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Discussion below this answer has got me thinking.

A circular orbit has eccentricity of 0, ellipses have eccentricity between but not including 0 and 1, and parabolic and hyperbolic orbits have eccentricity of 1 and greater-than 1.

But what if I let go of an object near a gravitational body and let it fall straight towards the body accelerating along a straight line?

  1. Is that a conic section? If so, what's it called
  2. What would be the eccentricity of this "orbit"?

I don't even know how to begin, or what this would be the limiting case of. It has a finite, non-zero apoapsis, but I suppose a periapsis of 0 if it didn't hit the body's surface.

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  • $\begingroup$ I'm not a good mathematician, but wouldn't that be infinite eccentricity? $\endgroup$ – BMF Sep 4 at 14:26
  • $\begingroup$ @BMF dunno. We can rule out 0 and all the finite positive reals, but infinity, negative, and complex numbers haven't been ruled out yet, nor has indeterminate eccentricity. $\endgroup$ – uhoh Sep 4 at 14:33
  • $\begingroup$ You would basically draw the orbit out "through the surface" and calculate eccentricity as if you weren't going to slam into the ground, right? Suborbital is still an "eccentric orbit" because it is non-circular but the orbit abruptly ends when you hit the ground as the periapsis is below the surface. So this would just be an orbit where your apoapsis is the highest point in the suborbital trajectory and your lowest would be the center of the orbited object. $\endgroup$ – Magic Octopus Urn Sep 4 at 14:37
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    $\begingroup$ It is a special kind of a parabola. A parabola has the eccentricity 1 and a special case of a parabola is a straight line. A parabola is defined by x*x = 4ay, if a is 0 then x is 0 and the straight line is the y axis. Any point with x = 0 and y between + and - infinity belongs to the straight line. $\endgroup$ – Uwe Sep 4 at 14:55
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    $\begingroup$ @uhoh: my mistake. Deleted $\endgroup$ – Michael Stachowsky Sep 4 at 18:11
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The eccentricity is 1.0.

The eccentricity $e$ of an orbit can be found from the radius of apoapse and periapse as:

$$e=\frac{r_a-r_p}{r_a+r_p}$$

and the semimajor axis $a$ can as well, from:

$$a=\frac{r_a+r_p}{2}$$

If you throw an object horizontally (velocity perpendicular to position vector) you will end up in a closed orbit if you throw at slower than escape speed, an open parabolic orbit if you throw it exactly at escape speed, or an open hyperbolic if you throw it greater than escape speed.

There is a speed which will result in a perfect circular orbit, with $e=0$. We might as well call this "circular orbit speed". In the eccentricity equation, if $r_a=r_p$ as it does in a circular orbit, we see that the numerator is zero, while the denominator is nonzero, so the whole fraction is also zero.

If you throw slower than circular velocity, the object will fall closer to the center before coming back up. The lower the object gets, the lower periapse is. In the eccentricity equation, as $r_p$ decreases, the numerator grows while the denominator shrinks, so the whole fraction increases. As we go slower, we increase eccentricity.

The limiting case of this is if you throw it at zero speed, IE you drop it.

For an object dropped in a gravity field around a true point mass, you will end up with the apoapse being the radius at which you dropped the object, and the periapse at zero. This is a very weird orbit, because the object will take a finite time to reach the center, but will reach infinite speed just as it passes the center where it will make a 180° turn and coast back up, until it reaches its original drop height at zero speed and starts another cycle. You can use Kepler's third law to figure the time of this orbit, since it still has a well-defined $a$.

A spherically-symmetric mass with a definite surface (density of zero outside a certain radius) has an identical gravity field to that of a centered point mass everywhere outside its surface. Therefore an object dropped above the surface on a more-realistic planet would follow an orbit identical to that dropped at the same radius above a point mass, until it hit the surface. If it were to pass through the surface (say you drilled a hole) the gravity field below the surface is not the same as that of a point mass.

Whenever I am at a baseball game and see a pop-fly, it always amuses me to think that the path the ball is following is not truly a parabola, just the end of a very stretched out ellipse, which if continued, would form the same shape near the center of the Earth.

In this case, $r_p=0$. The eccentricity fraction has its numerator equal to $r_a-0=r_a$, and denominator equal to $r_a+0=r_a$ as well. The eccentricity is exactly 1.0 .

"But Kwan!" I hear you shout. "If $e=1$, doesn't that make it a parabolic orbit?" In this case, no. A parabolic orbit has $e=1$ and $a=\infty$, while the drop orbit has $e=1$ but a decidedly non-infinite $a=r_a/2$.

This case is the limit of an ellipse becoming thinner and thinner as the foci move apart. In that limit, one focus is at the center, one at the drop point, and the ellipse has zero width but finite length.

