Marathon Runner: Newsletter Challenge (11/02/10)

Since we are dealing in velocity does it make sense to talk about a topographic velocity if the net work done is zero (assuming that the runner ended where they started 200 km ago)? Velocity also has a vector quantity The puzzle did not ask for speed, which averages to 8.333... km/h) and since this is an oval, it would average out to zero.

That being said, from a geocentric point of view the problem is the same. In 24 hrs you are at the same location relative to the distant stars (Earth makes one revolution and the equator is 40008.629 km), so again, the answer would be zero, not the rotational speed of the equator, which is 1667.03 km/h speed.

Now, we could talk about sidereal time and I guess that would add a small velocity component, but I think I digress.

Maybe I am all turned around? ;-0

I've been running around all day today!

I would agree if sidereal time was intended to be the key. Well put.

I'd be extra careful here, His average speed would be 8.33 km/hour, relative to the surface.

Also, 0.278 should read, 0.278% or 0.00278.

Speed relative to the center could be 1003mph, 24,067 miles/24 hours

Yes, thanks. I forgot the % symbol after the 0.278. S/B 0.278%.

Hi, Wade Cooper here, I always miss something with these questions, but if you want to know speed, it has to be relative to something, the center of the earth, the center of the sun, the center of the milkyway and add that to our retreating speed from the center of the universe, it all adds up, I would like to know what percentage of the speed of light we are traveling, being on the advancing side of the galaxies' arm, in the advancing zodiac, day verses night speeds etc. and then in the retracting sides of the orbits of all these things we are slowing down, shouldn't we feel pulled in the distant direction of away from the center of the universe? Thank God I am in a padded room! I think I feel a red shift coming on....

I agree with your answers in terms of velocity (GA). However, are you sure that the runner's average speed in geocentric coordinates is 8.33 km/h?

I think that the correct terminology for the question would be "geocentric-equatorial" coordinates, moving with earth's center, but not rotating. There are various types of geocentric coordinates systems, but the challenge question does define which one it means, as far as can gather.

SL

I suppose to be careful, I should have said his average instantaneous speed is 8.33 km/hr.

I taught Physics briefly in my career and one thing I learned in writing exams was that no matter how careful I was in writing the questions there was always a student who managed to find a valid alternate interpretation of one or two of the problems.

I know it's not in the question, but isn't the 8.33 km/h the average speed only in topocentric coordinates? In geocentric-equatorial coordinates the track itself has an instantaneous speed of about 40000/23.93 ~ 1672 km/h around the earth. I think the athlete's own 8.33 km/h will cancel out in geocentric coordinates, so his average speed should be the track's speed.

Or where am I missing the plot?

Speed is relative to whatever is chosen for the coordinate system. The 8.33 km/h value would be instantaneous, relative to the topocentric coordinates.

there is a 4 minute difference between civil time and sidereal time

?: is there a sideirreal or sidewrong time too?

Opposite of "real" would be fiction, or virtual, or imaginary, or counterfeit, or supernatural, or insincere.

Opposite of "wrong" is right.

I think you have the right solution approach but I agree with the numerical values in Randall's solution in post #19. One earth day is 24 hours. One sidereal day is 23.934 hours for a difference of 3.932 minutes. Average diameter of the earth is 12756.3 kilometers or a radius of 6378.2 kilometers. The earth rotates once in one sidereal day or 0.0172 radians in 3.93 sidereal minutes. 0.0172 radians time 6378.2 kilometers is 109.4 kilometers or 4.56 kilometers per hour (2.83 mph) average velocity.

Thanks,

Jim

positiv or negative (resp. forward or backward?)

earth surface based:

after 24h the runner stays at the same point of the surface of the earth he is starting from.

The marathon runner runs within 24h 40 laps each 5km, this is an average speed of 8 1/3 = 8.3333333... km/h

earth center based:

the marathon runner runs 12h in one direction and 12h in the opposite direction, he stays at the same place he is started; the oval are 2 parallel lines with a distance of zero. 24h is the time the earth needs for one turn around her axe. Now it depends on the northern/southern latitude of the place and the direction the marathon runner is running.

parallel to the equator:

On the Equator he is running 40000km in 24h, this is an average speed of nearly 1667km/h (or ~1000 miles per hour); at the north or south pole there is no place for a 5km lap but there is an average speed of 0 and the runner turns 40 times around his own axe; this is an average speed of 8.33333...km/h (or is this zero?). All other places are between these two velocities depending from the sine of the latitude:

v-average = 8.3333... + (1666.6666... - 8.3333...) * sin(latitude)

perpendicular to the equator:

v-average = sqrt((8.33333...)^2 + (1666.6666... * sin(latitude))^2)

and there are some places between parallel and perpendicular to the equator

Ithink he must have been very very tired!

