Is a free ballistic path a parabola or a cycloid?

[Zanket]

Feynman says that the path of projectile in a uniform gravitational field is a parabola, but the bottom of http://online.redwoods.cc.ca.us/instruct/darnold/CalcProj/Fall98/NateB/definition.htm says it's a cycloid. My calculation shows a parabola. Which is correct?

If parabola, why does a hypothetical Big Bang to Big Crunch scenario (a closed Friedmann model) plot, for the distance over time for a pair of galaxies, a cycloid? Is that because in GR in this scenario the galaxies are in free fall in signficiant expanding space (the expanding space paradigm) whereas for the projectile the expanding space is negligible?

 

[dextercioby]

In Newtonian physics is neither of them.It's an arch of en ellipse,with the center of the Earth in the closest focus.

Daniel.

 

[Zanket]

An arc of an ellipse is a parabola, no?

 

[dextercioby]

Nope,an arch of an ellipse is just an arch of an ellipse (pardon the tautology).

Daniel.

 

[Zanket]

You mean "arc"? What is the difference between arc and arch as you're using it? An arc of an ellipse looks like a parabola to me. How is it different? Feynman didn't know what he was talking about?

 

[dextercioby]

I don't know what he was talking about.Yes,"arc",if you prefer.

Give me a reference on Feynman to check him out.

Daniel.

 

[EL]

the path of projectile in a uniform gravitational field is a parabola

Yes.

By uniform gravitational field he means a field with the same amplitude and direction at all points.

The Earth's gravitational field is directed towards its centre (i.e. different directions at different points), and is getting weaker further away.

 

[Garth]

Feynman says that the path of projectile in a uniform gravitational field is a parabola, but the bottom of http://online.redwoods.cc.ca.us/instruct/darnold/CalcProj/Fall98/NateB/definition.htm says it's a cycloid. My calculation shows a parabola. Which is correct?

That link states "The "altitude" vs time of an object in gravitational free-fall is a cycloid"; that is not the same as the plot of altitude vs distance, which is a parabola on a flat Earth. As we are not on a flat Earth, although the real Earth can be approximated as such for small distances, the trajectory is part of an ellipse as stated by Daniel.

Garth

 

[Zanket]

That makes sense, thanks. Except, isn't part of an ellipse a parabola?

 

[Zanket]

Give me a reference on Feynman to check him out.

From Six Not-So-Easy Pieces, pg. 140: “...a parabola—the same curve followed by something that moves on a free ballistic path in the gravitational field”

 

[EL]

Except, isn't part of an ellipse a parabola?

Nope. An ellipse closes itself. For a parabola, say y=x^2, y goes to infinity when x does.

 

[Zanket]

That link states "The "altitude" vs time of an object in gravitational free-fall is a cycloid"; that is not the same as the plot of altitude vs distance,

Ah but wait--I had thought before about the altitude/time vs. altitude/distance difference. I thought they were synonymous plots because the horizontal velocity is constant. What am I missing?

 

[Zanket]

Nope. An ellipse closes itself. For a parabola, say y=x^2, y goes to infinity when x does.

OK. Is the path of the projectile an arc of a parabola on a (hypothetical) flat Earth and an arc of an ellipse on a round Earth? Edit: Never mind, I see that is just what Garth said. Maybe I'm getting this now.

 

[Zanket]

I don't know what he was talking about.

Looks like Feynman is correct, because he was talking about a uniform gravitational field. You are correct about a nonuniform field.

 

[EL]

OK. Is the path of the projectile an arc of a parabola on a (hypothetical) flat Earth and an arc of an ellipse on a round Earth? Edit: Never mind, I see that is just what Garth said. Maybe I'm getting this now.

You sure are!

 

[Garth]

Ah but wait--I had thought before about the altitude/time vs. altitude/distance difference. I thought they were synonymous plots because the horizontal velocity is constant. What am I missing?

The round Earth?

Garth

 

[dextercioby]

Incidentally,"ballistic" trajectories should keep account on many subtleties,on of them being the Earth's curvature,hence elliptical shape of orbit,neglecting rotatation & air motion & viscosity.

You know,ballistical missiles (so caled "strategical weapons") have ranges in thousands of kilometers...To assume Earth's flat is insanity.

Daniel.

 

[Zanket]

The round Earth?

Can you elaborate? Given a flat Earth, I'm thinking that if the horizontal velocity is 1 km/s, say, then a curve of altitude/time will be the same shape as altitude/distance. Given a paraboloidal curve for altitude/distance, I don't see how altitude/time will be cycloidal.

 

[selfAdjoint]

Garth's point is that absent atmosphere, on a spherical Earth the arc would be a section of an ellipse with a focus at the Earth's center. Geometrically if you allow that focus to move away "to infinity", i,.e. so far way the error is below your maxiumum accuracy, then the arc of the ellipse is indistinguishable from an arc of a parabola.

