Formula for Ellipse 'Change of Distance by Change of Theta'?

In summary, you need to find the change of distance $dr$ if you change $\theta$ by $d\theta$, and the relationship with $dt$ follows from Kepler's law.
  • #1
SquishyWeRRa
1
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I am working on planetary orbits in an ellipse where the sun is at the foci, not the centre of the ellipse.

I need a formula that describes the 'Change of Distance by Change of Theta angle' where Distance represents the distance between a planet to the sun (Planet is on parameter, Sun is at foci) and Theta represents the angle swept out when a planet moves from one position to the next.

E.g. Let Theta change by 1 degree, what is the change of Distance from the planet to the sun?

Since the orbit is ellipse, the change of distance by change of radius is not constant.

Furthermore, the distance is not calculated from the centre of the ellipse, but at the foci of the ellipse’

Kepler's 2nd law states that the area swept over a specific amount of time is constant.

Assuming that each sector's segment is negligible, the sector becomes a triangle

Area of triangle= 1/2 ab sin⁡C
dA=1/2 (Old Radius)(New Radius)sin⁡(dθ)
Using small angle approximation, sin(dθ)=dθ
dA/dt=(Old Radius)(New Radius)(dθ/dt)
Summation of dA is Area of ellipse πab
Summation of dt is Orbital period
πab/P=(Old Radius)(New Radius)/2 * (dθ/dt)

I know a, b and P. I can let dθ be increments of 1 degree. I need to know how New Radius changes to Old Radius with consideration of dθ. I need to find dt, which is the time taken to sweep that specific 1 degree.
Thank you
 
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  • #2
Hi SquishyWeRRa! Welcome to MHB! (Smile)

Here's how an ellipse with its foci looks like, including the relevant formulae.
\begin{tikzpicture}
\draw[gray, very thin,->] (-6,0) -- (6,-0); % x-axis
\draw[gray, very thin,->] (0,-4) -- (0,4); % y-axis
\draw[gray, very thin] (-5,-3) rectangle (5,3);
\draw[red,line width=2pt] (-4,0) -- (-4,1.8) node[right=1pt] {$\ell=\dfrac{b^2}{a}$}; % semi latus rectum
\draw[domain=-180:180,smooth,variable=\t,line width=2pt] plot ({5*cos(\t)},{3*sin(\t)});
\node at (1.2,-2) {$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2}=1$ };
\node at (-1.5,-2) {$r=\dfrac{b^2}{a - c \cos\theta}$ };
\node at (1.5,-3.5) {$(a \cos u, b \sin u)$ };
\draw (-4,0) -- (0,3) -- (4,0);
\node at (2,3pt) {c};
\node at (-4pt,1.5) {b};
\node at (2.2,1.65) {a};
\draw[<->, green] (-5,-0.3) -- (0,-0.3);
\node[green] at (-2.5,-5pt) {a};
\fill (-4,0) circle (0.1); % focus
\fill (4,0) circle (0.1); % focus
\end{tikzpicture}
It looks like you're looking for the formula:
$$r=\dfrac{b^2}{a - c \cos\theta}$$
The change in distance $dr$ if we change $\theta$ by $d\theta$ is given by the derivative.
And the relationship with $dt$ follows from Kepler's law.
 

1. What is the formula for finding the change of distance by change of theta in an ellipse?

The formula for finding the change of distance by change of theta in an ellipse is (a*b)/(a*cos(theta))^2, where a is the length of the semi-major axis and b is the length of the semi-minor axis.

2. How is this formula different from the formula for a circle?

This formula is different from the formula for a circle because it takes into account the difference in length between the semi-major and semi-minor axes. In a circle, these lengths are equal, but in an ellipse, they are different and therefore affect the calculation of change in distance as theta changes.

3. Why is the change of distance by change of theta important in an ellipse?

The change of distance by change of theta is important in an ellipse because it allows us to determine the change in distance between any two points on the ellipse as the angle (theta) between them changes. This can be useful in various applications, such as calculating the distance traveled by a planet in its elliptical orbit around the sun.

4. Can this formula be used to find the change of distance by change of theta in any ellipse?

Yes, this formula can be used to find the change of distance by change of theta in any ellipse, regardless of its size or orientation. As long as the values for the semi-major and semi-minor axes are known, the formula can be applied to any ellipse.

5. How is the change of distance by change of theta related to the eccentricity of an ellipse?

The change of distance by change of theta is related to the eccentricity of an ellipse because the eccentricity (e) is a factor in the calculation of the semi-major and semi-minor axes. The eccentricity determines the shape of the ellipse, with values closer to 0 resulting in a more circular shape and values closer to 1 resulting in a more elongated shape. Therefore, the change of distance by change of theta will be affected by the eccentricity of the ellipse.

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