \frac{2\pi \ \text{radian}}{86400 \ \text{seconds}}\\ \\ \omega & = 7.27 \times 10^{-5}\ \text{radians/second}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The linear velocity vv, and the angular velocity ω\omega are related by the formula

v=rωv = r \omega

From the given problem, we are given the following values: r=6.4×106 metersr=6.4 \times 10^{6} \ \text{meters}, and ω=7.27×10−5 radians/second\omega = 7.27 \times 10^{-5}\ \text{radians/second}. Therefore, the linear velocity at the surface of the earth is

v=rωv=(6.4×106 meters)(7.27×10−5 radians/second)v=465.28 m/s  (Answer)\begin{align*} v & =r \omega \\ \\ v & = \left( 6.4 \times 10^{6} \ \text{meters} \right)\left( 7.27 \times 10^{-5}\ \text{radians/second} \right) \\ \\ v & = 465.28 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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Problem 2-5: The Earth’s average speed and velocity

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PROBLEM:

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

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

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SOLUTION:

Part A

The average speed of the earth is equal to the distance traveled divided by time.

Average speed of the Earth=Distance TraveledTotal time of Travel=2πrt=2π(1.50×1011 m)365.25 days×1 day24 hours×1 hour3600 sec=2.99×104 m/s  (Answer)\begin{align*} \text{Average speed of the Earth} & =\frac{\text{Distance Traveled}}{\text{Total time of Travel}} \\ \\ & =\frac{2\pi \text{r}}{\text{t}} \\ \\ & =\frac{2\pi \left(1.50\times 10^{11}\text{ m}\right)}{365.25\:\text{days}}\times \frac{1\:\text{day}}{24\:\text{hours}}\times \frac{1\:\text{hour}}{3600\:\text{sec}} \\ \\ & =2.99\times 10^4\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

After a period of 1 year, the planet Earth has already returned to its original position with respect to the Sun. This means that we do not have any displacement. Therefore, the average velocity is zero.

Average velocity=0 m/s  (Answer)\text{Average velocity}=0\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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