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

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

Step-by-step 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((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* 10^(6)\,m` and
`\phi = 37.6819^(\circ)`, then the tangential speed at Livermore is:

`v = \left((2\pi)/(86160\,s) \right)\cdot (6.371* 10^(6)\,m)\cdot \sin 37.6819^(\circ)`

`v\approx 284.001\,(m)/(s)`

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