Assume a sun with mass $M$ and a planet with mass $m$. Assume at $t_0$ the planet is $r_0$ (distance) away from sun and has an initial velocity of $v_0$. Also, let’s assume the angle between the $r_0$ and $v_0$ is $\theta_0$ but I suppose if $\theta_0$ is $90$ degrees we can cover all the cases.

Are there any formulas and relations between $M, m, r_0, v_0$ that show wether the planet will make a circular, elliptical, parabolic or hyperbolic orbit ? (Or spiral orbit slowly moving towards the sun making smaller circles ) ?


3 Answers 3


Yes there is. Total energy is, in polar coordinates in the plane where the movement occurs:


A short rewrite using $C=\lVert\vec{L}\rVert/m$, the constant from Kepler's second law, yields:


The second and third terms form what's called the effective potential energy $V_\text{eff}$. Notice you necessarily have $V_\text{eff}(r)\leqslant E$, which can easily be interpreted on a graph:


In this example, $E<0$. Since $V_\text{eff}$ must remain below $E$, the only part of the plot that is physically accessible is $r\in[r_1,r_2]$. In other words, $r$ cannot go to infinity, it's a bound state. This is the elliptic case.

On the other hand, if $E>0$, $r$ cannot go to zero, but it can go to infinity. The movement isn't bounded, it's the hyperbolic case.

The special case $E=0$ is the parabolic one ($r$ can go to infinity, but then kinetic energy is zero).

The special case where $E$ is exactly at the bottom of the potential well is the circular movement (only one allowed value for $r$, here $r_0$).

In other words, the sign of $E$ will tell you exactly in which case you are. $E$ can be rewritten as:

$$E=\frac{1}{2}\,mv_0^2-G\frac{Mm}{r_0} \quad \begin{cases} >0 & \text{hyperbolic}\\ =0 & \text{parabolic}\\ <0 & \text{elliptic} \end{cases} $$

which should answer your question.


If a mass (like a planet) is orbiting a much larger mass (like the sun) under the action of gravity, it has kinetic energy (1/2)m$v^2$ and potential energy -GMm/r (negative because the zero is chosen to be at infinite r). If the total energy is negative, the orbit will be elliptical; zero, parabolic; and positive, hyperbolic. The path will be a spiral only if there is a mechanism which is causing a loss of energy. (The earth has been orbiting the sun for a very long time.)


For the planetary motion the total energy $$E=\frac{1}{2}mv^2-\frac{GMm}{r} \tag{1}$$ and the angular momentum $$L=mrv\sin\theta \tag{2}$$ will remain constant for all times $t$. Hence you can calculate them at time $t_0$ (with $r_0$, $v_0$, $\theta_0$).

From Newtonian mechanic the planetary orbit is known to be a conic section (never a spiral). Using the values for $E$ and $L$ from (1) and (2) you can calculate the eccentricity $\epsilon$ of the orbit. $$\epsilon=\sqrt{1+\frac{2EL^2}{G^2M^2m^3}}$$

From $\epsilon$ you can read off the shape of the orbit:

  • $\epsilon = 0$ : circular orbit
  • $0 < \epsilon < 1$ : elliptic orbit
  • $\epsilon = 1$ : parabolic orbit
  • $\epsilon > 1$ : hyperbolic orbit

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