Yea, me thinks your apple travels in a circle and if you are at the equator, at an average speed of about 1671 km/s. Quoted from the reference I gave above:

"Notice that the geocentric-equatorial system is fixed in space, and does not rotate with the earth. The unit vectors that we will be using with the geocentric-equatorial system are I (for the vernal direction), J for the second direction in the equatorial plane), and K (for the direction of the north pole)"

According to this, the I axis is not quite static relative to the distant stars, but shifts with the position of the vernal equinox due to Earth's precession, period about 26 thousand years. So one day's shift is irrelevant, I guess.

I'm not a marathon runner - I know why

The earth's center moves about the sun at a rate of 107,300 km/hr if we ignore eccentricity of 0.017.

So the runner will run at a rate of 8.33 km/hr based on surface topography. However the center of the earth will have traveled 107,300 x 24 =2,575,200 km. Based on the rotation of the geocentric coordinates of the earth about the sun the runner will be running 107,300 km/hr and zero to be added for the topocentric based running. The rotation of the earth about its axis is also zero as in 24 hours we are back to the topocentric origin. Damned fast running any way you want it stated. But all is relative.

If we are picky and want to add in eccentricity we would need to know earth's exact position in space. The earth will travel slightly faster at perihelion and thus we cannot predict this value without knowing position of the earth on its orbit.

and the sun (and the complete sun system) is turning around the galactic/milky way center .....

....and the galactic milky way is moving in expansion with the universe so at some measured point we may detect the Hubble shift in terms of the speed of light...In the end, the only relative motion of any consequence is the motion through space relative to the observer in the universe.

The velocity on the topographical coordinates is zero as is the velocity in geocentric movement. Instant speed on the earth's surface is 8.33 km/hr compared to an observer on the sideline. Compared to the geocentric coordinates the observer and the runner were basically standing still. We could add or subtract the runner moving about the oval at any instant time to the speed of the earth's rotation. I think it is useless to calculate geocentric unless the observer is trying to land on the earth. The runners motion from an observatory on mercury would be different. On the sun's surface (observers from hell) the runner would move according to the earth's orbit. I only threw in the space motion to indicate that we do need an observer point as all speed is relative. Now if the earth is flat...or the sun or the universe was only traveling about the earth, ouch I just hurt my noodle.

The earth does not rotate exactly 360° about its centre in 24 hours because it has moved an extra days worth of orbit around the sun.

The earth actually rotates ~366¼ times in a year.

I am going to really need an aspirin. Earth wobble, elevation above earth's surface, variation of oblateness of runner's position, earth rotation (366 1/4 times a year, what! we slowed in relation to the sun, this means we will live longer or is that less?), Milankovitch (sic?) cycles or changes of earth eccentricity. Other variations in the universe and galactic orbits likely exist. Chaos anyone?

My gut feeling is that the next most significant thing to take into account would be the height above sea level, and, that would be pretty tiny compared to the

circumference/365¼ distance.

Kevinm, I think the geocentric coordinate velocity question has been answered correctly in post 2 (Usbport) - its average velocity is not zero, as you imply.

Geocentric coordinates are not quite useless in astronomy, especially using satellite-telescopes. The distant stars all have fixed geocentric coordinates.

I liked Usports answer but I don't think sidereal time needs to be involved. The question states the runner was active "exactly 24 hours". Sidereal time would be involved if we wanted to calculate from exact points of the sun position each day (high noon, sunset, sunrise, etc). It would seem that the changes of the sun's position (change of sunset, sunrise, etc) are not important and in fact irrelevant to the question. The sun will rise or set a little later each day but the clock, if accurate, will still tick off 24 hours. That is the time the runner was running around the track ("exactly").

