finding linear speed?

Question

Assume the earth is spherical. Relative to someone on the rotation axis, what is the linear speed of an object on the surface if the radius vector from the center of the earth to the object makes an angle of 58.0° with the axis of rotation. The radius of the earth is 6.37×103 km.

Earths radius is 6.37x10^3 km

Answer 1

This is easy. You have to know the angular velocity of the earth, ω = 1 rev./day, and the distance away from the axis of rotation, x = r * sin(58.0°).

You have to convert ω into radians/sec.

1rev/day * 1day/24hr * 1hr/3600sec * 2πrad/rev = 7.27 x 10^-5 rad/sec

x = 6.37 x 10^6 m * 0.848 = 5.40 x 10^6 m

Thus v = ω * x = 7.27 x 10^-5 rad/sec * 5.40 x 10^6 m = 392.7 m/s

v = 393 m/s

Answer 2

Interesting geometry here, especially regarding "someone on the rotation axis." We'll come back to that one.

Draw a sphere, and a radius vector from the center out to a point P on the shell, with an angle of 58 degrees from the axis of rotation. The perpendicular distance r from P to the axis of rotation is r = R sin 58, where R = 6,370 km is the radius of the earth.

The circumference of the circle traversed by the point P as the earth rotates is 2 pi r = 2 pi R sin 58, and the linear speed of the point P is 2 pi R sin 58 km / 24 hrs = (pi R sin 58)/12 km/hr.

This works out to 1414.26 km/hr, and that might be the answer you're looking for.

Now let's consider the effect, if any, of someone "on the rotation axis." Where on the rotation axis? If at the center of the circle traversed by the point P, the observer is located at a distance R - R cos 58 = R(1 - cos 58) = 2994 km below the North Pole. Difficult to get there.

Suppose the observer is located on the surface of the earth, standing on the North Pole -- the axis of rotation. In that case, because of the curvature of the earth, the observer can't see the point P travel along its prescribed path.

So for the observer to see what's going on, he or she would have to be located somewhere directly above the North Pole out in space. How far up?

Draw a tangent from P to the axis of rotation somewhere above the North Pole. If the center of the earth is at O and the observer is above the Pole at Q, then OPQ is a right triangle, OQ is the hypoteneuse, and cos 58 = R/OQ, so the observer's height above the Pole is h = R/cos 58 - R = R(1 - cos 58) = 2994 km up in the air.

It's easier, I suppose, to go 3000 km up in a hot air balloon than to dig a 3000-km hole in the ground. Therefore I think the observer is up in the air.

No matter. The linear speed of the point P is still the same ... circular motion at 1414.26 km/hr.

Not a very hard problem, but fun to think about, especially regarding the observer.

[Edit: I just saw another answer expressed in m/s, so I thought I'd convert 1414.26 km/hr. Multiply by 1000 m/km and divide by 3600 sec/hr, and the answer becomes 392.85, or 393 m/s. End edit]

Answer 3

Assume the rotational speed is w radians per second
the exact speed can be calculated given that the Earth rotates 360 degrees every 24 hours. I'll let you do the math on that.

at the equator, the linear speed is
w/R where R is the radius of the Earth.

At an angle of 58 degrees from the equator (90 degrees is the pole), since the rotational speed is the same at w, the linear speed is
w/(R*cos(58))

j

Answer 4

The earth isn't spherical. The equatorial distance is 26, 500 miles, if we count the hours in a day--I'm guessing 24. So and henceforth, the earth spins at 1000 mph. Axis is 23.5 degrees off centre--volume is 6 sextillion tonnes. Solar system is traveling at 50 000 miles per second

The galaxy is travelling even faster

Answer 5

locate the circumference = 2 * pi * r = 2 * pi * 9.29 x 10^7 = 5.80 4 x 10^8 miles velocity consistent with day = 5.80 4 x 10^8 / 365 = a million.60 X 10^6 miles consistent with day velocity consistent with our = a million.60 X 10^6 /24 = 6.sixty six X 10^ 4 miles consistent with hour = 66600 miles consistent with hour Peter