Integration of the velocity in orbit is not easy. An easier method uses a reference circle center on the center of the elliptical orbit with a radius equal to the semi-major axis. The construction is shown here.

Equation of the Ellipse

** x ^{2}/a^{2} + y^{2}/b^{2} = 1**

with origin at the center (not the focus) of the ellipse.

with e = eccentricity of the elliptical orbit.

so....

Equation of the reference circle is

** x ^{2}/a^{2} + y^{2}/a^{2} = 1**

If the position of the orbiting body is projected onto the reference
circle as shown, the angular position of the reference position is
called the eccentric anomaly, **E**. This parameter can be related
to the area swept out by the radial line from the central body (focus
of the elliptical orbit) to the orbiting body. The
reference position when time = zero will be when the body is at
periapsis.

The angular position of the body from periapsis as measured about
the focus is the true anomaly,
**n**.

Note that the y value in orbit is just equal to **b/a = √(1 - e ^{2})**
of the y value of the reference point on the reference circle.

Derivation of the relationship of the areal area to the eccentric anamoly, E, procedes as follows

- Determine the area swept out by the radial line from the center to the object projection point on the reference circle (call it the reference area)
- Reduce the reference area by the ratio of y values of the ellipse to the circle
- Subtract off the extra triangular area due to the reference being to the center rather than the focus.

The diagram to the right can be used for reference: The reference area after reduced is just the sum of the green, orange and yellow areas. The areal area is the orange plus yellow area. The green area has to be subtract away. (π = 3.1415926535..)

Area 1 = Reference Area = πa^{2} E/2π = E a^{2}/2

Area 2 = Reference Area reduce by the ratio of ellipse height to
reference circle height = green + orange + yellow regions = E a^{2}√(1
- e^{2})/2

Area 3 = Area to be subtracted from Area 2 to give the Areal Area =
green region = triangle area = 1/2 base x height = a sin(E) √(1 - e^{2})
e a/2

Areal Area = Area 2 - Area 1 = E a^{2}√(1 - e^{2})/2
- a^{2} e sin(E) √(1 - e^{2})/2 = [a^{2}
√(1 - e^{2})/2][E - e sin(E)]

Note that the Areal Area has a relatively simple relationship with
Eccentric Anomaly compared to that for the True Anomaly (an elliptical
integral). This areal area is proportional to time. When the area
equals the area of the ellipse, then the time is the period of the
orbit, P.

Areal Area = K t = [a^{2} √(1 - e^{2})/2][E - e
sin(E)]

Where t is time from periapsis and K is a proportionality constant to
be determined

When E = 2π then Areal Area = πab = πa^{2} √(1 - e^{2})

The constant, K, can now be evaluated: K P = [a^{2} √(1 -
e^{2})/2][2π] since sin(2π) = 0

but the area of the ellipse is the same area = πab = πa^{2}
√(1 - e^{2})

So K = πa^{2} √(1 - e^{2})/P

and leads to the identity

πa^{2} √(1 - e^{2})/P = [a^{2} √(1 - e^{2})/2][E
- e sin(E)]

**2πt/P = E - e sin(E)**

This is called Kepler's Equation and gives a direct relationship between time and position on the eccentric reference circle. It is relatively easy to determine True Anomaly from Eccentric Anomaly.

The quantity 2πt/P is called the** Mean Anomaly** and represent
by the letter, **M**.

Kepler's Equation is usually seen written as

**M = E - e sin(E)**

The x value of the position can be expressed in terms of both E and
n.

x = a cos(E)

and

x = r cos(ν) + c = r cos(ν)
+ e a

where r is the equation for an ellipse with respect to the central
body.

r = (1 - e^{2})a/(1 + e cos(ν))

So......

a cos(E) = [(1 - e^{2})cos(ν)/(1
+ e cos(ν))]a + e a

cos(E)= [(1 - e^{2})cos(ν)/(1 +
e cos(ν))] + e

OR

**cos(E) = [cos(ν) + e]/[1 + e cos(ν)]**

Solving for cos(ν) in terms of cos(E) gives:

**cos(ν) = [cos(E) - e]/[1 - e cos(E)]**