Answer to Question #94538 in Astronomy | Astrophysics for Chris Harriss

Question

Prove that in a properly elliptic orbit, the angle between r ⃗ and v ⃗ is π/2 only at the apogee and perigee.

Clue: Do so by showing v ⃗⊥r ⃗ if and only if ε≠0, then τ is maximized (apogee) or minimized (perigee)when θ=0 or π

Answer 1

if $\vec{r}$ and $\vec{v}$ are the position and velocity vectors, assume that we put our elliptical orbit on an xOy plane with the semi-major axis $a$ along x-axis and semi-major axis $b$ along the y-axis. the center of the ellipse is in the origin of the coordinate system.

We know that the velocity vector is always directed along the tangent to the ellipse. Thus, we need to find where the position vector is located so that it is perpendicular to the tangent.

Write an equation of ellipse:

$$\Big(\frac{x}{a}\Big)^2+\Big(\frac{y}{b}\Big)^2=1.$$

Express $y$:

$$y=b\sqrt{1-\Big(\frac{x}{a}\Big)^2}.$$

The foci are located on either side of the origin. Their coordinates c can be found from:

$$c^2=a^2-b^2$$

At some point $S\space (m, n)$ lying on the ellipse the equation of a tangent (velocity) is

$$b^2mx+a^2ny=a^2b^2,\\ y_v=-\frac{b^2m}{a^2n}x+\frac{b^2}{n}.$$

The position vector with its beginning in the either of foci (but consider the right one located at x=c and y=0) and its end at the same point $S$ has equation :

$$y_r=k(x-c)=\frac{n-0}{m-c}(x-c)=\frac{n}{m-c}(x-c).$$

Now, since we have equations for velocity and position, and we know that the angle between them should be 90 degrees, find the coefficients before x in both equations which suffice the condition of perpendicularity, i.e. plug such m and n, which correspond to coordinates of perigee $P\space(a,0)$ and apogee $A\space(-a,0)$:

$$\text{tan}\theta_{[v\space\text{at P}]}=\lim_{n\to0}\Big(-\frac{b^2a}{a^2n}\Big)=-\infty,\\ \space\\ \text{tan}\theta_{[r\space\text{at P}]}=\frac{0}{a-c}=0,$$

or $\theta_{[v\space\text{at P}]}=-90^\circ,\space\theta_{[r\space\text{at P}]}=0^\circ.$ At the perigee velocity and position vectors are perpendicular.

$$\text{tan}\theta_{[v\space\text{at A}]}=\lim_{n\to0}\Big(-\frac{b^2(-a)}{a^2n}\Big)=\infty,\\ \space\\ \text{tan}\theta_{[r\space\text{at A}]}=\frac{0}{-a-c}=0,$$

or $\theta_{[v\space\text{at A}]}=90^\circ,\space\theta_{[r\space\text{at A}]}=180^\circ.$ At the apogee these vectors are perpendicular also.

Indeed, if we take a difference between the slopes of velocity and position vector lines, and substitute such m and n, which are defined by ellipse equation, we will never get values of d equal to negative or positive infinities unless m and n are the points of perigee and apogee:

$$d=\frac{n}{c-m}-\frac{b^2m}{a^2n}.$$

Here is an example with black velocity line and blue position line: