Por que as órbitas são elípticas em vez de circulares?

Por que os planetas giram em torno de uma estrela em uma órbita elíptica específica com a estrela em um de seus focos? Por que a órbita não é um círculo?

Suponha que o planeta tenha uma massa desprezível em comparação com a estrela, que seja esférica simétrica (de acordo com a lei da gravitação de Newton, mas isso normalmente acontece com uma aproximação muito boa) e que não há forças além da gravidade entre elas. . Se a primeira condição não se mantiver, a aceleração de cada uma será em direção ao baricentro do sistema, como se o baricentro os estivesse atraindo para uma força gravitacional com uma certa massa reduzida, de modo que o problema é matematicamente equivalente.

Pegue a estrela para estar na origem. Pela lei da gravitação de Newton, a força é , onderé o vetor para o planeta,mé a sua massa eμ=GMé o parâmetro gravitacional padrão da estrela.$\mathbf{F} = -\frac{m\mu}{r^3}\mathbf{r}$$\mathbf{r}$$m$$\mu = GM$

Leis de Conservação

Como a força é puramente radial , o momento angular L = r × p é conservado: ˙ L = $(\mathbf{F}\parallel\mathbf{r})$$\mathbf{L} = \mathbf{r}\times\mathbf{p}$ Se a velocidade inicial é diferente de zero e a estrela está na origem, então em termos de posição e velocidade iniciais, a órbita deve ser confinada ao plano de todos os pontos com vetoresxda origem que satisfaçamL⋅x=0. Se a velocidade inicial é zero, o movimento é puramente radial e podemos tomar qualquer um dos infinitos planos que contêm o baricentro e a posição inicial.

$$\dot{\mathbf{L}} = \frac{\mathrm{d}}{\mathrm{d}t}\left(\mathbf{r}\times\mathbf{p}\right) = m(\dot{\mathbf{r}}\times \dot{\mathbf{r}}) + \mathbf{r}\times\mathbf{F} = \mathbf{0}\text{.}$$ $\mathbf{x}$$\mathbf{L}\cdot\mathbf{x} = 0$

A energia orbital total é dada por onde a primeira parte do termo é a energia cinética e o segundo termo é a energia potencial gravitacional do planeta. Sua conservação, assim como o fato de invocar a energia potencial correta, podem ser comprovadas pelo teorema fundamental do cálculo para integrais de linha.

$$\mathcal{E} = \frac{p^2}{2m} - \frac{m\mu}{r}\text{,}$$

Defina o vetor Laplace-Runge-Lenz como Também é conservado: ˙

$$\mathbf{A} = \mathbf{p}\times\mathbf{L} - \frac{m^2\mu}{r}\mathbf{r}\text{.}$$ $$\begin{eqnarray*} \dot{\mathbf{A}} &=& \mathbf{F}\times\mathbf{L} + \mathbf{p}\times\dot{\mathbf{L}} - \frac{m\mu}{r}\mathbf{p} + \frac{m\mu}{r^3}(\mathbf{p}\cdot\mathbf{r})\mathbf{r}\\ &=& -\frac{m\mu}{r^3}\underbrace{\left(\mathbf{r}\times(\mathbf{r}\times\mathbf{p})\right)}_{(\mathbf{r}\cdot\mathbf{p})\mathbf{r} - r^2\mathbf{p}} - \frac{m\mu}{r}\mathbf{p} + \frac{m\mu}{r^3}(\mathbf{p}\cdot\mathbf{r})\mathbf{r}\\ &=& \mathbf{0}\text{.} \end{eqnarray*}$$

Finalmente, vamos também pegar , que tem as mesmas unidades que r , e como L ⋅ f = 0 , ela se encontra ao longo do plano orbital. Como é um vetor conservado dimensionado por um escalar conservado, é fácil mostrar que f também é conservado, desde que E ≠ 0 .$\mathbf{f} = \mathbf{A}/(m\mathcal{E})$$\mathbf{r}$$\mathbf{L}\cdot\mathbf{f} = 0$$\mathbf{f}$$\mathcal{E}\neq 0$

Simplificando

Empregando o produto triplo vetorial, podemos escrever cuja norma ao quadrado é fácil de arrancar: E2| f-r| 2=(E+m

$$\begin{eqnarray*} \frac{1}{m}\mathbf{A} &=& \frac{1}{m}\left[p^2\mathbf{r}-(\mathbf{p}\cdot\mathbf{r})\mathbf{p}\right] -\frac{m\mu}{r}\mathbf{r}\\ &=& \left(\mathcal{E}+\frac{p^2}{2m}\right)\mathbf{r} - \frac{1}{m}\left(\mathbf{p}\cdot\mathbf{r}\right)\mathbf{p}\\ \mathcal{E}(\mathbf{f}-\mathbf{r}) &=& \left(\frac{p^2}{2m}\right)\mathbf{r} - \frac{1}{m}\left(\mathbf{p}\cdot\mathbf{r}\right)\mathbf{p}\text{,} \end{eqnarray*}$$ $$\mathcal{E}^2|\mathbf{f}-\mathbf{r}|^2 = \left(\mathcal{E} + \frac{m\mu}{r}\right)^2r^2\text{,}$$ $\mathcal{E}$

Por que elipses?

