How fast/slow would one have to be travelling to circle the world in 80 hours(on earth) in a spaceship? How long would it seem like in the spaceship?
Woa, it takes 80 mintues typically? Hmm, well ok then :) How long would it seem for the people one earth and for the person in the spaceship? My Physics teacher says that both will experience different times. Didn't quite get that.Originally posted by LURCH
Typical orbit in the shuttle or other low orbit sattelite is closer to 80 minutes. To take 80 hours, you would have to orbit at about 1/60th that speed. This would put your speed at about 2km/sec. (a snail's crawl by orbital standards). But first you would have to get up to an altitude at which this orbit would keep you aloft. Now geosynchronous orbit is about 36,000 km. And that's just to get enough altitude to go around the world in twenty-four hours. To stay up for 80 hrs would take (at a rough geuss) something close to 100,000 km, I think.
He said "closer to" 80 minutes - its actually about 90.Originally posted by renedox
Woa, it takes 80 mintues typically? Hmm, well ok then :) How long would it seem for the people one earth and for the person in the spaceship? My Physics teacher says that both will experience different times. Didn't quite get that.
Opps, sorry.Originally posted by russ_watters
He said "closer to" 80 minutes - its actually about 90.
Oh. Thought it would be more noticible, guess not :)Originally posted by russ_watters
I read somewhere that the American astronaut with the most time in space (Story Musgrave) is several MILIseconds younger than he would be if he hadn't been in space.
Orbital velocity is 17,500 miles per hour. That seems fast, but its nothing compared to the speed of light, which is 186,000 miles per SECOND. Only when you get up to a decent fraction of the speed of light is time dilation noticeable on a human scale.Originally posted by renedox
Oh. Thought it would be more noticible, guess not :)
The formulae I am going to use are: ω = 2π/t F = mv^2/r A variation on F=ma where a=v^2/r, this is used when things are traveling in a circle F = mrω^2 Another variation on F=ma where a=ω^2 * r, this is used when things are traveling in a circle and ω is the angular velocity (radians per second) F = GMm Where F is the attraction of gravity between two objects of mass M and m --- and r is the distance from the point mass M (ie: the centre of the orbit). r^2 G is the gravitaion constant (6.67*10^-11) it is just included because otherwise the formulae don't work. First of all combine the last two formulae to give: mrω^2 = GMm --- r^2 and substitute in ω: mr * (2π/t)^2 = GMm --- r^2 which simplifies to: r^3 = GMt^2 ----- 4*π^2 Putting in values gives: r^3 = 6.67*10^-11 * 6.02*10^24 * 80*3600 80*3600 is 80 hours expressed in seconds ---------------------------------- 4 * π^2 r^3 = 8.4362*10^23 r = 94489000 m above the centre of it's orbit (the centre of the earth) radius of earth = circum/(2*π) circum of earth is 40,000 Km or 40,000,000 m. = 40000000/(2*π) = 6366200 m height above the earth = 94489000 - 6366200 = 88123000 m = 88.123 Km Now for finding v: mv^2 = GMm ---- --- r r^2 which simplifies to: v^2 = GM -- r v = [squ](GM) (--) (r ) v = 6.67*10^-11 * 6.02*10^24 ------------------------ 94489000 v = 2061.4 m/s = 4611.3 mph if you would prefer