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Around the world in 80 hours

  1. Oct 16, 2003 #1
    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?
  2. jcsd
  3. Oct 16, 2003 #2


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    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.
  4. Oct 17, 2003 #3
    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.
  5. Oct 17, 2003 #4


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    He said "closer to" 80 minutes - its actually about 90.

    At orbital velocity and below, there is very little time dilation - you need an atomic clock and sophisticated monitoring equipment to notice it.

    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.
  6. Oct 18, 2003 #5
    Opps, sorry.

    Oh. Thought it would be more noticible, guess not :)
  7. Oct 18, 2003 #6


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    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.
  8. Oct 22, 2003 #7
    To orbit the earth in 80 hours and remain in orbit without decaying, the satellite would have to travel at 70638 m/s at a distance of 80471 m above the earth's surface. These are both to too many significant figures (they should be to three not five), but I think it answers your question.

    If you want to know how I worked these figures out, let me know and I'll post the formulas and how I re-arranged and substituted in things etc.
  9. Oct 22, 2003 #8
    Yes please :D
  10. Oct 23, 2003 #9
    Since I posted the answer to your post, I have realised that I made a drastic answer in my working that threw the answer for height out a little and the velocity was miles out. So I've have found a simpler way of calculating these figures which hopefully are correct this time.

    Code (Text):

    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).

    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

    and substitute in ω:

    mr * (2π/t)^2 = GMm

    which simplifies to:

    r^3 = GMt^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

    v = [squ](GM)
         (r )

    v = 6.67*10^-11 * 6.02*10^24

    v = 2061.4 m/s
      = 4611.3 mph if you would prefer

  11. Oct 23, 2003 #10
    Oh cool! Thanks
  12. Oct 24, 2003 #11
    I think the minimum orbit length is about 85 minutes, but for that the satellite would need to be running along the surface of the earth.

    Happy to help. :smile:
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