Question

The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from equator, Longitude 121.

Answer 1

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed ($v$), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

$v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi$ (1)

Where:

$\Delta t$ - Rotation time, measured in seconds.

$R$ - Radius of the Earth, measured in meters.

$\phi$ - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that $\Delta t = 86160\,s$, $R = 6.371\times 10^{6}\,m$ and $\phi = 37.6819^{\circ}$, then the tangential speed at Livermore is:

$v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}$

$v\approx 284.001\,\frac{m}{s}$

The tangential speed at Livermore is approximately 284.001 meters per second.