Problem 6-18: The linear speed of an ultracentrifuge and Earth in its orbit

Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).

Solution:

Part A

We are given a linear speed of an ultracentrifuge of 0.50 km/s0.50\ \text{km/s}. We are asked to verify this value if we are given a radius of r=0.100 mr=0.100\ \text{m} and angular velocity of ω=50000 rev/min \omega = 50000 \ \text{rev/min}. We are going to use the formula

v=rωv = r \omega

Since we are given a linear speed in km/s\text{km/s}, we are going to convert the radius to km\text{km}, and the angular velocity to rad/sec\text{rad/sec}

r=0.100 m×1 km1000 m=0.0001 kmr=0.100\ \text{m} \times \frac{1\ \text{km}}{1000\ \text{m}} = 0.0001\ \text{km}
ω=50000 rev/min×2π rad1 rev×1 min60 sec=5235.9878 rad/sec\omega = 50000 \ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} =5235.9878\ \text{rad/sec}

Now, we can substitute these into the formula

v=rωv=(0.0001 km)(5235.9878 rad/sec)v=0.5236 km/s  (Answer)\begin{align*} v & = r \omega \\ \\ v & = \left( 0.0001 \ \text{km} \right)\left( 5235.9878 \ \text{rad/sec} \right) \\ \\ v & = 0.5236 \ \text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This value is about 0.500 km/s.

Part B

From Table 6.2 of the book

ParentSatelliteAverage orbital radius r(km)Period T(y)r3 / T2 (km3 / y2)
SunEarth1.496×1081.496 \times 10^{8} 13.35×10243.35 \times 10^{24}

Using the same formulas we used in Part A, we can solve for the linear velocity of the Earth around the sun. The radius is

r=1.496×108 kmr=1.496 \times10^{8} \ \text{km}

The angular velocity is

ω=1 revyear×2π rad1 rev×1 year365.25 days×1 day24 hours×1 hour3600 secω=1.9910×10−7 rad/sec\begin{align*} \omega & = 1 \ \frac{\text{rev}}{\text{year}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1 \ \text{year}}{365.25 \ \text{days}} \times \frac{1\ \text{day}}{24\ \text{hours}}\times \frac{1\ \text{hour}}{3600\ \text{sec}} \\ \\ \omega & = 1.9910 \times 10^{-7}\ \text{rad/sec} \end{align*}

The linear velocity is

v=rωv=(1.496×108 km)(1.9910×10−7) rad/secv=29.7854 km/s  (Answer)\begin{align*} v & = r \omega \\ \\ v & = \left( 1.496\times 10^{8}\ \text{km} \right)\left( 1.9910 \times 10 ^ {-7} \right) \ \text{rad/sec}\\ \\ v & = 29.7854\ \text{km/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The linear velocity is about 30 km/s.

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