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Low and high orbital energy differences (getting there and back)

  1. Nov 21, 2012 #1
    First of all am I right in thinking that ( comparitivly ) low orbits such as the ISS need very high velocities (in the region on 17,000 mph ) and that orbits further out require less speed ?

    Is there a "sweet spot" orbit that requires a minimum velocity ?

    what's the minimum velocity required to get into a minimum energy orbit ?

    and finally is there an orbit you can get back down from with a minimum velocity such as you don't need create a blazing meteor trail...
     
  2. jcsd
  3. Nov 21, 2012 #2

    rcgldr

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  4. Nov 21, 2012 #3
  5. Nov 21, 2012 #4

    rcgldr

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    Section 2 of that article, shows the math for the total energy of the satellite. I thought that part would help here.
     
  6. Nov 21, 2012 #5
    save for the fact that going from rest in an atmosphere isn't transferring from one orbit to another.

    I'm specifically interested / curious about what orbit is the "cheapest" to get to from the surface (at rest)

    And how to get back down without emulating a meteor !
     
  7. Nov 21, 2012 #6

    mfb

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    The lowest possible orbit, just above the dense parts of the atmosphere.

    With a massive rocket. Which is much more expensive than atmospheric braking.

    Higher orbits have a lower orbital velocity, but you need more energy (and more initial velocity) to reach them from earth.

    No.
     
  8. Nov 21, 2012 #7
    this is the bit I don't get, if the higher orbit is slower, why does it take more energy and velocity to get there?

    and conversely why is it the fastest orbit the "cheapest"
     
  9. Nov 21, 2012 #8

    jtbell

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    Potential energy. Higher orbits (larger radius) have higher gravitational potential energy. It's the same reason why it takes more work to lift a brick from the ground to a height of ten feet rather than one foot.
     
  10. Nov 21, 2012 #9

    mfb

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    This has a very interesting implication: If you are in a low circular orbit and accelerate (forward), you increase your distance to the central object - and if you accelerate (forward) again at apoapsis to get in a circular orbit, you are slower than you were before. Going upwards slowed you down.
     
  11. Nov 21, 2012 #10

    rcgldr

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    Note that if the goal is to establish a circular orbit, then the worst case total delta-v occurs at some specific radius, and beyond that the total delta-v decreases. This is because the first impulse approachs a limit of sqrt(2) times the velocity of a circular orbit, since sqrt(2) x orbital velocity (sqrt(G M / r) = escape velocity (sqrt (2 G M / r)), while the second impulse approaches zero. Again a reference to that wiki article about this:

    maximum_delta_v.htm

    The other issue is that aerodynamic drag limits the practical amount of velocity achieved while still within the bounds of the atmosphere, so much of the increase in velocity occurs after a rocket has gone beyond the boundary of the main atmoshpere (there is still a very sparse atmosphere even in low orbits, such as the space station, enough to require occasional bursts to avoid orbital decay). In the case of the space shuttles, they would be going less than 4000 mph as they left the main part of the atmosphere, and increased speed to about 17,200 mph afterwards.
     
    Last edited: Nov 21, 2012
  12. Nov 21, 2012 #11

    K^2

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    Higher orbits are slower, but the energy required to get to higher obit is still higher due to gravity. Look at Space Shuttles maximum useful payload for LEO and compare it to much higher Geo-stat orbit.

    Edit: And it appears that people have already pointed this out.
     
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