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I understand that the parabolic trajectory is an approximation of a more elliptical trajectory, since acceleration due to gravity is taken to be a constant for a projectile. However I'm intrigued to know that what changes in kinetic and potential energy contribute to the question? What I mean is I read that for a trajectory under a central force, the reciprocal of radial distance is $$\frac{1}{r}=-\frac{mK}{L^2}+A \cos\theta$$ where $$A^2 = \frac{m^2K^2}{L^4}+\frac{2mE}{L^2}.$$

For an ellipse clearly the first term should be greater than $A$ but somehow that doesn't seem to make a lot of sense to me. Further for a parabola of I equate the first term with $A$, which again, seems a little weird. Where am I going wrong?

Please feel free to criticise me but it's a request not to vote close the question unnecessarily. My question regarding fusors were vote closed simply because a few didn't know about the existence of such a thing.

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    $\begingroup$ This question could be improved by defining the terms involved. Presumably $m$ is mass, $L$ is angular momentum, and $E$ is energy. $\endgroup$ – Paul T. Jun 15 '20 at 17:46
  • $\begingroup$ Is it possible that for A , the energy term is negative? I can't remember, it's been a while since I looked at my CM books, and I don't have any to hand, but it was in my head. $\endgroup$ – MC2k Jun 15 '20 at 17:58
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    $\begingroup$ I may be wrong, but it appears to me that none of the three answers that are posted at the time that I write answers the question. Also: I agree with the first comment. $\endgroup$ – garyp Jun 15 '20 at 21:37
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Technically a projectile under the effects of the Earth's gravity is an ellipse. It is under a central force which leads to that trajectory. However, this ellipse is tremendously big compared to the actual path taken by a projectile. It is a reasonable assumption to replace the central force (and all of its complex equations) with a "downward" force that's simpler to work with. The discrepancies between these trajectories will be far smaller than the discrepancies caused by aerodynamic effects surrounding a small scratch on the projectile.

In reality, you're just making a second order model of the absolute tip of a several thousand kilometer long ellipse. Near that tip, ellipses are similar in shape to that of parabolas.

Once the projectile starts going hundreds of kilometers, that assumption starts to get problematic. However, when dealing with projectiles on this scale, we typically have some guidance attached to them which, once again, creates a much larger effect than the central force.

When we get to satellites, where we can start to ignore the aero forces, it becomes more feasible to create good predictions of where the satellite is going over thousands of kilometers. In this case, we must give up on our simple "gravity pulls things down" model, and use the more accurate "gravity pulls things towards the center" model.

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  • $\begingroup$ Cort Ammon could you please suggest what is the change in the energy terms that give rise to the parabola and ellipse? What I mean is that can we prove rigorously just by the initial conditions when the trajectory would be parabolic or elliptic? $\endgroup$ – Richard Kiddman Jun 15 '20 at 18:02
  • $\begingroup$ @MC2k any book you'd suggest with clear elucidation? $\endgroup$ – Richard Kiddman Jun 15 '20 at 18:02
  • $\begingroup$ @anna v can prediction about the shape of trajectory be made mathematically if only the initial condition is given? $\endgroup$ – Richard Kiddman Jun 15 '20 at 18:05
  • $\begingroup$ @RichardKiddman yes, that is the way satellites are put in orbit $\endgroup$ – anna v Jun 15 '20 at 18:22
  • $\begingroup$ all conic sections are solutions of trajectories in a gravitational potential. If it is an open trajectory it is a different solution than the ellipse., onefocal point only $\endgroup$ – anna v Jun 15 '20 at 18:26
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It is simpler to think in terms of the solution of 1/r two body potentials, these are conic sections.

One application is that a moving particle that is subjected to an inverse square law force like gravity or Coulomb's law will follow a path described by one of the conic section:

conic

Each of the conic sections can be described in terms of a semimajor axis a and an eccentricity e. Representative values for these parameters are shown along with the types of orbits which are associated with them.

See this for the history.

Closed Curves:

Ellipses Circles, which are a special case of an ellipse with e=0 These orbits are bound: objects will orbit forever around the parent body.

Open Curves:

Hyperbolas Parabolas, which are a special case of a hyperbola These orbits are unbound: objects will pass by the parent body only once and then escape from the parent body's gravity.

link goes on to describe circular speed and escape speed.

Which of these orbits you will be in is determined by your orbital speed. There are two special speeds of particular interest.

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We consider the mechanical energy of the object in orbit due to a central force to be $$E=\frac{1}{2}mv^2 + U(r)$$ where $m$ is the reduced mass of the object, $v$ is the instantaneous speed, and $U(r)$ is the instantaneous potential energy of the system due to the central force.

There are three general values for the mechanical energy:

E>0

This energy results in a hyperbolic orbit, which is an open path because no matter how far away you go there is always some kinetic energy to keep moving away.

E=0

This energy results in a parabolic orbit, again, an open path. There is always kinetic energy to keep moving until $r\to \infty.$

Note that this is a different energy/orbit calculation than for the constant force situation (F=mg) because the potential energy zero is arbitrary for F=mg.

E<0

This is the elliptical orbit, which is a closed path. There are turning points because there is a minimum value of kinetic energy related to the (constant) angular momentum, and the kinetic energy is always less than the magnitude of the potential energy.

If the kinetic energy, $K$, is constant, the orbit will be circular and the total mechanical energy will be $E= - K$.

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