Fazer Owners Club - Unofficial

General => General => Topic started by: purplebear7 on 07 April 2012, 08:06:08 am

Title: Serious Daft Question Time
Post by: purplebear7 on 07 April 2012, 08:06:08 am
Hi Guys,     
OK ... it's Daft Question Time again .. only this time I'm serious  :)

My mate Mr Google has a trucking load of info on this topic .. but I still can't get an answer .. and it's starting to bug me just a bit. I am NOT into astronomy, the NASA fan club or Virgin space flight .. I'm a Pro Diver, Trucker .. retired  :lol

The International Space Station (ISS) is visible from Earth as it orbits the planet. (but they don't wave back)

I have ben able to work out the flight path direction and angle of altitude and visually 'tracked' it .. for six minutes .. as it passed over the Costa Blanca area of southern Spain .. one dark night. That's a very long 'exposure' time .. most nights you will only get one or two minutes .. according to the predicted tables and your location.   

It looks like a very bright star travelling at very high speed across the sky .. bit like a large aircraft .. or UFO  :eek . The flight path changes so we can not see it every night, even when there are no clouds.   

It's average speed is said to be 17.500 mph, or thereabouts and it's altitude is around 240 miles above sea level (I think there may be some slight 'wobble' variations with all these figures, but not a lot ?)

The distance around the equator is .. lets say .. 30.000 miles. So the total distance of the ISS orbit loop must therefore be much further than that!

They recon the ISS does a full lap once every 90 minutes .. as you see these 'sums' don't add up .. so my question is:  HOW can that be?

I look forward to your thoughts and calculations. Cheers  ;)      As always .. Stay Safe.   Polar Bear.





 
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Title: Re: Serious Daft Question Time
Post by: Davidjw on 07 April 2012, 08:27:13 am
well I was under the impression that the distance round the equator was more like 25k not 30 and if it passes over you its going a shorter distance round the earth perhaps , only my thoughts but I would suggest patrick moore or an astrological type may answer it better    :)
Title: Re: Serious Daft Question Time
Post by: Bracechenko on 07 April 2012, 08:48:15 am
There are many different types of orbit that it could be doing though. It doesn't necessarily have to be around the fattest part of the Earth?
Title: Re: Serious Daft Question Time
Post by: Phil TK on 07 April 2012, 09:30:05 am
A mate of mine has a phone app that uses gps to 'bleep' him every time the space station is about to come over the horizon and and also tells him which direction to look, it works brilliantly. Might be worth looking at this app PB, I'm afraid I don't know what it's called.
 One thing to note regarding the SS orbit time is that the earth is spinning at 600mph, so perhaps when the orbit is counter to the earths spin it manages an orbit of 90 minutes point to same point. Maybe.
Title: Re: Serious Daft Question Time
Post by: Fazerider on 07 April 2012, 09:42:33 am
The equatorial circumference is 24,900 miles.
The ISS at an altitude of 240 miles would then be travelling roughly an extra 2.π.240 miles on top of that... roughly 1500 miles for a total orbital track of 26,400 miles.
Which, at an average 17500mph results in an orbit of 90.5 minutes. A near enough approximation since the orbit isn't perfectly circular.
Title: Re: Serious Daft Question Time
Post by: DryRob on 07 April 2012, 09:58:51 am
The short answer is there may be some commas mixed up with decimal points and the speed is reasonable.
 
The long answer is:
r = c/(pi*2), c = pi*2*r, s = d/t, t = d/t
r = radius, c= circumference, s = speed, d = distance, t = time
 
c(earth) = 24901.55 miles @ equator, 24859.82 miles pole to pole ==> average 24880.69 miles
h(orbit) = 240 miles
s = 17239.2 miles per hour
 
c(orbit) needs calculating:
 
r(orbit) = r(earth) + h(orbit) ==> c(earth)/(pi*2) + h(orbit)
r(orbit) = 24880.69/(pi*2)   240 ==> 3959.88   240
r(orbit) = 4199.88
c(orbit) = r(orbit)*pi*2 ==> 4199.88*pi*2
c(orbit) = 26388.65 miles
 
t = d/s ==> t(orbit) = c(orbit)/s
t = 26388.65 / 17239.2 ==> 1.5307 hours
t = 1.5 hours
 
Due to the earth not being perfectly spherical, variations in the height of orbit and errors in the measured quantities this value can only be considered an approximation.
Title: Re: Serious Daft Question Time
Post by: Phil TK on 07 April 2012, 10:17:21 am

 
Due to the earth not being perfectly spherical, variations in the height of orbit and errors in the measured quantities this value can only be considered an approximation.

Also, at that speed, the time taken for an astronaut on the SS to do one orbit will take slightly less time than an observer watching the same orbit on the ground.
Title: Re: Serious Daft Question Time
Post by: Fazerider on 07 April 2012, 10:29:20 am
A rope is made that fits snugly round a planet at the equator.
It's then decided that it would be much better if it was suspended 1m above the surface.
The inhabitants know the diameter of the planet is 12,000km so how much longer does the rope need to be? :lol


FWIW, a useful website for predictions of visible ISS passes is this one:
http://www.heavens-above.com (http://www.heavens-above.com)
Just be aware that the times given are GMT!
The site also gives predictions for other bright satellites including Iridium flares.
Title: Re: Serious Daft Question Time
Post by: pilgrim on 07 April 2012, 09:04:31 pm
Heavens Above is a great website.
Not that I ever use it.
I'm not a nerd, honest :o
P.S Their clock is an hour fast at the moment.
Or so I am told.
The ISS orbit is tilted at an angle which takes it from south west to north east. Its currently somewhere of the eastern tip of South America heading for North Africa and the middle of the Med. 8)