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Orbital mechanics

  1. Oct 23, 2015 #1
    I'm having trouble with orbItal mechanics, I'm trying to determine the total potential energy of an orbiting satellite, what I've done so far is this:
    I know m*g*h is potential energy, but I also know that gravity deceases with distance avoiding to the inverse square law. I know I Just can't use the satellites current Ag because that will increase as it falls towards the earth, and I'm having trouble determining exactly what that is.
    This is giving me serious trouble because I want to calculate periapsis or apoapsis given just one of those two, and the velocity at that point.
     
  2. jcsd
  3. Oct 23, 2015 #2

    SteamKing

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    The gravitational potential energy of an orbiting body is different from mgh, which applies only to something located close to the surface of a planet, like earth.

    http://hyperphysics.phy-astr.gsu.edu/hbase/gpot.html

    U = -GMm / r

    where

    M - Mass of the planet, kg
    m - mass of the orbiting body, kg
    G - Universal Gravitational Constant = 6.67408 × 10-11 m3/kg-s2
    r - distance between the centers of mass of M and m, in meters
    U - gravitational potential energy (= 0 when r → ∞)
     
  4. Oct 23, 2015 #3

    BobG

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    I'd make one adjustment to SteamKing's equation. Use specific energy per unit of mass.

    F= -GMm/r^2
    F=ma
    therefore, a=GM/r^2

    All objects accelerate at the same rate (due to gravity).

    Since it's the motion you're interested in, you can do the same with the energy (divide out the mass). And, if you do need to know the energy (to determine how much fuel to do a delta V, for example), you just multiply the specific energy (or the change in specific energy) by the mass when you need that info.

    Likewise, your specific kinetic energy would be (v^2)/2. And your total specific energy would be (GM)/(-2a), where a is the semi-major axis of your orbit.

    Given that you mentioned Earth, I assume you're talking about a satellite orbiting Earth? The universal gravitational constant and the mass of the Earth are constant. Once you've multiplied them together once, you can just remember the answer. You can even look up the answer in a book. It's your geocentric gravitational constant (3.986 x 10^5 km^3/sec^2).
     
  5. Oct 28, 2015 #4
    Is it really that simple? It makes sense but It seems too simple. I came up with that about two weeks ago but I didn't believe myself when I did it. I guess I'll just have to launch a scientific mission in KSP to verify it.
     
  6. Oct 28, 2015 #5
    Wait, that doesn't make sense, by getting higher your potential energy decreases? And in a higher orbit you move slower so your kinetic energy decreases too, that can't be right.
    I'm looking for Σ specific potential energy, our the whole of all the specific kinetic energy that would be gained from a fall of such altitude.
     
  7. Oct 28, 2015 #6

    SteamKing

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    First, you were looking for orbital elements; now, you're looking for the K.E. a satellite gains by falling out of orbit.

    This article discusses orbital mechanics in some detail:

    http://www.braeunig.us/space/orbmech.htm

    This article talks about the relationship between gravitational P.E. into K.E. for an orbiting body:

    https://en.wikipedia.org/wiki/Specific_orbital_energy
     
  8. Oct 28, 2015 #7
    No. Notice the minus sign in front of the expression fro PE.
    The PE increase as you go higher, the maximum value being zero. This is so because the reference is chosen to be at infinite distance.
     
  9. Oct 28, 2015 #8
    I don't see how my request has changed. if an object has 25,000,000 Joules of kinetic energy when it impacts, then it must have had 25,000,000 Joules of potential energy when it began to fall. I am just looking for the potential energy of an object in freefall beginning at a distance where the gravitational inverse-square law becomes relevant, as I always have been. maybe I was sloppy with my wording before, but I truly do not see a difference between what I was originally asking for and the description you gave here.
     
  10. Oct 28, 2015 #9

    SteamKing

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    The gravitational inverse square law is always relevant, regardless of distance. After all, this law still works between Pluto and the Sun, even though these two bodies are separated by a distance of some 4.5 billion kilometers.

    If you want to calculate the periapsis or apoapsis of an orbiting body, then application of Newton's laws of motion can tell you this.

    If you want to calculate the impact energy of a body falling out of orbit, a different application of the same laws is required.

    Depending on the problem you want to solve, you have to tailor the analysis to obtain the solution. It is not immediately obvious how knowing the periapsis of an orbiting body will necessarily convert into the K.E. of that body hitting the ground.
     
  11. Oct 28, 2015 #10

    jbriggs444

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    No, it must have had 25,000,000 more Joules of potential energy when it began to fall than when it impacts. Zero is 25,000,000 more than -25,000,000.
     
  12. Oct 28, 2015 #11

    Janus

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    The total orbital energy of your satellite is the sum of it's kinetic and potential energies.
    KE = mv2/2
    and
    GPE = - GMm/r
    (note since r is measured from the center of M, working out the potential energy difference between the satellite in orbit vs on the ground means solving the following
    [tex]\Delta GPE = \frac{GMm}{r_E}- \frac{GMm}{r_o}[/tex]

    Where rE and ro are the Earth's radius and orbit radial distance

    Now, to get to your orbital question:

    We start with:

    (1)[tex]E= \frac{mv^2}{2}- \frac{GMm}{r}[/tex]

    If we assume a circular orbit, we know that [itex]v= sqrt{\frac{GM}{r}}[/itex]

    subbing for v, and reducing the equation gives

    (2)[tex]E=- \frac{GMm}{2r}[/tex]

    It also turns out that this equation holds for elliptical orbits if we use a, the semi-major axis for r:

    (3)[tex]E=- \frac{GMm}{2a}[/tex]

    If we equate equation 1 and 3 and solve for a, we get

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

    which gives us the semi-major axis if we know the velocity at r
    (This last equation is a re-arrangement of the vis viva equation)

    Now its just a simple case of knowing that

    [tex]a= \frac{r_{per}+r_{app}}{2}[/tex]

    to find either apogee or perigee if you know the other and the velocity at it.
     
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