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Speed of International Space Station

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  1. Apr 29, 2015 #1
    Why ISS need to run on such high speed 7.7km/s ?
    What will affect if it made to run on very slower speed.

    thanks
     
  2. jcsd
  3. Apr 29, 2015 #2

    NascentOxygen

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    Hi Israr. :smile: http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

    Good luck with your study of science.

    Technical questions belong in the technical forums. I've moved your thread to a suitable forum where it can be appropriately discussed.
     
    Last edited by a moderator: May 7, 2017
  4. Apr 29, 2015 #3

    e.bar.goum

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    Hi Israr, welcome to PF.

    The speed of any object in orbit in a gravitational field (planets around stars, moons around planets, the ISS around the Earth) is determined by the orbital height of the object.

    For a circular orbit, and something of negligable mass compared to the other body (e.g. the ISS around the Earth, or the Earth around the sun) this is roughly equal to

    ##v \approx \sqrt{\frac{G M}{r}}##.

    Where G is the gravitational constant, M the mass of the body being orbited around, and r the radius. If you plug in the mass of the earth, and the radius of orbit of the ISS (~412 km above sea level) into that equation, you see that you get 7.7 km/s!

    So, you see that if you wanted the ISS to go at a much slower speed, you'd have to push it much further away from the Earth!
     
  5. Apr 29, 2015 #4
  6. Apr 29, 2015 #5

    Drakkith

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    If the ISS were to fire its thrusters and slow down, it would fall into the Earth's atmosphere.
     
  7. Apr 29, 2015 #6
    The speed to completely escape Earth's gravity is something very different from the speed to be in a stable orbit.

    @Israr , the ISS *could* fly slower, but in order to still have a stable orbit (i.e. neither falling towards Earth nor flying away from it) the station would have to be in a higher orbit. The higher the orbit, the slower the speed necessary (as evidenced by the formula e.bar.goum quoted)
     
  8. Apr 29, 2015 #7
    You're absolutely right, I erroneously thought they were the same thing. Why do you use ≈?
     
  9. Apr 29, 2015 #8
    I don't, that was e.bar.goum.
     
  10. Apr 29, 2015 #9

    davenn

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    Did you read the sentence immedialy prior to that equation that e.bar.goum wrote ?
    it answers that question

    Dave
     
  11. Apr 30, 2015 #10
    The equation is derived by equating the centrifugal force to the gravitational force. The small mass m cancels out so I don't why it needs to be "negligible mass".

    For a circular orbit, which the ISS is pretty close to, this is exact.
     
  12. Apr 30, 2015 #11

    e.bar.goum

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    Yes, but you can only make that equality for stable, well behaving orbits. If the masses of the bodies are similar, and the orbits are eccentric you'll get into trouble.

    As you say, for the ISS it's pretty much exact, but I didn't want to give the impression that this is true in general for all orbits, hence the caveats.
     
  13. May 4, 2015 #12
    Thanks
     
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