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Problems on Kepler's First Law

Problems on Kepler's First Law

Problems on Kepler's First Law

Problems on Kepler's First Law

Problems on Kepler's First Law

Problems on Kepler's First Law

Problem : Calculate the eccentricity of an ellipse with one focus at the origin and the other at $(-2k, 0)$, and semimajor axis length $3k$.

It is easiest if we draw a diagram of the situation:
Ellipse with semimajor axis length $3k$.
We need to calculate $b$, the semiminor axis length. This is given by applying Pythagoras' theorem to the right triangle: $ b = \sqrt{(3k)^2 - k^2} = 2\sqrt{2}k$ The eccentricity is then given by: \begin{equation} \epsilon = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{8}{9}} = \frac{1}{3} \end{equation}

Problem : For an ellipse with its major axis parallel to the $x$-direction and its rightmost focus at the origin, derive the position of the other focus in terms of its eccentricity $\epsilon$ and $k$, where $k$ is defined as $k = a(1- \epsilon^2)$.

The $y$-coodinate of the other focus is the same--zero. The other focus is a distance $2\sqrt{a^2 – b^2}$ in the negative x-direction, so the coordinates are $(-2\sqrt{a^2-b^2},0)$. But $\epsilon = \sqrt{1 - \frac{b^2}{a^2}}$ so we can write $-2\sqrt{a^2-b^2} = -2a\sqrt{1 – \frac{b^2}{a^2}} = -2a\epsilon$. We are given that $k = a(1 - \epsilon^2)$, so $a = \frac{k}{1 - \epsilon^2}$, and $- 2a\epsilon = \frac{-2k\epsilon}{1 – \epsilon^2}$. Thus the coordinate of the other focus is $(\frac{-2k\epsilon}{1\epsilon^2},0)$.

Problem : The general equation for orbital motion is given by: \begin{equation} x^2 + y^2 = k^2 – 2k\epsilon x + \epsilon^2 x^2 \end{equation} Where the $k$ is the same $k$ as in the last problem: $k = a(1-\epsilon^2) = \frac{L^2}{GMm^2}$. Show that when $\epsilon = 0$, this reduces to an equation for a circle. What is the radius of this circle?

Clearly, when $\epsilon = 0$ the second and third terms on the right hand side go to zero, leaving: \begin{equation} x^2 + y^2 = k^2 \end{equation} This is the equation for a circle of radius $k$. Since $\epsilon$ is dimensionless and $k = a(1 - \epsilon^2)$, $k$ has the correct units of distance.

Problem : Derive the formula for the area of an ellipse by integration.

The equation for an ellipse is given by: \begin{equation} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \end{equation} Solving for $y$ yields, $y = b\sqrt{1 - \frac{x^2}{a^2}}$. We can take an ellipse centered at the origin and look at the area in the first quadrant. Because of symmetry, the total area will be four times this. Thus, \begin{equation} \frac{A}{4} = b\int_0^a \sqrt{1 – \frac{x^2}{a^2}} dx \end{equation} We can trigonometric substitution, $x = a\sin\theta$, then $dx = a\cos\theta d\theta$. The limits of integration can now be chosen as $\theta = 0$ and $\theta = \pi/2$. Then: \begin{equation} \frac{A}{4} = b\int_0^{\pi/2} \sqrt{1 - \sin^2\theta} a\cos\theta d\theta = ab\int_0^{\pi/2} = cos^2\theta d\theta = \frac{\pi ab}{4} \end{equation} So the total area is $A = \pi ab$.

Problem : Prove that for a point on an ellipse, the sum of the distances to each foci is a constant.

We can say without loss of generality that the ellipse is centered at the origin and then the coordinates of the foci are $(\pm\sqrt{a^2 – b^2},0)$. Then a point on the ellipse with coordinates $(x,y)$ will be a distance: \begin{equation} ((x-\sqrt{a^2-b^2})^2 + y^2)^{1/2} \end{equation} from one foci and distance: \begin{equation} ((x + sqrt{a^2-b^2})^2 + y^2)^{1/2} \end{equation} from the other focus. Thus the total distance is just the sum: \begin{equation} D= ((x-\sqrt{a^2-b^2})^2 + y^2)^{1/2} + ((x+\sqrt{a^2-b^2})^2 + y^2)^{1/2} \end{equation} But the equation for an ellipse tells us that $y^2 = b^2(1 - \frac{x^2}{a^2})$, and we can substitute this in: \begin{equation} D = ((x-\sqrt{a^2-b^2})^2 + b^2(1 -\frac{x^2}{a^2}))^{1/2} + ((x-\sqrt{a^2-b^2})^2 + b^2(1 -\frac{x^2}{a^2}))^{1/2} \end{equation} We can then square this to find: \begin{equation} D^2 = 2x^2 + 2(a^2 – b^2) +2b^2(1 - \frac{x^2}{a^2}) - 2\sqrt{(x-\sqrt{a^2-b^2})^2 + b^2(1 -\frac{x^2}{a^2}))^2 – 4x^2(a^2-b^2)} \end{equation} Expanding out the terms under the square root we find: \begin{equation} D^2 = 2x^2 + 2a^2 – 2b^2 + 2b^2 - \frac{2b^2x^2}{a^2} – 2x^2 + 2a^2 + \frac{2b^2x^2}{a^2} = 4a^2 \end{equation} Therefore the total distance is independent of the coordinates $x$ and $y$, and is $2a$, as we would expect, since it is obvious that the distance has to be this at the narrow endpoints of the ellipse.