In this image, the circle that isn't moving represents a sphere with radius 1, and the ellipse that is moving represents an orbit with a constant apoapse radius of 2.0 but a varying eccentricity. One quick Python script later, and voila!

Note that this is not what would happen if you drilled a hole through the earth and dropped an object. This only applies to a true point mass, and doesn't take into account relativity (a true point mass would be a singularity, and the object would pass the event horizon on the way down and never come back up).

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  • $\begingroup$ This is a great answer, thank you! It provides a good intuitive framework to understand what's going on, and that this is indeed just another "normal, well-behaved" orbit, at least outside whatever region the central body includes. Nicely done. $\endgroup$ – uhoh Sep 4 at 23:36
  • $\begingroup$ +1 Oh, wow! I forgot that orbital velocity approaches infinity as it nears the point mass, which is why my answer was wrong. The math shows it all, though. Great answer! $\endgroup$ – BMF Sep 4 at 23:50
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The eccentricity of a radial orbit is $1$, regardless of its energy.

This is a class of orbits where the type of orbit cannot be inferred from the eccentricity alone. With a "traditional" parabolic orbit of $e=1$, the angular momentum $L$ has a well defined value, but the semi-major axis $a$ is not defined. In the case of a vertical bounded free-fall orbit, the semi-major axis $a$ is a well defined value, and the angular momentum $L$ is $0$.

The equation relating angular momentum to eccentricity and semi-major axis shows how the eccentricity can be calculated given the angular momentum. $$l = \sqrt{\mu a(1-e^2)}$$ For a parabolic orbit, $a\rightarrow\inf$, so defining the angular momentum via this formula does not work. However, for radial orbits where the object is not travelling at exactly escape velocity, this formula is well defined. It would imply that the eccentricity for a radial trajectory, no matter the energy, is $1$. The wikipedia article on orbital eccentricity confirms this.

Radial trajectories are classified as elliptic, parabolic, or hyperbolic based on the energy of the orbit, not the eccentricity. Radial orbits have zero angular momentum and hence eccentricity equal to one. Keeping the energy constant and reducing the angular momentum, elliptic, parabolic, and hyperbolic orbits each tend to the corresponding type of radial trajectory while e tends to 1 (or in the parabolic case, remains 1).

Multiple definitions allow the radial orbit to have an eccentricity of one as well.

$$e = \frac{r_a-r_p}{r_a+r_p}$$

$$e = \sqrt{1 + \frac{2\epsilon l^2}{\mu^2}}$$

with $\epsilon$ being the specific orbital energy. However, using other with eccentricity as the input may lead to undefined results. Namely the parametric definition fails at $\theta=0$:

$$ r = \frac{a(1-e^2)}{1-e\cos\theta}$$

So the take home is that a radial orbit has an eccentricity of $1$ no matter its energy. However attempting to use the eccentricity for orbital calculations may lead to undefined results.

I had previously mentioned that the conic section is a line, but I think this is wrong, because the orbit can be bounded (and is in the question), and this is a line segment. So I am curious as to the conic section from a bounded orbit (and for that matter a positive energy radial orbit).

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    $\begingroup$ I disagree. A line is a degenerate conic, it is not the same as a parabola, which is not a degenerate conic (en.wikipedia.org/wiki/Degenerate_conic). The eccentricity of a straight line is undefined because it can take on multiple values at once. Dr. Math explains this better than I can mathforum.org/library/drmath/view/72756.html $\endgroup$ – Michael Stachowsky Sep 4 at 14:59
  • $\begingroup$ @MichaelStachowsky from your wikipedia link : "Parabolas can degenerate to two parallel lines: x 2 − a y − 1 = 0 {\displaystyle x^{2}-ay-1=0} x^{2}-ay-1=0 or the double line x 2 − a y = 0 , {\displaystyle x^{2}-ay=0,} x^{2}-ay=0, as a goes to 0; but, because parabolae have a double point at infinity, cannot degenerate to two intersecting lines." $\endgroup$ – Uwe Sep 4 at 15:18
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    $\begingroup$ @uhoh I updated the answer substantially. There were 2 major corrections, the first being that it isn't a parabolic orbit and correcting myself on the line conic section, and second linking to evidence that the eccentricity can be defined from an orbital mechanics standpoint. Not that it is terribly useful for calculating other orbital parameters in the radial orbit case. $\endgroup$ – Quietghost Sep 4 at 20:44
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    $\begingroup$ Wow, this has really developed into a fascinating answer, thank you! I've read through a few times and will now contemplate further. I like that $r_p$ is zero, which suggests that this is fundamentally different than the "passing through a hole in the Earth" problem. Nicely done. $\endgroup$ – uhoh Sep 4 at 23:37

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