At intervals of \(15^{\circ}\) we measure its distance r from the Sun. The true anomaly is decreasing as it is moving closer, so we subtract 15, 30, 45 and 60 degrees from \(\theta_1\).
| label | distance r (AU) |
angle (degrees) |
|---|---|---|
| $$r_1$$ | $$100$$ | $$\theta_1$$ |
| $$r_2$$ | $$86.15$$ | $$\theta_1 - 15$$ |
| $$r_3$$ | $$75$$ | $$\theta_1 - 30$$ |
| $$r_4$$ | $$66.41$$ | $$\theta_1 - 45$$ |
| $$r_5$$ | $$60$$ | $$\theta_1 - 60$$ |
The equation of ellipse
$$ r= \frac{a(1-e^2)}{1+e \cos \theta} $$ has four unknowns:distance r
semimajor axis a,
eccentricity e,
true anomaly θ1.
To find the values, we need at least four simultaneous equations. Maybe even five. They are:
$$ r_1 = \frac{a(1-e^2)}{1+e \cos \theta_1 } = 100 $$ $$ r_2 = \frac{a(1-e^2)}{1+e \cos (\theta_1 - 15^\circ)} = 86.15 $$ $$ r_3 = \frac{a(1-e^2)}{1+e \cos (\theta_1 - 30^\circ)} =75 $$ $$ r_4 = \frac{a(1-e^2)}{1+e \cos (\theta_1 - 45^\circ)} = 66.41 $$ $$ r_5 = \frac{a(1-e^2)}{1+e \cos (\theta_1 - 60^\circ)} = 60 $$We can start by \(r_1 \div r_3\) $$ \frac{100}{75}= \frac{1 + e \cos (\theta_1 -30^o)}{1+ e \cos \theta_1} $$ with the use of $$ \cos(A-B) = \cos A \cos B + \sin A \sin B $$ we arrive with Eq A: $$ 1+ 1.4 e \cos \theta_1 = 1.5 e \sin \theta_1 $$
Similarly \(r_3 \div r_5 \), we get Eq B:
$$ 1+2.33 e \cos \theta_1 = 0.96 e \sin \theta_1 $$Dividing Eq A with Eq B and rearranging, eventually we get:
$$ e \cos \theta_1 = -0.25 $$Substituting back into r1:
$$ r_1 = \frac{a(1-e^2)}{1+e \cos \theta_1 } = 100 $$ $$ a(1 -e^2) = 75 $$Substituting \(e \cos \theta_1= -0.25\) into r4 and combining them
There are two possible values of θ1:
$$ \theta_1 = 120^{\circ}, 300^{\circ} $$.The value that makes sense is $$ \theta_1 = 120^{\circ} $$
With θ1 known, everything else falls into place. We can use r1 and r3
$$ r_1 = \frac{a(1-e^2)}{1+e \cos 120^{\circ} } = 100 $$ $$ r_3 = \frac{a(1-e^2)}{1+e \cos 90^{\circ} } = 75 $$ and find that eccentricity is $$ e = 0.5$$ and semimajor axis is $$ a = 100 $$With a and e known, we have described the elliptical orbit of the object. $$ r = \frac{75}{1+ 0.5 \cos \theta } $$
Periapsis/Perihelion is at θ= 0, giving us rper = 50 A.U.
Apoapsis/Aphelion is at θ = 180o, giving us rap = 150 A.U.The semi-minor axis b is obtained $$ e^2 = 1 - \frac{b^2}{a^2} $$ $$ b =\sqrt{ 100^2( 1 - 0.5^2 )} $$ $$ b = 86.6 \mathrm{ A.U.} $$