8
$\begingroup$

Let's assume I shoot an object from a high tower horizontal to the earth's surface. As far as I understand, depending on the velocity I will get different types of orbits. With decreasing velocity I will go from

  1. hyperbolic orbit where the focal point is the earth's center
  2. parabolic orbit where the focal point is the earth's center
  3. elliptic orbit where the earth's center is that focal point that is nearest to the tower
  4. circular orbit where the earth's center is the center of the circle
  5. crash orbit

My question is about the last orbit. If earth was transparent for the thrown object,...

Would orbit 5 be an ellipse with the center of earth being at the focal point furthest from the tower?

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ It's worth noting that a circular orbit is just a special case of an elliptical orbit. $\endgroup$
    – David White
    Nov 2, 2020 at 23:22
  • $\begingroup$ The answer does seem to be yes, but at first it may seem strange to go down the progression from hyperbolic to circular then "turn around" and go back to elliptical. That the cannon is horizontal means that at this point in any orbit motion is perpendicular to the radial direction; we can say that it is tangential motion. For 1, 2, and 3 this is periapsis, for 4 all points are equivalent. The ellipse of #5 differs from the ellipse of #5 because you've flip-flopped from periapsis to apoapsis, and the other end of the ellipse will be closer to rather than farther from the center. $\endgroup$
    – uhoh
    Nov 3, 2020 at 10:55
  • $\begingroup$ The eccentricity goes from >1 to 1, to 0<e<1 to 0 at the circle, then goes back up to 0<e<1 for #5. And by the way, when the kick from the cannon drops to zero and the projectile drops straight down, the eccentricity reaches 1 again. See answers to What's the eccentricity of an orbit (trajectory) falling straight down towards the center? $\endgroup$
    – uhoh
    Nov 3, 2020 at 10:57
  • $\begingroup$ Not a full answer, but when exploring Keplerian orbits, its worth remembering that the last valid conic shape that an orbit can take is a line. Not that this orbit occurs in any real situation, but it fills in a "hole" that might be confusing when thinking about "crashing" orbits. $\endgroup$
    – Cort Ammon
    Nov 3, 2020 at 15:34

2 Answers 2

Reset to default
13
$\begingroup$

The answer is yes, even though it doesn't make sense to talk about "orbit" in your case as the object crashes on the surface of the planet.

As a thought experiment, though, you can think of the Earth as a point particle, and your object being shot in space far from the surface. Then, as its trajectory starts "bending" downwards, it doesn't hit the surface of the Earth and can propagate.

This is best exemplified in the following image from here, where the blue planet would be my "point-like" Earth: enter image description here

Out of completeness, it might be worth mentioning that for spherically symmetric vacuum solutions where GR effects are important, there is a correction to the Newton's potential that actually results in an innermost stable circular orbit ($R_{\text{ISCO}}$), below which there are plummeting orbits.

Improve this answer
$\endgroup$
4
  • $\begingroup$ :-) it is very closer to my question: physics.stackexchange.com/questions/590597/… $\endgroup$
    – Sebastiano
    Nov 2, 2020 at 22:07
  • $\begingroup$ Excellent answer! I've added a supplementary comment. $\endgroup$
    – uhoh
    Nov 3, 2020 at 10:59
  • 5
    $\begingroup$ It doesn't seem so counterintuitive to go ellipse-circle-ellipse when you note which of the ellipses' foci are closest to the origin. The 2 elliptical foci can be seen to cross each other, meeting at the center of the circle. $\endgroup$
    – Cristobol Polychronopolis
    Nov 3, 2020 at 15:21
  • $\begingroup$ If we're talking about point-planets, might as well consider the case where tangential velocity is zero and the "orbit" is a line. $\endgroup$
    – Phil Frost
    Nov 3, 2020 at 17:42
7
$\begingroup$

transparent for the thrown object

Worth mentioning an alternative to point-mass orbits here. Assuming the Earth was literally transparent to the object, but all the mass was still there, distributed over the full volume.

Assuming uniform density, we now have a central force field where gravity is proportional to $r^1$ due to the shell theorem, unlike the point-mass case where gravity is proportional to $r^{-2}$.

Such force fields also have stable periodic orbits (actually the only other exponent than $r^{-2}$ that has them), but with some notable differences.

  • All orbits are elliptical with no exception, as there can be no escape trajectories.
  • Elliptical orbits have the centre of the Earth at their geometric centre, not at a focal point, meaning you end up at the surface again at the opposite side of the planet.

In practise, however, the density of the Earth is not uniform, so we don't have a neat $r^1$ force field for tunnel orbits.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Excellent answer, as a term like "crash orbit" implicitly forbids treating earth as a point particle. $\endgroup$
    – Ralf Kleberhoff
    Nov 3, 2020 at 13:22
  • $\begingroup$ And, of course, in practice, the Earth is prone to cause Rapid Unexpected Deconstruction in spacecraft which get the bright idea that the Earth might be transparent in this way! $\endgroup$
    – Cort Ammon
    Nov 3, 2020 at 15:32
  • $\begingroup$ @Cort Aka Rapid Unintentional Disassembly or Rapid Unplanned Disassembly. ;) $\endgroup$
    – PM 2Ring
    Nov 3, 2020 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.