Question

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

Answer 1

Answer:

a) Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

Explanation:

The linear speed of the particle ($v$), measured in kilometers per second, rotating in a circular pattern is calculated by the following formula:

$v = R\cdot \omega$ (1)

Where:

$R$ - Radius, measured in kilometers.

$\omega$ - Angular speed, measured in radians per second.

Now we proceed to calculate the linear speed of each element:

a) Ultracentrifuge

If we know that $\omega \approx 5235.988\,\frac{rad}{s}$ and $R = 1\times 10^{-4}\,km$, then the linear velocity is:

$v = (1\times 10^{-4}\,km)\cdot \left(5235.988\,\frac{rad}{s} \right)$

$v = 0.524\,\frac{km}{s}$

Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Earth

The Earth is 150 million kilometers away from the Sun and takes 365 days to complete one revolution around the Sun. First, we calculate angular speed of the planet:

$\omega = \frac{2\pi}{T}$ (2)

Where $T$ is the period, measured in seconds.

If we know that $T = 31536000\,s$, then the angular speed of the Earth is:

$\omega = \frac{2\pi}{31536000\,s}$

$\omega = 1.992\times 10^{-7}\,\frac{rad}{s}$

Now, we determine the linear speed:

$v = (1.5\times 10^{8}\,km)\cdot \left(1.992\times 10^{-7}\,\frac{rad}{s} \right)$

$v = 29.88\,\frac{km}{s}$

Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.