Science a GoGo's Home Page Forums General Discussion Physics Forum Speed of Earth's Rotation

Here's a solution to a problem I thought of:

How fast does a point (or a person) on the Earth's surface rotate?

Find: v
Given: r(E)=6.37E6 m, t=24 h x 60 min/h x 60 s/min = 86400 s
s=C(E)=2(pi)r(E)=2(pi)(6.37E6)=4.00E7 m
s=vt
v=s/t=4.00E7 m/86400 s = 463 m/s

If you want the answer in km/h, here's how it goes:

s=C(E)=4.00E7 m/1000 m/km=4.00E4 km, t=24 h
v=s/t=4.00E4 km/24 h = 1667.6621 km/h

Since you're standing on the Earth, and the Earth is moving at the same speed as you are, it feels as if you're standing still, but in fact you're moving at 1668 km/h!

Close but not explicitly right lets drill the motions down more.

First:

Your latitude relative to the axis poles changes your speed. If you are standing right on the pole you have no rotational speed.

Wolfram has an applet for it choose your latitude for your speed:
http://demonstrations.wolfram.com/SpeedOfAPointOnTheSurfaceOfTheEarth/


Second:

That is only one of the motions of the earth, the earth orbits the sun the average of this speed over a year is around 29.7 Km/sec which is around 1700 km/hr


Third:

We orbit the sun but the sun orbits the galactic centre of the milky way. The average speed of that orbit is around 250 km/sec or 15000 km/hr.


Fourth:

The milky way is flying off into the universe and we estimate that speed to be 630 km/sec or 37800 km/hr.

Reference for you: http://en.wikipedia.org/wiki/Milky_Way




Putting it all together:

Your absolute universe speed is constantly varying throughout time as the 4 motions compound together to create you absolute spaceframe motion.

However yes you are blisssfully unaware of all the motions :-)
.

Just to confuse the issue further; if Mazur and Chapline are right, the Universe is rotating, so there would be another, unknown, element to consider.

Since the game seems to be, for the sake of interest, to add further complications, here is another.

Climb to a top of a mountain and you are necessarily travelling faster than when you were at the bottom.

Or, consider that tidal effects - even on 'solid' ground will raise and lower you by a small amount, also changing your speed.

And those same tidal effects will cause distortions in the body of the planet that will cause subtle variations in rotational speed.

Mike Kremer

Oph, the important thing is not to be intimidated.

It's great to see a relatively new poster getting involved. stick at it! As Paul pointed out in another thread, there are comparatively few of us. He also intimated that we regulars tend to post a lot of crackpot stuff, but I can't think where he got that idea from.

I am far too arrogant and self possessed to be initmidated by a poster from London. Perhaps Mike would like to comment on the application of Special Relativity in this context?

Why London? Where would a poster have to live to stand a better chance of intimidation?

Issac



total distance traveled divided by the time duration

v or velocity = d/t ( distance / time )

v = velocity
r = the points distance from the center of rotation
pi = 3.14159

v = (r x 2) x pi / time

now find the velocity that you would need to be traveling at
in order for you to begin to move away from the center of earths rotation.

the problem is that the earths gravity doesnt seem to be the same , you found
this out by dropping a 1/2 pound baseball from a elevation of 100 ft.

and the baseball required only 1 second before it hit the ground.

the moon seems to be getting closer and closer

somehow the earths mass must have increased and you need to calculate a new escape velocity for your spaceship.