Area of an Elliptic Sector

Constrained only by gravity and having insufficient energy to fly free of its companion, a low-mass body traces an ellipse about the much more massive body that attracts it. We select two points on that elliptic orbit, call them α and β, and ask how much time the small body takes in going from one point to the other. We can calculate an answer to that question by way of Johannes Kepler’s second law of planetary motion.

A form of the law pertaining to conservation of angular momentum, Kepler’s second law tells us that a straight line extending from the prime focus of the ellipse (the point occupied by the large gravitating body) to the small body sweeps over equal areas within the ellipse in equal intervals of time. The points α and β and the prime focus f define three lines (an elliptic arc and two straight lines, from f to α and from f to β) that bound an elliptic sector. If we can calculate the area enclosed within that sector and divide it by the entire area enclosed within the ellipse, then by Kepler’s law we will have calculated the time that the small body takes to traverse the elliptic arc, expressed as a fraction of the time the small body takes to traverse the whole ellipse.

We can calculate the area of an elliptic sector by exploiting the fact that we can represent an ellipse as the result of applying an affine transformation to a circle. We can calculate the area of a circular sector easily and we can use that calculation as the basis for calculating the area of an eccentric circular sector, a sector whose vertex does not coincide with the circle’s center. We can then apply the affine transformation to the result and then convert the angles of the circular sector into the azimuths of the elliptic sector.

Area of an Eccentric Circular Sector

In Figure 1 the letters A and B on the circle label the
points that will become the points
α and
β
on the ellipse when we apply the affine transformation to the circle. We
represent the angle AOf with ψ_{A}
and the angle BOf with ψ_{B}
and measure those angles in radians.

drawn with Geogebra (see appendix)

(Fig 1)

We then calculate the area of the circular sector AOB as

(Eq’n 1)

in which R represents the radius of the circle. We now want to calculate the area of the eccentric circular sector AfB. Both sectors include the area between the arc AB and the chord AB, so we set that area aside and concentrate our attention on the triangles AOB and AfB.

One of the first theorems that we learn in our study of plane geometry tells us that two triangles that have the same base and whose vertices touch the same line drawn parallel to the base contain the same area. In order to exploit that theorem we draw on Figure 1 the line D, extending the chord AB; the line E, parallel to D and passing through the origin O; and the line F, parallel to D and passing through the focus f. Next we draw a line perpendicular to D such that it passes through O and cuts the chord AB at its midpoint, the point we label C. Then we extend that line OC to cross the line F at g.

We have the altitude of triangle AOB as h, the length of the line OC. We have the lengths of lines OA and OB as equal to R, the radius of the circle, so we have

(Eq’n 2)

We also have the length of chord AB as

(Eq’n 3)

The area covered by the triangle AOB thus corresponds to

(Eq’n 4)

Applying the trigonometric formula sinα=2sin(α/2)cos(α/2), we convert that equation into the more convenient form

(Eq’n 5)

With that information we can determine the area enclosed within the segment defined by the chord AB and the arc AB. We simply subtract Equation 5 from Equation 1 and obtain

(Eq’n 6)

Assume the existence of the triangle AgB (not drawn out in Figure 1). We calculate its area as

(Eq’n 7)

To calculate the value of δ we take eR as the length of the hypotenuse of the right triangle Ogf. The length of the side Og equals eR multiplied by the sine of the angle Ofg. By simple inspection of Figure 1 we see that

(Eq’n 8)

Applying the appropriate trigonometric identity thus gives us

(Eq’n 9)

With that result we get Equation 7 as

(Eq’n 10)

But triangles AgB and AfB have the same base and their vertices both lie on the same line drawn parallel to the line AB, so, by the theorem mentioned above, they must have the same area.

We now calculate the area of the eccentric circular sector
AfB by adding S_{3} and S_{4} together. Adding Equations 6 and
10 then gives us

(Eq’n 11)

in which I have made use of the fact that the first term in Equation 10 negates the second term in Equation 6.

Area of an Elliptic Sector

We now imagine transforming the eccentric circular sector AfB into an elliptic sector αfβ by applying to Figure 1 the affine transformation that converts the circle into an ellipse of eccentricity e and with a semi-major axis of length a=R.

drawn with Geogebra (see appendix)

(Fig. 2)

We accomplish that transformation by shrinking all distances from the x-axis
in the y-direction by the ratio b/a=(1-e^{2})^{½}. That
transformation diminishes all areas on the diagram in the same proportion.

