# Elliptical Orbits ($$0 < e < 1$$)¶

If the eccentricity is between 0 and 1, then the radius of the orbit varies with the true anomaly. However, the magnitude of the product $$e \cos\nu$$ is never greater than one. This means that the bottom of the fraction in the orbit equation, Eq. (113), is never zero and the orbit is an elliptical shape. The minimum value of $$r$$ occurs at periapsis where $$\nu = 0$$ and the maximum value of $$r$$ is at apoapsis where $$\nu = \pi$$. The distance to periapsis and apoapsis are given by Eq. (116) and Eq. (117), respectively.

## Major and Minor Axes¶

The geometry of the ellipse is shown in Fig. 37.

Fig. 37 The geometry and definition of distances in an elliptical orbit.

Let $$2a$$ denote the total distance from periapsis to apoapsis along the apse line. Then:

(128)$a = \frac{h^2}{\mu} \frac{1}{1 - e^2}$

where $$a$$ is the semi-major axis of the ellipse. We can then write the orbit equation, Eq. (113) in terms of the semi-major axis:

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

From the definition of the parameter of the orbit in Eq. (119), we can see that:

(130)$p = a \left(1 - e^2\right)$

Then, the distance to periapsis, in terms of the semi-major axis is:

(131)$r_p = a\left(1 - e\right)$

and the distance from $$m_1$$ to the center of the ellipse is:

(132)$c = ae = a - r_p$

If $$B$$ is the point on the orbit directly above $$C$$, the center, then $$b$$ is the distance of the semi-minor axis. The semi-minor axis can then be found in terms of the semi-major axis and the eccentricity:

(133)$b = a\sqrt{1 - e^2}$

## Specific Energy¶

The specific energy of the orbit is found in terms of the semi-major axis as:

(134)$E = -\frac{\mu}{2a}$

which is the same formula as the circular orbit, with the radius equal to the semi-major axis. In addition, the specific energy depends only on the semi-major axis, and is independent of the eccentricity.

## Period¶

The period of the elliptical orbit can be found in terms of the semi-major and semi-minor axes. The area of an ellipse is given by:

(135)$A = \pi a b$

From Kepler’s second law (equal areas in equal times), given by Eq. (109), we find:

(136)$A = \frac{h}{2} \Delta t$

If $$A$$ is the complete area of the ellipse, then $$\Delta t$$ is the period $$T$$:

(137)$T = \frac{2\pi ab}{h}$

Subbing the expressions for $$a$$ and $$b$$, we find:

(138)$T = \frac{2\pi}{\sqrt{\mu}}a^{3/2}$

which is also the same formula as a circle, with the semi-major axis in the role of the radius. Note that this formula is also independent of the eccentricity. This equation represents Kepler’s third law:

Note

Kepler’s Third Law: The period of a planet is proportional to the 3/2 power of the semi-major axis of its orbit. This is shown in Fig. 38.

Fig. 38 Log-log plot of orbital period in Earth years vs orbit semi-major axis in AU of some Solar System bodies. Crosses denote values used by Kepler. Data is from NASA. Credit Cmglee CC BY-SA 4.0, via Wikimedia Commons

## Eccentricity¶

The eccentricity of the orbit can be determined if the distances to periapsis and apoapsis are known. By dividing the equations for $$r_p$$ and $$r_a$$, we find:

(139)$\frac{r_p}{r_a} = \frac{1 - e}{1 + e}$

Solving Eq. (139) for the eccentricity:

(140)$e = \frac{r_a - r_p}{r_a + r_p}$

One final interesting parameter of the orbit is the average distance from $$m_1$$ to $$m_2$$ over the orbit. This is:
(141)$\overline{r}_{\nu} = a\sqrt{1 - e^2} = b = \sqrt{r_p r_a}$
where the $$\nu$$ subscript indicates that this average is performed over all the values of the true anomaly. Thus, we can see that the average orbital radius is equal to the semi-minor axis, or the square root of the product of the periapsis and apoapsis.