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The mean eccentricity of an object is the average eccentricity as a result of perturbations over a given time period. Neptune currently has an instant (current epoch ) eccentricity of 0.011 3 , [ 13 ] but from 1800 to 2050 has a mean eccentricity of 0.008 59 .
Consider the ellipse with equation given by: + =, where a is the semi-major axis and b is the semi-minor axis. For a point on the ellipse, P = P(x, y), representing the position of an orbiting body in an elliptical orbit, the eccentric anomaly is the angle E in the figure.
Note that for a given amount of total mass, the specific energy and the semi-major axis are always the same, regardless of eccentricity or the ratio of the masses. Conversely, for a given total mass and semi-major axis, the total specific orbital energy is always the same. This statement will always be true under any given conditions. [citation ...
If the eccentricity is less than 1 then the equation of motion describes an elliptical orbit. Because Kepler's equation M = E − e sin E {\displaystyle M=E-e\sin E} has no general closed-form solution for the Eccentric anomaly (E) in terms of the Mean anomaly (M), equations of motion as a function of time also have no closed-form solution ...
Gives the equation of the center to order e 10. Morrison, J. (1883). On the computation of the eccentric anomaly, equation of the centre and radius vector of a planet, in terms of the mean anomaly and eccentricity. Monthly Notices of the Royal Astronomical Society, Vol. 43, p. 345. Gives the equation of the center to order e 12. Morrison, J ...
For a parabolic orbit this equation simplifies to = For a hyperbolic trajectory this specific orbital energy is either given by ε = μ 2 a . {\displaystyle \varepsilon ={\mu \over 2a}.} or the same as for an ellipse, depending on the convention for the sign of a .
The formula for an escape velocity is derived as follows. The specific energy (energy per unit mass) of any space vehicle is composed of two components, the specific potential energy and the specific kinetic energy. The specific potential energy associated with a planet of mass M is given by =
Inversely, for calculating the distance where a body has to orbit in order to have a given orbital period T: a = G M T 2 4 π 2 3 {\displaystyle a={\sqrt[{3}]{\frac {GMT^{2}}{4\pi ^{2}}}}} For instance, for completing an orbit every 24 hours around a mass of 100 kg , a small body has to orbit at a distance of 1.08 meters from the central body's ...