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Consider a projectile body launched from the earth's surface with a speed lower than the escape velocity.

The orbit of the body under Newtonian gravity traces an ellipse which intersects with the earth's surface, resulting in a collision. However, if we approximate the motion as a simple projectile instead, using the flat earth assumption (with position invariant vertical gravitational field), then the trajectory recovered is a parabola. Please see this physics.se post and answers therein for a detailed discussion of why the trajectory is a parabola.

From the above analysis, can we conclude that a segment of an ellipse represented in the local coordinates (say with origin at one end of the segment and a coordinate axis along the segment) a parabola?

Note: The referenced parabola is not limited to the curve trajectory traced by the projectile from start to end point but also includes all segments along the elliptic trajectory. For instance, in the figure below, the ellipse segment from the 'booster burnout, 180 seconds after launch' point to 'end bussing, 650 seconds' point can be approximated as a parabola by considering constant gravity equal to, say, the average gravity over the elliptical trajectory.

enter image description here

enter image description here

Qmechanic
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kbakshi314
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  • You might compute the eccentricity of the ellipse and compare to the $e=1$ parabola that way; see https://physics.stackexchange.com/a/373256/44126 – rob Apr 09 '21 at 14:45
  • Your question is missing a verb, but anyway you already know that the parabola is an approximation, so honestly I'm not sure what the question is. – Javier Apr 09 '21 at 15:42

2 Answers2

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If you zoom in close enough on any continuous curve such that $(t-t_{0})^3 \approx 0$, then, the curve will be indistinguishable from a parabola. And since you can taylor expand around any point, you can do this anywhere.

This is basically just rediscovering Runge-Kutta integration, where you pick some step size $\Delta t$, and an initial value for $x$ and $\dot x$ (add y and z if you want, the procedure is the same, with more dimensions). Then, at each timestep,

  1. compute $F(x)$ using whatever force law you're integrating (in your case, $F = mg\left(\frac{R^2}{r^{2}}\right)$, where $R$ is the radius of the earth, and $g$ is the acceleration due to gravity at the surface)
  2. evaluate the position and velocity at the next timestep:
    1. ${\dot x}_{t+1} = {\dot x}_{t} + \frac{F\Delta t}{m}$
    2. $x_{t+1} = \frac{1}{2}\frac{F}{m}\left(\Delta t\right)^{2} + {\dot x}_{t}\Delta t + x_{t}$
  3. repeat until evolved for the desired amount of time, if accuracy is bad, make $\Delta t$ smaller

Which is precisely pretend the motion is a bunch of parabolas, and just recalculate the forces at every point. This is, in fact, the easiest way to make a semi-reasonable 3-body problem numeric solver in Newtonian gravity.

Zo the Relativist
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  • @JerrySchirmer thanks for the answer. It is interesting to note the Runge-Kutta connection with parabola approximation of the numerical model of a trajectory. However, the answer does not connect with the parabola which the OP references. The OP has been edited to more accurately identify the parabolas being considered. – kbakshi314 Apr 10 '21 at 03:12
  • @kb314: even with teh update to the question, aren't you basically segmenting the curve into a bunch of parabolas, or saying "pick a point, and around that point, the ellipse is close to some parabola?". How is that different than the above? – Zo the Relativist Apr 10 '21 at 16:55
  • @JerrySchirmer if we look at the motion in a small time interval from the start point, indeed your answer and preceding comment are accurate. However, even over large time intervals, the trajectory is locally a parabola governed by constant (time-invariant) flat earth gravity model. I'm not sure that the answer or preceding comment addresses the curve over a large time intervals from the start point. – kbakshi314 Apr 11 '21 at 03:13
  • @kb314: It's not a parabola over any curve where $g$ varies (unless the initial velocity is exactly at escape velocity, I guess). If you break the curve into small enough windows so that $g$ is approximately constant, then you can do the above on each of those windows. The fact that you can break this up into a bunch of local parabolas is exactly the Runge-Kutta trick. – Zo the Relativist Apr 11 '21 at 12:47
  • @JerrySchirmer the trajectory of a projectile under constant gravity is indeed, also a parabola. This is the trajectory shown in the gif simulation in the second figure in the OP. A segment of the ellipse is therefore a local parabola under the flat earth (or constant gravity assumption with a value dependent on the radial location of the local region considered) assumption. The described situation therefore goes beyond small time intervals during which, as your answer explains, we can approximate any trajectory as a parabola. – kbakshi314 Apr 12 '21 at 17:24
  • @kb314 What do you mean by local parabola? A segment of an ellipse is a segment of an ellipse, and can always be distinguished from a segment of a parabola if viewed with sufficiently high resolution. Parabolas and ellipses are different curves, full stop. The fact that the latter can be approximated by the former doesn't change that. – J. Murray Apr 13 '21 at 03:37
  • @J.Murray if you please check this answer to a related post, then hopefully we agree that the statement 'For instance, in the figure below, the ellipse segment from the 'booster burnout, 180 seconds after launch' point to 'end bussing, 650 seconds' point can be approximated as a parabola by considering constant gravity equal to, say, the average gravity over the elliptical trajectory.' in the OP is accurate. Hence, we should be able to construct the ellipse by approximating large segments of it by parabolas as in the linked answer. – kbakshi314 Apr 13 '21 at 03:41
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    @kb314 I guess I’m not following. If you break up the ellipse into pieces, you can approximate the pieces by parabola segments. They are not parabola segments, but they can be approximated as such as long as the pieces aren’t too big (especially near the apex), as per Jerry’s answer. However, the answer to “can we conclude that [a segment of an ellipse is] a [segment of a] parabola?” is no. – J. Murray Apr 13 '21 at 03:53
  • @J.Murray let's look at simple projectile motion of a material particle launched from the earth's surface under constant gravity. The trajectory is a parabola. At any location $S$ of the elliptical trajectory show in the OP, note the launch angle and speed and compute the parabolic trajectory using constant gravity, the value of which is the gravity at the location $S$. The trajectory will then intersect the tangent (to the earth's surface) at say point $E$. The segment $SE$ is the parabola I'm referring to and should approximate the elliptic orbit segment corresponding to the end-points $SE$. – kbakshi314 Apr 13 '21 at 05:01
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The ellipse and the parabola are tangent to each other at the launch site, and their curvatures (determined by local acceleration at the launch site, which is the same for "flat earth" and "spherical earth" by definition) are locally the same. Further away from the launch site, they will diverge.

oliver
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  • Thanks for the answer. However the OP addresses a more general interpretation. Consider any segment of the ellipse, with end points which do not coincide with the launch site, for instance. If we knew the launch speed and angle w.r.t. the tangent plane to the earth's surface passing through the end points of the segment, then that segment will also be an ellipse in the flat earth interpretation. The question is does this mean that the elliptic orbits are generalized curved earth trajectories corresponding to localized flat earth parabola trajectories. – kbakshi314 Apr 09 '21 at 06:22