This 24 hour clock gets us back to geocentric and its importance to the question. The center will move along the earth's orbit a large distance (2575.2 km) as stated earlier. Unless the earth is not moving along its orbital path, then we are back to some point on the track in 24 hours compared to the geo-center. That is the reason I have said the geocentric comparison is useless unless we can say zero. At least that is my take on the question or am I missing my understanding of sidereal time?

I do not think geocentic coordinates are useless but not all telescopes are aligned along such coordinates. In the case of binary stars or other objects of irregularity, barycentric may be more important. I am sorry if I implied anything different.

Depends on the season for the Earth because the Earth's orbit is not a true circle, but an ellipse. So the distance traveled per 24 hours depends where along the Earth's orbital path the Earth is when the runner started. This means the Earth' orbital speed is not constant.

Thanks for pointing that out. Agreed, but I thought I covered myself in response #14 so I didn't want to reiterate it.

Kevin, the way I understand the sidereal day (23 hrs 56 min), it is what makes for the one degree extra rotation (361 degrees) that the earth turns every civil day (24 hrs). In 23 hrs 56 min the "oval track" of the question returns to its original geocentric coordinates, zero average velocity. In "exactly 24 hrs", earth has to turn about one extra degree (geocentric) to bring the sun to its zenith at 12 noon local (on average). Thus the track has moved an additional 1 degree, which is some 110 km at the equator, giving an average geocentric velocity of around 4.6 km/h.

I also think this velocity is totally independent of where in the elliptic orbit around the sun we are. After all, the rotation rate of earth cannot change during the course of a year, can it?

I believe sidereal day means the same as you state. Except that the rotation is meant to bring the rotation in a line parrallel with the stars not necessarily the sun. However, we are not concerned with bringing the clock back to high noon or any other equated time. Sidereal time would be so concerned but the atomic clock is simply ticking and that is time we are to be concerned in the question.

And yes the earth will move a different speeds in its path around the sun. It will move faster at perihelion and slower at aphelion. The question posed is loaded with red herrings. The observers position is also important.

we are not concerned with bringing the clock back to high noon or any other equated time.

True, but, 24 hours is the average time it takes from noon to noon in any place over the whole year: we measure time consistently ignoring the changes due to the earths distance from the sun.

The ~4 minute lag in sidereal time versus solar time is the result of the earth orbiting the sun and the change of position with the earth in relation to the distant stars. A 24 hour clock on earth will rotate 4 minutes past the parallel point. Wiki has a good diagram for reference. High noon will also change a little each day if we use solar time. A parsec is a unit of locating distances of near stars and takes advantage of the apparent parallel effect and the Earth's orbit.

I still contend that the runner is on solar time not sidereal time as the clock is stated as 24 hours not 23 hours and 56 minutes. The position of the observer seems more important. If the observer is near the track the geocentric change will only be observed as where he is on the track after 24 hours in relation to where he started at hour zero (a small amount based on the earth's radius). If the observer is on Mercury (the planet) he may notice the Earth's orbital change of position but not likely the small change of the runner's position on the track.

Kevin, maybe we should just determine if you agree with the answer: the geocentric coordinate average velocity of the runner, as measure in civil time is about 4.6 km/h. If yes, then our argument is simply noise in the system. If no, I would like to know how you work it out.

I think that is the crux of our differences. If we use sidereal time your average is correct. I contend we don't use sidereal as it is only 23'54" and not civil time of 24 hours. The position on the earth is the same and velocity would be zero. Why do we have to add to the time, the geoentric coordinates haven't moved in relation to the runner other than the small change of position on the track. The runners position will change dramatically (+ 2.5 million km) if observed from a position above the solar plane.

I haven't figured if I am stubborn or dense but my logic seems sound to me. But my last course in astronomy was in the early 70's and I am oozing gray matter. Alas.

The earth is spinning once every ~23 hours 56 minutes: so if he sat on the start finish line for that long he would not have moved relative to the centre of the earth. But he didn't he sat/ran for an extra ~4 minutes.

Click! the light goes on. You are of course right and I stand down.

Thanks for entertaining this stubborn dense astronomer with his false logic.

Kevin: "Sidereal time would be so concerned but the atomic clock is simply ticking and that is time we are to be concerned in the question."

