|WikiProject Physics||(Rated Start-class, Low-importance)|
|WikiProject Astronomy||(Rated Start-class, Mid-importance)|
- For a given μ, and a given r and v at a single point of the orbit, one can compute ε, allowing an object to be classified as having not enough energy to be orbital, hence being "suborbital" (a ballistic missile, for example), having enough energy to be "orbital", but possibly being suborbital anyway due to the shape of the orbit, or having enough energy to be extraterrestrial (as a meteor, for example).
Is ε specific orbital energy? Although there's a link to that at the end of the article, the symbol isn't explicitly defined anywhere in this article, and eccentricity can also be denoted by epsilon. Dependent Variable (talk) 06:15, 29 April 2010 (UTC)
Actually, I have always seen specific (orbital/mechanical) energy as ξ (xi), not ε (epsilon). Not sure if this is a mistake or just someone using different notation. Tfr000 (talk) 21:48, 3 December 2015 (UTC)
I believe b may mean semiminor axis but it is not clear in the article. Could someone please clear this up? This is at the end of the section "Derivation for elliptic orbits (0 ≤ eccentricity < 1)"Circuitboardsushi (talk) 21:33, 4 April 2017 (UTC)
We read here that "It is the direct result of the principle of conservation of mechanical energy which applies when the only force acting on an object is its own weight." ITS OWN WEIGHT? OMG! What is the "weight" of a satellite in orbit at 100,000 km above Mars?! (will it matter whether the orbit is circular, elliptical, parabolic, or hyperbolic???) Why is mechanical energy not conserved when other external forces are applied???? Garbage! Also, we will learn that "v is the relative speed of the two bodies" !!! MORE GARBAGE! NOWHERE is it mentioned that the equation is about only two bodies, nowhere is it mentioned that one body is in orbit above the other. Nowhere is the definition of the "relative" speed to be found. Really horrible development of the concept.22.214.171.124 (talk) 17:58, 21 July 2016 (UTC)
- What is the "weight" of a satellite in orbit at 100,000 km above Mars? The short answer is: Nobody knows. The long answer is: The weight of a planet, natural satellite or artificial satellite supplies the total force acting on the body. (True, artificial satellites sometimes experience forces resulting from the firing of maneuvering rockets or discharges of gas for the purpose of maneuvering, but those forces are applied only very briefly so I am ignoring them.) For a planet, or a satellite, there is no external force other than its own weight – no friction, no rockets, no springs, no electro-magnetic forces etc. The weight of the planet or satellite is equal to the vector representing its acceleration, multiplied by the mass of the planet or satellite (N2L).
- Why is mechanical energy not conserved when other external forces are applied? The principal of conservation of mechanical energy applies if, and only if, all the applied forces are conservative. Gravity is a conservative force field so the weight of any object is a conservative force acting on that object. Perfect springs also exert conservative forces. Most other forces that we are familiar with on a day-to-day basis are not conservative. The principle of conservation of mechanical energy is explained in all basic physics text books. For example, in PHYSICS by Resnick and Halliday, "Potential Energy", section 8-3. Consequently, if the only force acting on a body is its own weight, the mechanical energy of that body will be conserved. Planets and satellites are good examples. An object falling freely in a vacuum is also a good example of conservation of mechanical energy. If the falling object bounces off a perfect spring its mechanical energy is conserved. Dolphin (t) 01:12, 22 July 2016 (UTC)
This step in the derivation does not seem to follow
In the derivation, we go from:
It is clear that each cancels on the RHS, but how does one cancel the with one of the ?? Note that the expression is and not , which is necessary to get that latter result. I am willing to believe that the Vis-viva equation works, but we missed an assumption in the derivation here.