$k$ here is a constant dependent on the gravitational constant, and the masses of the two objects. If I transform it into cartesian coordinates:

$$\ddot{X}(t)=-\frac{k X(t)}{\left(X(t)^2+Y(t)^2\right)^{3/2}}$$

$$\ddot{Y}(t)=-\frac{k Y(t)}{\left(X(t)^2+Y(t)^2\right)^{3/2}}$$

I can not solve this system of equations. Perhaps it would require some special functions like the elliptic function etc. Maybe I should get rid of the time dependency and represent it as an implicit function but I do not know how. I realize that solving an elliptic orbit is know, I am curious about how one would solve it in the fundamental f=ma kind of way, without any other assumption.

  • 2
    $\begingroup$ You may find this interesting: johncarlosbaez.wordpress.com/2015/03/17/… $\endgroup$
    – CDCM
    Commented Aug 22, 2017 at 22:34
  • $\begingroup$ Set $Z(t)=X(t)+iY(t)$ then the two equations becomes $\ddot{Z}(t)=-\frac{k Z(t)}{|Z(t)|^{3}}$. Next set $Z(t)=\rho(t)e^{i\phi(t)}$, and separate the real/imaginary parts, then you will obtain two equations. You can continue from there. $\endgroup$
    – mike
    Commented Aug 23, 2017 at 0:28
  • $\begingroup$ @mike this is what i ended up with $$\rho (t)^2 \left(\rho ''(t)+2 i \rho '(t) \phi '(t)-\rho (t) \left(\phi '(t)^2-i \phi ''(t)\right)\right)=-k$$an inhomogeneous nonlinear equation. I dont really know what to do now. $\endgroup$ Commented Aug 27, 2017 at 17:48
  • $\begingroup$ @grdgfgr: This is the real part of the complex ODE. What is the imaginary part? $\endgroup$
    – mike
    Commented Aug 28, 2017 at 1:19

1 Answer 1


First of all, it is best to solve the equation in polar coordinates rather than Cartesian coordinates: this works well with the fact that the force vector - and therefore the acceleration vector - is always pointing in a radial direction.

While it is possible to express the equations of motions as differential equations in time, it is not possible to obtain closed form solutions of the orbital position as a function of time, and so these time differential equations cannot be solved as they as. The strategy is to reformulate the time differential equations into "positional" differential equations, i.e. the derivative terms are differentiated with respect to the orbital angular position. Details for this are provided below.

Formulation of polar equation of motion in time

$r(t)$ and $\theta(t)$ are the polar coordinates. It is useful to be able to relate the acceleration vector to the polar coordinates and their derivatives. The position vector can be represented as:

$$\mathbf{r}(t) = \begin{bmatrix} r\cos{\theta}\\ r\sin{\theta} \end{bmatrix}$$

The acceleration is obtain by double differentiation in time:

$$\ddot{\mathbf{r}}(t) = \begin{bmatrix} \ddot{r}\cos{\theta} - 2\dot{r}\dot{\theta}\sin{\theta} - r\dot{\theta}^2\cos{\theta} - r\ddot{\theta}\sin{\theta}\\ \ddot{r}\sin{\theta} + 2\dot{r}\dot{\theta}\cos{\theta} - r\dot{\theta}^2\sin{\theta} + r\ddot{\theta}\cos{\theta} \end{bmatrix} = \left(\ddot{r} - r\dot{\theta}^2\right) \begin{bmatrix} r\cos{\theta}\\ r\sin{\theta} \end{bmatrix} + \left(2\dot{r}\dot{\theta} + r\ddot{\theta}\right) \begin{bmatrix} -r\sin{\theta}\\ r\cos{\theta} \end{bmatrix} $$

Note how $\ddot{r} - r\dot{\theta}^2$ is the radial component of acceleration and $2\dot{r}\dot{\theta} + r\ddot{\theta}$ is the tangential component. However, we know that gravity only imparts a radial acceleration on the body, and so two simultaneous equations of motion are obtained:

$$\ddot{r} - r\dot{\theta}^2 = -\frac{k}{r^2}$$

$$2\dot{r}\dot{\theta} + r\ddot{\theta} = 0$$

These equations cannot be immediately solved as they are, since both equations contain $r$ and $\theta$ terms together. However, it can be shown that the second equation amounts to the conservation of angular momentum. Multiply the second equation by $r$ to yield

$$2 r \dot{r}\dot{\theta} + r^2\ddot{\theta} = 0$$

Noting that $\frac{\mathrm{d}}{\mathrm{d}t}r^2 = 2r\dot{r}$, the equation can be re-expressed as

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(r^2\right)\dot{\theta} + r^2\frac{\mathrm{d}}{\mathrm{d}t}\left(\dot{\theta}\right) = 0 \longrightarrow \frac{\mathrm{d}}{\mathrm{d}t}\left(r^2 \dot{\theta}\right) = 0$$

This means that

$$r^2\dot{\theta} = \textrm{constant}$$

We can represent this constant quantity with $h = r^2\dot{\theta}$, and $h$ is the specific angular momentum. Now we can substituted $\dot{\theta} = h/r^2$ into the first equation of motion, yielding

$$\ddot{r} - \frac{h^2}{r^3} + \frac{k}{r^2} = 0$$

This is the polar equation of motion in time, but this form is still not sufficient in determining the shape of an orbit.