Yes, but that's why the position of earth around the sun does not influence the answer to the question. I suppose we could have considered velocity and gravitational time dilation that are different for earth being farther or closer to the sun, but I think the differences are utterly ignorable over one year (likely to be microseconds, but haven't checked). In the correct answer we are talking about 4 minutes in time and over 100 km in distance.

I think that's what Randall also said.

Kevin, what you worked out, correctly I think, is for heliocentric coordinates, not the geocentric coordinates that the 2nd part of the question asked for. Earth's center is not moving in geocentric - it sits fixed at 0,0,0.

Well, if you ended where you started, it would have to be zero.

Now that that's over - I have often pondered; is there more or less HP expended if you run up an escalator - as opposed to riding 'standing still'? <relative to escalator tread that is>

Assume it's otherwise vacant and in normal operation mode.

HP is total of your's and it's expenditure, for either condition.

I think the HP would be more while you were running up, but for a shorter period of time. The total energy kWh would probably be the same for both?

Work done is the same. HP is the rate at which it is done, so running uses more HP.

Harrumph (GA)

zero and ~40075/(365.25*24) km/hr = 4.57 km/hr

what is time - if a day is a year

Which is it? Is he running at the equator or in an oval? He cannot do both at the same time....

The equator runs through the oval somewhere:-

It won't make any difference to your calculations whether you chose E1 E2 or something like E3.

The height of the track above sea level will make a much more significant difference.

The answer is 0 (zero)

as the displacement is zero for both topocentric and geocentric cooordinates.

Ok.. I think I've learned something here!!

The average speed and velocity in local/terrestrial coordinates is understood. To get the average velocity in the geocentric system I set up parametric equations to describe the position as a function of time, differentiated to get a velocity vector function, and then integrated it over a 24 hour period. It might be the difficult way to do the problem but I got 2.88 miles/hr, which is reasonably close to Usbport's (in post #2).

Big thanks to Usbport and Scruffy for helping me learn a bit here.

This made me think... is this type of system (earth centered, but non-rotating) the most convenient/practical way to do problems on a stellar scale or is it simply a system carried forward from a time when the Earth was the center of the universe?

Earth is still the center of the universe, at least of the observable universe

I also realized that geocentric-equatorial is essentially the Cartesian form of the right-ascension-declination polar form used by astronomers, with the only difference that they express the right-ascension in hours, minutes and seconds, with one hour = 15 degrees. You will find all star coordinates given in the latter form.

Hi Scruffy, you wrote: "I also realized that geocentric-equatorial is essentially the Cartesian form of the right-ascension-declination polar form used by astronomers, with the only difference that they express the right-ascension in hours, minutes and seconds, ..."

This in not quite correct. The geocentric coordinates that the question refers to are used to define a set of vectors (I,J,K in the link you gave in #7). The RA-Declination coordinates used by astronomers only give the direction of a vector, not it's magnitude. I guess that's because we do not know the distances to stars to the same accuracy than we know directions. Besides, all we need to point a telescope are the angles.

-J

Earth surface :

- modulus answer : 40*5 km in 24 hours : 8.3 km/h

- vectorial answer : 0 (going back to departure point)

Geocentric coordinates:

- modulus answer : 40 000 km in 23 hours 56 minutes : 1671 km/h

- vectorial answer : 0 (going back to departure point)

23h 56mn is non rotating relative to stars, it would be 24h relative to sun

Correction to my previous answer (11/16/2010 at 10:14 AM)

0 for vectorial answer / geocentric coordinates if the good answer relative to sun.

Relative to stars, the distance is the Earth rotation in 4 minutes : 111 km.

Average velocity : 4,6 km/h

Hello "jmb"

If this is your first "official post" to CR4, welcome!

Your 4.6 km/h geocentric sounds more in line with the consensus answer so far than your previous 0.

SL

velocity=displacement/time=0 because displacement is 0 - vector unit.

However the speed of this runner (scalar)=0.12km/h REALLY slow. This is definitely not a marathon runner -

A: 40 Laps x 5 km = 200 km / 24 hr = 8-1/3 km/hr.

B: "A" +/- the difference between a Standard day (24 hr) and Siderial day plus the starting time during the day, the time of year (the position of Earth in it's orbit) and a little Trigonometry thrown in for good measure