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Delta V as a Function of Altitude

  1. Jun 7, 2013 #1
    Greetings everyone!

    I have a question on how the delta-V required to reach different orbits is determined. I refer to lift-off delta-V.

    I'm curious to find the relationship between altitude and delta V required to get to the height.

    From what I have found out, the lift-off delta v to a 100km altitude is in the range of about 1.4km/s for an ideal system. I would like to know how this is determined and how would it scale for something like a 150km altitude. (Apparently spaceshipone had a delta V of 1.7km/s to reach a 112km altitude).

    It seems like the available information focus mainly on calculating orbital transfer Hohmann transfer delta-vs, and lift-off to orbit delta v (determined by gravity, mass and radius of the planet). The trouble is, while there are certain delta-v maps (e.g. 9.2 km/s from earth to LEO etc.. ), I haven't found any way to derive the figures at intermediate altitudes.

    Would really appreciate any insight. Thank you for your time
     
  2. jcsd
  3. Jun 7, 2013 #2

    mfb

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    If you can neglect air drag, it is just energy conservation. The sum of potential and kinetic energy is conserved.
    For circular orbits, you need a kinetic energy of 1/2 of the (negative) potential energy.
    For non-circular orbits, the formulas are a bit more complicated.
     
  4. Jun 8, 2013 #3
    Thank you very much.
     
  5. Jun 14, 2013 #4

    BobG

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    Actually, mfb answered a slightly different question than you asked.

    To just reach a height of 150km at least once, you don't need to achieve a true orbit (one where perigee misses the Earth).

    The size of your orbit depends on the specific energy of the orbit:

    [tex]\epsilon = \frac{GM}{-2a}[/tex]
    where a is the semi-major axis, G the universal gravitational constant, M the mass of the Earth, and epsilon the specific energy. You could multiply the universal gravitational constant by the mass of the Earth once and use it as a constant. In fact, it is a constant and has a name, the geocentric gravitational constant, and is equal to 398600 km^3/sec^2 (assuming you don't want to convert your altitudes, radii, and semi-major axis to meters).

    The semi-major axis is half the long axis. You need the long axis to be at least 150 km above the surface of the Earth (with the Earth's radius being 6378 km). So your a needs to be about 3265 km.

    Your total energy is the sum of kinetic energy and potential energy. Your potential energy depends on your distance from the center of the Earth:

    [tex]U = -\frac{GM}{r}[/tex]

    where U is potential energy (per unit of mass) and r is your distance from the center of the Earth.

    At the surface of the Earth, r is equal to the Earth's radius.

    The difference between your total specific energy and your potential energy has to equal your kinetic energy, which is:

    [tex]T=1/2 v^2[/tex]

    where T is your kinetic energy (per unit of mass) and v is your speed.

    1.7 km/sec should get you to an altitude of 150 km.

    You can combine all of the above into one equation and only have to determine the proper semi-major axis to make it work:

    [tex]v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}[/tex]

    Your r will always equal the Earth's radius, while a will be affected by what altitude you want to achieve.

    Now mind you, your 'orbit' will virtually hit the center of the Earth, so you'll only achieve the desired altitude once. If you want to actually orbit at some given altitude, your semi-major axis needs to big enough for perigee to miss the Earth, as well (which is the question mfb actually answered).
     
    Last edited: Jun 14, 2013
  6. Jun 14, 2013 #5

    mfb

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    My post covered both the orbital and non-orbital motion, as both were mentioned in the first post.
     
  7. Jun 15, 2013 #6
    Thank you both very much. I can appreciate the math and physical implications better now.
     
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