Obtaining the "positional" equation of motion

A useful step to perform is to defined a new quantity $q = \frac{1}{r}$, and substitute this into the equation of motion. In the process, we can convert time derivatives of $r(t)$ into positional derivatives of $q(\theta)$

Note how $\dot{r}$ can be expanded using the chain rule:

$$\dot{r} = \frac{\mathrm{d}r}{\mathrm{d}t} = \frac{\mathrm{d}r}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}\theta}\left(\frac{1}{q}\right)\dot{\theta} = -\frac{1}{q^2}\frac{\mathrm{d}q}{\mathrm{d}\theta}hq^2 = -h\frac{\mathrm{d}q}{\mathrm{d}\theta}$$

Likewise, $\ddot{r}$ becomes

$$\ddot{r} = \frac{\mathrm{d}\dot{r}}{\mathrm{d}t} = \frac{\mathrm{d}\dot{r}}{\mathrm{d}\theta}\frac{\mathrm{d}\theta}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}\theta}\left(-h\frac{\mathrm{d}q}{\mathrm{d}\theta}\right)\dot{\theta} = -h^2 q^2\frac{\mathrm{d}^2 q}{\mathrm{d}\theta^2}$$

Therefore, the equation of motion can be re-expressed as:

$$-h^2 q^2 \frac{\mathrm{d}^2 q}{\mathrm{d}\theta^2} - h^2 q^3 + k q^2 = 0$$


$$\frac{\mathrm{d}^2 q}{\mathrm{d}\theta^2} + q = \frac{k}{h^2}$$

This equation can be solved, and is in fact a linear differential equation with a general solution:

$$q(\theta) = A\cos{\theta} + B\sin{\theta} + \frac{k}{h^2}$$

Solving the equation of motion

$A$ and $B$ are yet to be known, but can be determined with two boundary conditions. $h$ is also unknown, and can be determined with a third boundary condition.

Let's say that $r_0$ is the initial distance of the body from the orbited planet, $u_0$ is the initial (outward) radial velocity, and $v_0$ is the initial (anticlockwise) tangential velocity. With no loss of generality, we can assume the orbiting body is initially at $\theta = 0$.

The first boundary condition is that $r = r_0$ at $t=0$. In terms of $q$, this becomes $q = \frac{1}{r_0}$ at $\theta = 0$.

The second boundary condition involves the radial velocity: $\dot{r} = u_0$ at $t = 0$. This becomes $\mathrm{d}q/\mathrm{d}\theta = -u_0/h$ at $\theta = 0$.

As for the final boundary condition, note that $h = r^2\dot{\theta}$. Then $h$ can be determined from the initial conditions, such that $h = r_0 v_0$.

With these conditions, the expression for $q$ becomes:

$$q(\theta) = \left(\frac{1}{r_0} - \frac{k}{r_0^2 v_0^2}\right)\cos{\theta} - \frac{u_0}{r_0 v_0}\sin{\theta} + \frac{k}{r_0^2 v_0^2}$$

Then, since $r = 1/q$:

$$r(\theta) = \frac{r_0}{\left(1 - \frac{k}{r_0 v_0^2}\right)\cos{\theta} - \frac{u_0}{v_0}\sin{\theta} + \frac{k}{r_0 v_0^2}}$$

This equation is the polar equation of the orbit, and is the equation of an ellipse for certain choices of initial conditions $r_0$, $u_0$ and $v_0$. This equation will also account for parabolic and hyperbolic "orbits".

  • $\begingroup$ Nice! But I don't see how your derivative $d/d\theta(-hdq/d\theta)\dot\theta$ works. $\endgroup$
    – PM 2Ring
    Commented Dec 25, 2018 at 19:33
  • $\begingroup$ @PM2Ring The $h$ is constant - and so can be pulled out of the derivative - and the $\dot{\theta}$ is outside of the derivative $\endgroup$
    – Involute
    Commented Dec 25, 2018 at 20:46
  • $\begingroup$ Ah, of course. But now I get $\ddot r=-h^2q^2d^2q/d\theta^2$. What have I done wrong? $\endgroup$
    – PM 2Ring
    Commented Dec 25, 2018 at 22:07
  • $\begingroup$ Actually, you have done nothing wrong! I’ve spotted my typo and fixed it now. :) $\endgroup$
    – Involute
    Commented Dec 26, 2018 at 0:08
  • $\begingroup$ Phew! It all makes sense now. $\endgroup$
    – PM 2Ring
    Commented Dec 26, 2018 at 0:40

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