Talk:Specific orbital energy

Regarding the Sign of semi-major axis

I have never seen a different equation for hyperbolic SE. Hyperbolic trajectories are normaly associated with negative semi-major axes, which correctly gives a positive energy when using the general form of the energy equation. Why present it this way?

Here is what is currently in the article:

For a hyperbolic trajectory this specific orbital energy equation takes form:

$\epsilon ={\mu \over {2a}}\,\!$

--Keebler71 18:09, 2 June 2006 (UTC)

Characteristic energy

The article says "For a hyperbolic orbit, it is equal to the excess energy compared to that of a parabolic orbit. In this case the specific orbital energy is also referred to as characteristic energy." Our article on characteristic energy says characteristic energy is twice specific orbital energy, which also matches the formulas given.--agr (talk) 03:51, 22 April 2012 (UTC)

Earth Orbits example: LEO

The specific orbital energy for low Earth orbit appears to be incorrect. Given m1 >> m2 and GM for Earth is 398,600 km^3s^-2; given a 300km circular orbit (therefore a=r=6,678km) we get:
e = -398,600 / (2 * 6,678)
e = -398,600 / 13,356
e = -29.84 Mj/kg

Alternatively, potential energy -u / r is roughly -59.69 Mj/kg for r=6,678km and kinetic energy v^2 / 2 is roughly 29.85 Mj/kg for v = 7.726 km/s, yielding -29.84 Mj/kg. Freshgroundcoffee (talk) 23:56, 28 February 2013 (UTC)

Semi-major axis

There's a potential point of confusion that's worth pointing out. The equation "epsilon = - mu / 2a" is correct, but for an elliptical orbit of a small body around a large one (e.g., earth) it's easy to confuse the semi-major axis of the orbit with the radius at apoapsis from the center of the large body. Under that (wrong!) interpretation, specific orbital energy would be independent of the radius at periapsis.

But r_sub_a is not the semi-major axis. It's "(r_sub_p + r_sub_a) over 2". That's probably worth pointing out. Agnostic Engineer (talk) 00:50, 26 January 2015 (UTC)

Assessment comment

The comment(s) below were originally left at Talk:Specific orbital energy/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

Comment(s)

Press [show] to view →

In astrodynamics the specific orbital energy (or vis-viva energy) of an orbiting body traveling through space under standard assumptions is the sum of its potential energy () and kinetic energy () per unit mass. According to the orbital energy conservation equation (also referred to as vis-viva equation) it is the same at all points of the trajectory: Can the spinning kinetic energy of the Earth be neglected in the calculation of its specific orbital energy which is derived to be GMc/2a in the current literature? Let's take a look of the value of GMc/2a where a, Earth semi-major axis, is 1.50E+11 meters, and GMc for the Sun is 1.33E+20 mmm/ss. Then GMc/2a is 4.44E+08 mm/ss, where the Earth spinning kinetic energy per unit mass is neglected. We know the Earth spining kinetic energy per unit mass is 0.5*(0.4*Radius of Earth*Radius of Earth)*(Angular velocity of Earth spin*Angular velocity of Earth spin) where the Radius of Earth is 6.37E+06 meters, and the Angular velocity of Earth spin is 2*3.1416/(24*60*60) radians/second. Upon calculation we have the Earth spinning kinetic energy per unit mass to be 4.29E+04 mm/ss. Therefore, the question now is whether we can ignore 4.29E+04 as compared with 4.44E+08. Critiques and comments and advices are solicited. Thanks. Additional questions: During our augmentation we continued to find out most recently some very interesting results as indicated below: EARTH MOON VEUS At furthers point i.e. Apogee,etc Centripetal force 3.43E+22 1.78E+20 5.44E+22 Centrifugal force 3.38E+22 1.69E+20 5.40E+22 At nearest point i.e. Perigee, etc Centripetal force 3.67E+22 2.22E+20 5.65E+22 Centrifugal force 3.72E+22 2.34E+20 5.61E+22 For EARTH & MOON, at Perigee each's Centrifugal force is greater than Centripetal force. But for Venus, its Centrifugal force is less than Centripetal force. Why does such a difference exist? Remember that the orbital direction and the spin direction of EARTH & MOON are all counter clockwise. But the spin direction of VENUS is Clockwise, or retrograde, although its orbital direction is counter clockwise as that of EARH & MOON. Why does such a difference exist? Would you, Honorable Board Members, please advise us whether it is worthwhile for further study? Thanks. Praise be to our LORD JESUS CHRIST. God bless you, your families, and your careers, Sincerely yours, Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang —Preceding Dear Honorable Board Members: On 5/15/08, Wednesday, we continued to find planet data for other Planets to calculate and compare. Surprisingly, we found out the spinning kinetic energy per mass of Jupiter and Saturn to be in the same order of magnitude of the value of their GMc/2a. For Jupiter its GMc/2a is 8.52E+07, its spinning kinetic energy per unit mass is 3.16E+07. For Saturn its GMc/2a is 4.63E+07, its spinning kinetic energy per unit mass is 1.95E+07. Therefore, it is assured that the spinning angular momentum and kinetic energy of the planets cannot be neglected as in the existing literature in the derivation of their orbital equations. Praise our Lord Jesus Christ. What next steps should we take? Please further critique and advise. Thanks. Sincerely, Elder Dr. & Mrs. Hsien-Lu & Hui-Lien Peng Huang P.S. Jupiter Saturn Earth Length of Day (hours) 9.9259 10.656 24 Orbital Period (days) 4330.595 10,746.94 365.256 Semi-Major Axis a (m) 7.79E+11 1.43E+12 1.50E+11 unsigned comment added by 70.241.96.76 (talk) 14:34, 13 May 2008 (UTC)

Last edited at 16:06, 15 May 2008 (UTC). Substituted at 06:40, 30 April 2016 (UTC)

🌈 Ice.Queen (talk) 04:36, 6 February 2017 (UTC)