We confirm the truth of that statement to mathematics by representing the calculation of the area contained within any closed curve or array of lines as a Riemannian sum. Imagine slicing our eccentric circular sector into strips with minuscule width dx, slicing parallel to the y-axis. Each strip contains an area equal to the product of its length, L(x), and its width. When we apply the affine transformation, the length of each strip shrinks in accordance with the proportion above but the strip’s width remains unchanged, so we have the new area of the strip as

(Eq’n 12)

When we add those contributions together to calculate the total area enclosed within the figure, we can simply factor out the diminution factor (the square root) before carrying out the summation, which means that the area of the transformed figure equals the area of the untransformed figure multiplied by that factor. So to calculate the area of the elliptic sector we need only multiply the area of the corresponding eccentric circular sector by the diminution factor, which gives us

(Eq’n 13)

Now we can calculate the time t that an orbiting body on
the ellipse takes to go from
α
to β.
We simply divide S_{6} by the total area enclosed by the ellipse,

(Eq’n 14)

and then multiply the quotient by the orbital period P. We then obtain

(Eq’n 15)

In order to use that equation we must confront the fact that we do not directly use the eccentric anomaly, ψ, in astrogation: we use azimuth, because that’s what we measure directly or derive from our specifications of where the trajectory must go (as in the following example). To bridge the gap between our astrogator’s input and Equation 15, then, we need a transformation equation that will convert azimuth to eccentric anomaly. To that end we need two separate descriptions of some feature on our diagram, one expressed in terms of the eccentric anomaly only and the other as a function of the azimuth only. For that feature we take the distance between the origin of the coordinate grid and some arbitrary point on the x-axis. For clarity in our derivation we want the cosines of both the eccentric anomaly and of the azimuth to have positive values, so we want to determine the distance between the origin and the projection of the point K or κ (kappa) on the x-axis (see Figure 2).

As a function of the eccentric anomaly we have that distance as

(Eq’n 16)

As a function of the azimuth we calculate that distance by adding the
distance between the origin and the focus (ea) to the distance from the focus to
the projection of the point kappa onto the x-axis (rcosθ_{κ}):

(Eq’n 17)

In the second step of that derivation I have replaced the radial distance r
between the focus and the point kappa by applying the standard orbit equation
(see Equation 19 below). Equating those two representations of x_{K} and
x_{κ}
and dividing out the length of the semi-major axis gives us

(Eq’n 18)

We could use that equation to rewrite Equation 15 as a function of the azimuths of the ends of our elliptic arc, but we would get a highly complicated function that would greatly ease the making of errors. Alternatively, we can use Equation 16 to translate azimuths into the equivalent eccentric anomalies and then use those in Equation 15.

Example: An Enhanced Hohmann Trajectory

Let’s consider sending a spaceship from Earth to Mars. On the conventional Hohmann transfer ellipse (with perihelion touching Earth’s orbit and aphelion touching Mars’ orbit), the transfer time spans a little over 0.7 year (roughly 8.4 months). We would like to take less time, so let’s assume that we have available to our ship sufficient rocket power to manifest some rather large delta-vees and thus put our spaceship onto a trajectory that coincides with an ellipse that has its perihelion 0.5 AU from the sun’s center and its aphelion 4.0 AU from the sun’s center, somewhere on the outermost fringe of the Asteroid Belt. How much time will elapse as our spaceship goes from Earth to Mars on an arc of that ellipse?

From the transfer orbit’s perihelion and aphelion we
determine the orbit’s other parameters. For the length of the semi-major axis we
calculate a=(1/2)(0.5+4.0)=2.25 AU. We then get the eccentricity as
e=(2.25-0.5)/2.25=0.7778. Applying Kepler’s third law of planetary motion (P^{2}/a^{3}=1yr^{2}/1AU^{3})
gives us the period as P=3.375 years.

We then use the orbit equation,

(Eq’n 19)

to calculate the azimuths of the points of Earth departure and Mars arrival. For departure from Earth we have r=1.00 AU, so we get

(Eq’n 20)

For arrival at Mars we have r=1.52 AU, so we get

(Eq’n 21)

Next we translate those azimuths into the equivalent eccentric anomalies by way of Equation 18. We get

(Eq’n 22)

and

(Eq’n 23)

And then we plug those angles into Equation 15, using radians in the first
term and using either degrees or radians (depending upon what our calculators
take more readily) in the second term. We thus calculate **t=0.1628 year (59.5
days)**, taking over six months off the Hohmann transfer time.

As for the delta-vees needed to put a spaceship onto and off of that high-speed trajectory, those provide the subject of another essay.

Appendix: Geogebra

I created the two figures in this essay with GeoGebra, a
free program that you can find on the Internet at
__www.geogebra.org.__
The program was created by Judith Hohenwarter and Markus Hohenwarter and posted
on the Internet for public use. The program consists of a manual, an exercise
book, and the interactive mathematics program itself, all of which you can copy
into your computer (indeed, the creators encourage you to do so). This is a
great program; it’s easy to use; and I recommend it to anyone who needs to
create mathematical diagrams for their Internet files.

gabh