Talk:Semi-major axis

The relation between energy and semimajor axis

Since the semimajor axis becomes infinite when the orbital energy equals the escape velocity it can be seen that the relation between the two cannot be linear. - Farred 18:29, 4 February 2013 (UTC)

You are correct. The relation is linear for elliptic orbits; I misspoke. Since we're on the nitpicking topic, you must mean orbital velocity equal to escape velocity . ChristiaanK 16:39, 5 February 2013 (UTC)

Well I knew what I was thinking. The semimajor axis becomes infinite when the orbital energy per unit mass equals 1/2 times the square of the escape velocity. And when the orbital energy is 0.99999 times that there is a very long semimajor axis elliptical orbit, but nowhere near infinite energy. I tink I have it all correkt now. - Farred 21:44, 5 February 2013 (UTC)

After I made my response on this talk page, I saw that you changed the article. What you have written is simply wrong. I do not have the elementary physics texts handy to give a reference for this but my memory of the correct formulae can be checked with published orbital data.

The orbital energy per unit mass equals one half the square of the velocity minus (gravity times radius). Conventionally the energy is set to zero when an object is at escape velocity. Now let us plug numbers into the formula and see how it compares to published data. According to the CRC HANDBOOK of CHEMISTRY and PHYSICS, 64th EDITION, (C)1983, pages F-128 through F-143, the gravitational constant of the sun (gravity times the square of the radius) equals 1.3256 E +11 km^3/sec^2 (or 1.3256 E +20 meters cubed per second squared). The semimajor axis of the Earth's orbit = 1.4957 E +11 meters. With the radius set equal to Earth's semimajor axis, that gives us a gravity for the sun equal to 5.925 E -3 meters per second squared at the location of Earth's semimajor axis. The orbital velocity of the Earth at that distance is given as 2.9771 E +4 meters per second. So energy per unit mass equals ((2.9771 E +4 m/s)^2)/2 - ((5.925 E -3 m/s^2) * (1.4957 E +11 m) equals (8.8631 E +8 m^2/s^2)/2 - 8.8620 E +8 m^2/s^2 equals -4.430 E +8 m^2/s^2. That negative energy is what binds the Earth to the sun. Now if the tabulated solar escape velocity from the position of Earth's orbit, 4.210 E +4 m/s, is substituted in the formula for energy, this yields 2.7 E +3 m^2/s^2 which as a difference between two numbers raised to the eighth power of ten equals zero within the significance of the values in the computation.

The semimajor axis is proportional to the two thirds power of the period, which also is born out by the tabulated figures in the cited tables. Earth's orbital period is 3.155815 E +7 seconds. Mars' period is 5.93553 E +7 seconds. Mars' semimajor axis is 2.2784 E +11 meters. So, (5.93553 E +7 / 3.155815 E +7)^2/3 * 1.4957 E +11 meters equals 2.2790 E +11 meters which is within the significance of the tabulated figures equal to the tabulated value for Mars' semimajor axis. So the semimajor axis is proportional to the two thirds power of the period, not proportional to the energy which is evaluated by quite a different formula.

The semimajor axis is not proportional to the energy in eliptical orbits or any other kind of orbit. - Farred 01:34, 6 February 2013 (UTC)

From the new article that you posted about specific energy is the formula epsilon = -mu/2a The formula you have as energy per unit mass = V^2/2 - mu/r can be correct only if mu = (local gravity * r^2) which is a constant of the central body. Specific energy being equal to mu/2a does not in any way indicate that specific energy is proportional to the semimajor axis. The specific energy mu/2a here refers to the energy lost compared to an object traveling at escape velocity. What you have is a negative energy proportional to the multiplicative inverse of the semimajor axis. This is not at all the same as the semimajor axis being proportional to the energy. - Farred 02:53, 6 February 2013 (UTC)