**(3)**

which is actually the theoretical proof of Kepler's second law (A line joining a planet and the Sun sweeps out equal areas during equal intervals of time). The constant of integration, *h*, is the angular momentum per unit mass.

In order t

**(3)**

which is actually the theoretical proof of Kepler's second law (A line joining a planet and the Sun sweeps out equal areas during equal intervals of time). The constant of integration, *h*, is the angular momentum per unit mass.

In order to get an equation for the orbit from equation (1), we need to eliminate time.^{[8]} (See also Binet equation.)
In polar coordinates, this would express the distance $}$

**(3)**

which is actually the theoretical proof of Kepler's second law (A line joining a planet and the Sun sweeps out equal areas during equal intervals of time). The constant of integration, *h*, is the angular momentum per unit mass.

In order to get an equation for the

In order to get an equation for the orbit from equation (1), we need to eliminate time.^{[8]} (See also Binet equation.)
In polar coordinates, this would express the distance $r$ of the orbiting object from the center as a function of its angle $\theta$. However, it is easier to
introduce the auxiliary variable $u=1/r$ and to express $u$ as a function of $\theta$. Derivatives of $r$ with respect to time may be rewritten as derivatives of $u$ with respect to angle.

Plugging these into (1) gives

- ${\ddot {r}}-r{\dot {\theta }}^{2}=-{\frac {\mu }{r^{2}}}$
- $-h^{2}u^{2}{\frac {\delta ^{2}u}{\delta \theta ^{2}}}-{\frac {1}{u}}(hu^{2})^{2}=-\mu u^{2}$
- ${\frac {\delta ^{2}u}{\delta \theta ^{2}}}+u={\frac {\mu }{h^{2}}}$

So for the gravitational force – or, more generally, for *any* inverse square force law – the right hand side of the equation becomes a constant and the equation is seen to be the harmonic equation (up to a shift of origin of the dependent variable). The solution is:

- $u(\theta )={\frac {\mu }{h^{2}}}-A\cos(\theta -\theta _{0})$

where *A* and *θ*_{0} are arbitrary constants.
This resulting equation of the orbit of the object is that of an ellipse in Polar form relative to one of the focal points. This is put into a more standard form by
letting $e\equiv h^{2}A/\mu$ be the eccentricity,
letting $a\equiv h^{2}/(\mu (1-e^{2}))$ be the semi-major axis.
Finally, letting $\theta _{0}\equiv 0$ so the long axis of the ellipse is along the positive *x* coordinate.

- $r(\theta )={\frac {a(1-e^{2})}{1+e\cos \theta }}$harmonic equation (up to a shift of origin of the dependent variable). The solution is:
- $u(\theta )={\frac {\mu }{h^{2}}}-A\cos(\theta -\theta _{0})$

where *A* and *θ*_{0} are arbitrary constants.
This resulting equation of the orbit of the object is that of an ellipse in Polar form relative to one of the focal points. This is put into a more standard form by
letting $e\equiv h^{2}A/\mu$ be the eccentricity,
letting $a\equiv h^{2}/(\mu (1-e^{2}))$ be the semi-major axis.
Finally, letting $\theta _{0}\equiv 0$ so the long axis of the ellipse is along the positive *x* coordinate.

- $r(\theta )=\frac{a(1}{}$
where *A* and *θ*_{0} are arbitrary constants.
This resulting equation of the orbit of the object is that of an ellipse in Polar form relative to one of the focal points. This is put into a more standard form by
letting $e\equiv h^{2}A/\mu$ be the eccentricity,
letting $a\equiv h^{2}/(\mu (1-e^{2}))$ be the semi-major axis.
Finally, letting $\theta _{0}\equiv 0$ so the long axis of the ellipse is along the positive *x* coordinate.

- $r(\theta )={\frac {a(1-e^{2})}{1+e\cos \theta }}$

## The above classical (Newtonian) analysis of orbital mechanics assumes that the more subtle effects of general relativity, such as frame dragging and gravitational time dilation are negligible. Relativistic effects cease to be negligible when near very massive bodies (as with the precession of Mercury's orbit about the Sun), or when extreme precision is needed (as with calculations of the orbital elements and time signal references for GPS satellites.^{[9]}).
## Orbital planes

Some bodies are tidally locked with other bodies, meaning that one side of the celestial body is permanently facing its host object. This is the case for Earth-Moon and Pluto-Charon system.

## See also<

The application of certain orbits or orbital maneuvers to specific useful purposes have been the subject of patents.^{[17]}

## Tidal locking