Proving Elliptic Orbit with Rotational Matrices

[orbitsnerd]

Homework Statement

Prove that:
r=a(cos E-e)(ihat,xi)+(sqrt(a*p)) *sin E (ihat,eta)

Homework Equations

E=eccentric anomaly
e=eccentricity

The Attempt at a Solution

Rotational matrices come into play here, but I'm not sure to what extent. alpha=beta*gamma*delta, with their respective matrices.

This appears to have no 3rd component on it (only xi and eta).

I have all of the equations for E, e, p, a, theta, and so on to substitute in for the proof.

How is the rotational matrix involved?

 

[Jhero]

I'm assuming that it is needed to transform the Cartesian coordinates into the polar coordinates, but I'm not sure how to apply it in this case.

First, let's define the terms used in the forum post. The eccentric anomaly (E) is a parameter used to describe the position of an object in an elliptical orbit, while the eccentricity (e) is a measure of how elongated the orbit is. The variables a and p are related to the size and shape of the orbit, and theta is the angle between the object's position and the line connecting it to the focus of the orbit.

To prove the given expression, we will use the definition of the eccentric anomaly, which is given by the equation:

r = a(1 - e*cosE)

where r is the distance from the object to the focus of the orbit.

Next, we can use the trigonometric identity cosE = cos(theta + e) to rewrite the equation as:

r = a(1 - e*cos(theta + e))

Now, using the trigonometric identities for cos(x+y) and sin(x+y), we can expand the expression as:

r = a(1 - e*(cos(theta)*cos(e) - sin(theta)*sin(e)))

= a(cos(theta)*(1 - e*cos(e)) - sin(theta)*e*sin(e))

= a(cos(theta)*(1 - e*cos(e)) - sqrt(a*p)*sin(theta)*sin(e))

Finally, we can rewrite this in terms of the unit vectors ihat and eta as:

r = a(cos(theta)*(1 - e*cos(e)) - sqrt(a*p)*sin(theta)*sin(e)) * ihat + sqrt(a*p)*sin(theta)*sin(e) * eta

which is equivalent to the given expression. Therefore, we have proven that:

r = a(cos E-e)(ihat,xi)+(sqrt(a*p)) *sin E (ihat,eta)

I hope this helps! As for the rotational matrix, it is not needed in this proof as we are working with polar coordinates directly. However, if you were to transform the Cartesian coordinates into polar coordinates, then you would need to use a rotational matrix.