Consider 2D motion of a particle,

For a particle to move in a circle there must be a force of constant magnitude acting on it always pointing towards the centre of the circle.

Similarly for an ellipse there must be a force acting on it with a magnitude inversely proportional to the square of the distance from one of its focus and always pointing towards that focus.

For a parabolic path there must be a force with a constant magnitude and direction acting on it.

However I do not know the conditions for a hyperbola and a rectangular hyperbola. Any help will be greatly appreciated. Also correct me if I'm wrong about any of my assumptions.

Edit: I tried to find the required acceleration of the particle by differentiating $x=a\sec(ωt)$ and $y=b\tan(ωt)$

I got the acceleration as $ax=(ω^2)x(2((x^2)/(a^2)) - 1)$ and $ay=2(ω^2)y(((y^2)/(b^2)) + 1)$

This seems to suggest that the force acting on the particle increases with its distance from focus. Is this correct?

  • $\begingroup$ Why don't you try differentiating an equation of hyperbola and find the relation? $\endgroup$ Commented Jul 9, 2020 at 7:45
  • $\begingroup$ Thank you for the suggestion. I tried it out and have edited my question. Please check it and see if it's correct. $\endgroup$ Commented Jul 9, 2020 at 8:31
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    $\begingroup$ I have a recollection that knowing the shape of one trajectory is not sufficient to determine the force law that produced it; different parametrizations with respect to time imply different force laws for the same curve. I’ll see if I can dig up a reference later today. $\endgroup$ Commented Jul 9, 2020 at 12:05
  • $\begingroup$ @MichaelSeifert OK, that would be great $\endgroup$ Commented Jul 9, 2020 at 12:06
  • $\begingroup$ @MichaelSeifert That is correct. $\endgroup$ Commented Jul 9, 2020 at 12:31

2 Answers 2


If you know that a particle's trajectory $r(\theta)$ in polar coordinates is produced by a central force law, and you know the particle's angular momentum $L$, then easiest way to "back out" that force law is probably via the Binet equation:
$$ F\left( r \right) = - \frac{L^2}{m r^2} \left( \frac{d^2 (1/r)}{d\theta^2} + \frac{1}{r} \right), $$ where $u \equiv 1/r$, $m$ is the mass of the particle, and $L$ is its angular momentum. If you know $r(\theta)$, then you can take its multiplicative inverse, differentiate it twice, and plug it into the above expression. For example, for an ellipse, we have $r = r_0/(1 + \epsilon \cos \theta)$ in polar coordinates, with $0<\epsilon<1$. Applying this logic yields $$ F\left( r \right) = - \frac{L^2}{m r^2 r_0} \left( - \epsilon \cos \theta + 1 + \epsilon \cos \theta \right) \propto -\frac{1}{r^2}. $$ I'll leave it to you to do the similar calculation for the hyperbola.

On the other hand, if you do not know that the trajectory is due to a central force, then you can't really say anything about the force that created it. The reason is that different parametrizations of the curve $\{ x(t), y(t) \}$ may correspond to the same path in space; but the particle will have different accelerations at a given point on the path for each parametrization, meaning that the force must be different for each parametrization. In particular, the parametrization that you used for the hyperbola, $$ x(t) = a \sec(\omega t), \qquad y(t) = b \tan (\omega t) $$ can't correspond to a central force, since $L_z = m (x \dot{y} - y \dot{x})$ is not constant along the trajectory (if I'm not mistaken it works out to be $L = - a b \omega \sec (\omega t)$.)

  • $\begingroup$ Thank you for providing the Binet equation. Now I understand my mistake in choosing the parameters without considering that angular momentum due to a central force would be constant. So by applying the formula it seems that the force for all 4 paths would be inversely proportional to the square of the distance from the focus of the path. Is this correct? $\endgroup$ Commented Jul 9, 2020 at 14:37

Inverse square forces can also produce hyperbolic trajectories if the body has enough energy.

Keep in mind that the shape of a trajectory does not determine the forces that cause the trajectory (so your statements of "there must be a force" should really be "it is sufficient to have a force with the appropriate initial conditions"). For example, in circular trajectory with increasing speed (like a rocket on a string) the centripetal force will not be constant, and there will also be a force component along the circle.

If you want to find the force for a path, just do what you did at the end of your post. Parametrize the path somehow and call the parameter time $t$ so you have $\mathbf x(t)$. Then the force is just proportional to $\ddot{\mathbf x}$. Then you will have a force producing the desired trajectory.

  • $\begingroup$ Thank you for the explanation. I have gone through the Wikipedia article. Would it be right to say that the path of the particle will be a hyperbola if the sum of its potential and kinetic energy is positive, a parabola if the sum is zero and either an ellipse or a circle if the sum is negative? $\endgroup$ Commented Jul 9, 2020 at 13:50

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