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If you are orbiting below the surface of a planet, and ignore drag from rocks (or orbit in a vacuum tunnel), what is the shape of the orbit? Usually, the proportionality law for gravity is $\frac{1}{r^2}$, but in this case, it scales with the radius $r$, increasing in strength when you move away from the centre. The potential energy is simple to calculate, as is the angular momentum, but the overall shape is not completely obvious to me. From simulating this using time increments, it seem to me that the shape is an ellipse, with the centre of the planet as the centre (not focus). However, that is just speculation on my part as such a method does not give any reliable results.

Simplifications made: Uniform density, planet not rotating, orbiting body's mass negligible to main, and orbiting object not influenced by other forces like drag. Of course I also appreciate answers covering those factors too, and also how an object orbiting both inside and outside the planet behaves. But that is just a bonus

Example of an orbit with gravity proportional to $r$:

enter image description here

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  • $\begingroup$ I had asked a similar question: math.stackexchange.com/questions/834143/… $\endgroup$ – HopDavid Jan 2 '16 at 22:34
  • $\begingroup$ @ HopDavid But is is an ellipse? $\endgroup$ – Hohmannfan Jan 2 '16 at 22:37
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    $\begingroup$ @Hohmannfan I don't know. $\endgroup$ – HopDavid Jan 2 '16 at 23:55
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – TildalWave Jan 3 '16 at 17:14
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You are right, it's a perfect ellipse with the center of the planet in the geometrical center of the ellipse.

A little physics:

The force on a body with mass $m$ in the gravitational field of a planet with radius $R$ and gravity $g$ at its surface is:

$$\vec{F}(\vec{r})=\begin{cases}-m\cdot g\cdot \frac{\vec{r}}{R} &\text{for} &|\vec{r}|\le R\\ -m\cdot g\cdot R^2\cdot\frac{\vec{r}}{|\vec{r}|^3} &\text{else}\end{cases}$$

Outside of the planet, you have to solve the differential equation

$$m\ddot{\vec{r}}=-m\cdot g\cdot R^2\cdot\frac{\vec{r}}{|\vec{r}|^3}$$

which is a bit difficult due to the $\frac{1}{|\vec{r}|^3}$. (In fact, one uses other approaches)

However, inside the planet, it's simply

$$m\ddot{\vec{r}}=-m\cdot g\cdot \frac{\vec{r}}{R}$$

Or with $\vec{r}=\begin{pmatrix}x\\y\end{pmatrix}$ in the plane of the moving body:

$$\begin{align}\ddot{x}(t)&=-\frac{g}{R}x(t) \\ \ddot{y}(t)&=-\frac{g}{R}y(t)\end{align}$$

This is are two independent linear differential equations, the solution is just

$$\vec{r}(t)=\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}A\sin(\omega t+\varphi)\\B\sin(\omega t+\psi)\end{pmatrix}$$

This is an ellipse with semi-axes A and B. $\omega=\sqrt{g/R}$ defines the period

$$T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{ R}{g}}$$

Notes:

  • Of course, this assumes the body is always inside the planet, i.e. $A,B\le R$. It would be interesting to see trajectories where the body leaves the planet.
  • It's remarkable that the period is constant for a given planet and does not depend on the orbit parameters.
  • If you set e.g. $A=0$ and $B=R$, you get the special case @hopDavid mentioned in his comment
  • If you have trouble to understand / imagine this: The equations for a pendulum are identical (with $\omega=\sqrt{g/l}$), so a pendulum and the body behave the same way.
  • Like for normal orbits, body and planet will both have an elliptical trajectory with the geometrical centers in the barycenter.
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