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Question for from my son


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1,045mph (I think) based on needing to move 15 degrees around the globe in an hour and a circumference of 24,800 miles. This ignores the movement round the sun as that will have a tiny impact on the calculation. Happy to be corrected though !!!!

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Id disagree, it would depend on your latitude. At the equator the circumference of the earth is indeed about 25000miles as has been said, but the further north you go the smaller the rotational circumference so by the time you reach the pole you only need to turn on the spot once every 24 hours.

Cheers,

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Id disagree, it would depend on your latitude. At the equator the circumference of the earth is indeed about 25000miles as has been said, but the further north you go the smaller the rotational circumference so by the time you reach the pole you only need to turn on the spot once every 24 hours.

Cheers,

Agreed, but if you wanted to be really accurate, you would probably also need to consider the relative postion of the point in space to our own galaxy, although you would probably need to keep looking for a few million years to notice any difference in position.

Phil

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You have to move at an angular velocity that translates to a linear velocity.

The angular velocity is 360 deg in 24 hours, 15 arc sec per second or whatever time unit you want.

As said the latitude is therefore highly relevant, at the poles you don't move, just slowly rotate on the spot.

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Id disagree, it would depend on your latitude. At the equator the circumference of the earth is indeed about 25000miles as has been said, but the further north you go the smaller the rotational circumference so by the time you reach the pole you only need to turn on the spot once every 24 hours.

Cheers,

Good point. Too much haste, not enough thought !!

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Good point. Too much haste, not enough thought !!

Should have occurred to me, too. I shall blame lack of sleep and the drugs I'm taking to keep the germs my children have shared with me at bay. Generous to a fault with their lurgies, my children :(

James

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I agree it all depends on your location, to work it out multiplying the speed at the equator by the cosine of the latitude of your location, calculators at the ready!!!!!

standing at my hometown the speed will be 639.59

Edited by scoobee
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If you wished to stay facing the same point in space allowing for Earths spin and rotation around the Sun how fast would you need to travel ?

What do we mean by "the same point in space"? There would need to be a fixed universal co-ordinate system, and general relativity tells us this does not exist. But we can get something close to it thanks to the cosmic microwave background. We define a "fixed" point in space as being one that is stationary with respect to the overall expansion of the universe (i.e., given a bunch of such points, each would see all the others receding at exactly the same rate).

The question then is how fast the Earth is moving with respect to the cosmic microwave background. This motion makes the background radiation look blue-shifted in one direction and red-shifted in the other, hence the speed and direction can be found. The sun is found to have a speed of about 370km/s, in the direction of a point on the sky with co-ordinates RA 11.2h, Dec -7 deg (in the constellation Virgo). This speed is much greater than that of Earth's rotation or orbit round the sun, which are negligible by comparison.

So to keep looking at the "same" point in space (fixed with respect to universal expansion) you would need to move at about 370km/s in the opposite direction to the sun.

Edited by acey
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I think there are many ways to interpret this one. This is how I would have answered it if asked by my (5 year old) niece.

I'd have said "you barely have to move at all". Simply stand and look at Polaris (or indeed any circumpolar star). A "fixed" star would fit the definition of "the same point in space" as far as most people would mean it, and you can face it 24 hours a day without moving an inch!

To be fair, at this point my niece would probably have gone "<SIGH!> That's not what I MEEEAN Bob!" and flounced off. ;)

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1,045mph (I think) based on needing to move 15 degrees around the globe in an hour and a circumference of 24,800 miles. This ignores the movement round the sun as that will have a tiny impact on the calculation. Happy to be corrected though !!!!

I tend to work from 6,400km as the radius of the Earth and came out to the same answer within two mph, so I'm happy to agree with you :)

James

The radius of the Earth is also 6378.1km (at the equator) and 6356.8km (at the poles)

(Source: http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html)

To be honest, this is all sort of over complicating the question, even though it is being precise, it would be roughly 1,000 miles per hour according to Freddie's calculation.

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Id disagree, it would depend on your latitude. At the equator the circumference of the earth is indeed about 25000miles as has been said, but the further north you go the smaller the rotational circumference so by the time you reach the pole you only need to turn on the spot once every 24 hours.

Cheers,

If you want to find out the exact speed at your latitude in Blackpool which has a lattitude of 53.8° N you would have to use the formula C=2 pi r cos(x), where x=latitude to find the circumfrence of the earth at you latitude and then calculate for the speed you would need to travel. Of course the forula above assumes the earth is a perfect sphere (which it isn`t) and therefore the answer will still be an approxiamtion, but it will be relatively accurate.

Using 6378km as the radius of the earth and plugging in for 53.8° N I got a circumfrence of 23668km at Blackpool, from there it`s pretty easy to calculate the speed of 986km/h or 612 mph.

More in the way of over complicating things but sort of fun none the less.

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I agree it all depends on your location, to work it out multiplying the speed at the equator by the cosine of the latitude of your location, calculators at the ready!!!!!

standing at my hometown the speed will be 639.59

Oh that's a ,uch easier way to do it, guess I missed this comment first time round...

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