How fast is a stationary brick moving relative to a man floating in outer space? The earth revolves a full 360 degrees in 24 hours.. and the Earth orbits the sun 360 degrees in 365.25 days.

How fast is a stationary brick moving relative to a man floating in outer space? The earth revolves a full 360 degrees in 24 hours.. and the Earth orbits the sun 360 degrees in 365.25 days.
When you say the man is floating in outer space it sems to imply that there is some fixed reference frame in space. There is no preferred reference frame; everything is relative. Are you asking how fast some point on the earth's surface is moving relative to the sun?Originally Posted by DivideByZero
I'm guessing he means floating in space at one relative point, perhaps a few hundred thousand miles away. But the brick if he's orbiting will of course be only moving slowly relative to him. If you could have a complete stationary position in space DBZ, then the brick would move at close to the speed of light, and also then universe speed past you.
I think he means the brick is on earth and the man is stationary relative to the sun. Another schoolwork question :?
Given the facts about earths rotation and orbit, we can figure out how fast the brick is traveling relative to the sun, right? Just divide the distance the brick traveled by the time it took. But it gets tough to calculate it because there are 2 things occurring at the same time (earth rotating on its own axis, and earth orbiting the sun). So what would be the speed of the brick relative to the sun. (not relative to a floating man because I don't think thats what I meant earlier)
and no this isn't any hw. I'm just curious.
You cannot get an answer to that question because you have not specified the time slice  thanks to the earth's rotation and orbit, at different times a brick(/man in space/what have you that is fixed relative to a point on the earth's surface) will have different 'speeds' relative to a 'fixed point' on the sun.Originally Posted by DivideByZero
For instance, the 'speed' of a point on the earth's equator at any time is only about 1/30th of the 'speed' at which the earth is moving around the sun. Factor that in, and any point on the earth's surface is moving, relative to a notional fixed point on the sun at 14 km/sec give or take (1/30th).
Lets say the brick on earth is on the equator.You cannot get an answer to that question because you have not specified the time slice  thanks to the earth's rotation and orbit, at different times a brick(/man in space/what have you that is fixed relative to a point on the earth's surface) will have different 'speeds' relative to a 'fixed point' on the sun.
For instance, the 'speed' of a point on the earth's equator at any time is only about 1/30th of the 'speed' at which the earth is moving around the sun. Factor that in, and any point on the earth's surface is moving, relative to a notional fixed point on the sun at 14 km/sec give or take (1/30th).
But that's the answer then: 14km/sec give or take a km/sec (actually, as I said, if it were accurately calculated, then give or take 0.32 km/sec). And you can reduce that 'give or take' by speifying the time of day  add at midnight, subtract at midday, and make zero change at 0600 and 1800 hours (all in local time).Originally Posted by DivideByZero
The answer in this case is 28.86km/s  30.29 km/s (sunshinewarrio's value for the orbital velocity of the Earth was too small by a factor of 2)Originally Posted by DivideByZero
There are two reasons for the variation in the range of velocities. the first, already mentioned, is due to the rotation of the Earth.
The second is due ot the fact that the Earth's orbital velocity, due to its eccentricity, varies as it orbits the Sun. The rotation of the Earth accounts for +/0.46 km/s, and the variation in orbital velocity accounts for +/0.225 km/s.
So at midnight on the day of the Winter soltice the answer would be 30.29 km/sec and at noon on the day of the Summer soltice the answer would be 28.86 km/sec.
Also take into account the galaxies rotation.Originally Posted by DivideByZero
Why? It does not form part of the frame of reference.Also take into account the galaxies rotation.
Bother. Back to the drawing board for me.Originally Posted by Janus
Thanks for that.
cheer
shanks
I don't do this often (I'm just having fun) so don't persicute me.Originally Posted by DivideByZero
If the brick's "stationary", than everything else is moving. The brick (relative to the brick) remains stationary. :P
DON'T TAKE IT SERIOUSLY...
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