$\mathcal{E}$$\mathcal{E}<0$$|\mathbf{f}-\mathbf{r}| = -\mathcal{E}^{-1}\left(\mathcal{E}r + m\mu\right)$

$$|\mathbf{f}-\mathbf{r}| + |\mathbf{r}| = -\frac{m\mu}{\mathcal{E}}\text{,}$$ which defines an ellipse with foci $\mathbf{0},\,\mathbf{f}$ and major axis $2a=-m\mu/\mathcal{E}$.

Why Not Circles?

The circle is a special case where the foci are the same point, $\mathbf{f} = \mathbf{0}$, which can be restated as

$$\mathcal{E} = -\frac{1}{2}\frac{m\mu}{r} = -\frac{p^2}{2m}\text{.}$$ In other words, circular orbits require the orbital energy to be the negative of the kinetic energy. This is possible, but almost certain not to hold exactly. Since any values of $\mathcal{E}<0$ are allowed for bound orbits, there are many more ways to have elliptic orbits. (Although some of them would actually crash because the star and planet have positive size.)

Note that hyperbolic orbits have $\mathcal{E}>0$, and we can still find the foci using the above method, though being careful with the signs. For $\mathcal{E}=0$, the second focus $\mathbf{f}$ is undefined because this is a parabolic orbit, and parabolas only have one focus within a finite distance from the center.

Additionally, the eccentricity vector $\mathbf{e} = \mathbf{A}/(m^2\mu)$ is an alternative choice for the LRL vector; as the name suggests, its magnitude is the orbital eccentricity.

It is posible for a planet to have a circular orbit, a circle, after all, is an ellipse where both foci are in the same place; this is known as having an eccentricity of 0. Eccentricity is defined in the following way:

$$e = \frac{r_{a} - r_{p}}{r_{a} + r_{p}}$$ where $r_{a}$ is the apoapsis (farthest point in the orbit from the center of mass), and $r_{p}$ is the periapsis (the closest distance). Just to build some intuition here, if the apoapsis is twice the distance of the periapsis, the eccentricity will be $e=0.333$.

From all the planets of the solar system, Venus, with an eccentricity of 0.007 has the most circular orbit.

As to why all orbits aren't round, it comes down to kinetic energy. The kinetic energy is proportional to the square of the speed. In the orbital plane and in polar coordinates about the star, we can decompose this is into a combination of radial velocity $\dot{r}$ and angular velocity $\dot{\phi}$:

$$v^2 = \dot{r}^2 + r^2\dot{\phi}^2\text{.}$$ Since circles have constant radii, for the orbit to be circular around the star, the planet's radial velocity must be exactly zero. Additionally, the angular speed must be such that the centrifugal force in the corotating frame exactly balances the gravitational force--a little more or a little less, the imbalance will change the radial velocity, spoiling the circle.

Given the fact that velocities vary for a large number of reasons, it's no wonder that only a few orbits end up being circular, and considering that actual orbits change with time, we know they can't stay this way for long.

If you're looking for a mathematical proof, this link shares some details about it.

Here's an image showing eccentricity of some bodies in the solar system extracted from here:

I always prefer answers which try to avoid any formula and reply on argumentation instead. Concerning the part of the question why not all orbits are circular, an argumentation would be like this:

Consider a stationary star and a moving planet. For each impulse the planet can have, a curve for its further movement can be predicted. If this impulse is directed exactly orthogonal to the line from the star to the planet, and if the velocity has the exact amount, then this curve of movement can be an exact circle.

But for every deviance of this one exact impulse, the resulting curve cannot be a circle:

• If the speed is too low, the planet will fall towards the star (in the extreme case of an impulse of zero, this fall will be in a straight line).
• If the speed is too high, the planet will gain distance from the star (similar to a slingshot).
• If the impulse is not directly orthogonal to the line to the star, the first movement will move the towards or from the star, so again the curve will not be a circle.

So, one can simply argue, a circle is a very special case for the curve a planet can take around a star.