 

[Zanket]

Thanks, that agrees with what I learned above. What I'm asking now is, for a hypothetical uniform gravitational field, why is the path paraboloidal for an altitude/distance plot but cycloidal for an altitude/time plot (as told by the link in the first post in this thread)? Given that the horizontal velocity is constant, I would think these plots would have the same shape.

 

[EL]

Anyway, Feynman was of course correct in saying

that the path of projectile in a uniform gravitational field is a parabola

I would think these plots would have the same shape.

So would I.

 

[selfAdjoint]

Thanks, that agrees with what I learned above. What I'm asking now is, for a hypothetical uniform gravitational field, why is the path paraboloidal for an altitude/distance plot but cycloidal for an altitude/time plot (as told by the link in the first post in this thread)? Given that the horizontal velocity is constant, I would think these plots would have the same shape.

The cycloid is the brachistochrone, the path of quickest descent. If the particle is constrained to follow it, the time to fall along it will be less than for any other curve connecting the first and last points. This is not the same thing as a geodesic because the hypotheses for the brachistochrone include constrained fall, not free fall.

 

[Garth]

Thanks, that agrees with what I learned above. What I'm asking now is, for a hypothetical uniform gravitational field, why is the path paraboloidal for an altitude/distance plot but cycloidal for an altitude/time plot (as told by the link in the first post in this thread)? Given that the horizontal velocity is constant, I would think these plots would have the same shape.

The gravitational field is not uniform because of the Earth's curvature therefore the horizontal velocity is not constant. There is a horizontal gravitational gradient.

Garth

 

[chronon]

The website you refer to means the plot of the altitude of an orbiting body against time. (The shape of the orbit itself is an ellipse). Except that this plot isn't actually a cycloid, at least not a normal cycloid, rather it's what is known as a curtate cycloid.

 

[robphy]

Thread-readers might be interested in
http://arxiv.org/abs/physics/0310049
Ballistic trajectory: parabola, ellipse, or what?
Authors: Lior M. Burko, Richard H. Price
(Now appearing in American Journal of Physics, Vol. 73, No. 6, pp. 516–520, June 2005 )

 

[Crosson]

The path of least time between two points in a uniform gravitational field is a CYCLOID.

 

[jdavel]

The path of least time between two points in a uniform gravitational field is a CYCLOID.

Right, but the referenced web page said that the altitude vs time of an object in gravitational free-fall is a cycloid. That's not the same thing.

The path of an object moving in the Earth's gravitational field (assuming no other forces) is an ellipse. That's Kepler's 1st law. The path of an object moving in a uniform gravitational field (again, assuming no other forces) is a parabola.

So, how can the referenced statement be correct?

Edit: Ok, I reread Garth's post #23, and now I get it. Nice insight Garth!

 

[Zanket]

The gravitational field is not uniform because of the Earth's curvature therefore the horizontal velocity is not constant. There is a horizontal gravitational gradient.

Makes sense. Then in a hypothetical uniform gravitational field, would both a plot of altitude/distance and a plot of altitude/time have the same shape?

I appreciate everyone's input so far. I haven't digested all this yet. The arxiv article looks helpful.

 

[arildno]

Interesting article, robphy!

We don't need to bound two parameters in order to get a parabolic approximation, though.
It is sufficient that $$\gamma=\frac{V_{0}^{2}}{gR}<<1$$
where $$V_{0},\sqrt{gR}$$ are the launch speed and escape velocity, respectively (R is the radius of the Earth, g the acceleration due to gravity).

We have:
1. 2-D Coordinate choice:
Let y measure the radial distance above the Earth's surface, whereas $$x=R\theta$$ measures the arclength along the Earth's surface, where $$\theta$$ is the angular displacement.

2. Projectile description:
a) Position vector: $$\vec{r}(t)=(R+y(t))\vec{i}_{r}$$
where $$\vec{i}_{r}$$ is the radial unit vector
(with $$\vec{i}_{\theta}=\frac{d\vec{i}_{r}}{d\theta}$$ orthogonal to $$\vec{i}_{r}$$)


We also define $$\epsilon=\frac{y(t)}{R}$$
(which clearly must be greater than 0)
b) Velocities:
bi) $$\dot{y}$$ radial velocity component of projectile
bii) $$\dot{x}=R\dot{\theta}$$ arc-length velocity
biii) $$(1+\epsilon)\dot{x}$$ projectile's velocity component in direction $$\vec{i}_{\theta}$$

c) Geometric trajectory description:
We introduce $$Y(x(t))=y(t)$$, so that Y(x) is the traversed curve as seen in the 2-D coordinate system.

d) Initial conditions:
$$y(0)=0,x(0)=0,V_{y,0}=\dot{y}(0)=\dot{y}_{0},\dot{x}(0)=\dot{x}_{0},V_{x,0}(0)=(1+\frac{y(0)}{R})\dot{x}_{0}=\dot{x}_{0}$$

3) Conservation of angular momentum
Angular momentum is conserved with respect to the Earth's center, i.e we have (cancelling the mass):
$$V_{x}=\frac{V_{x,0}}{1+\epsilon}=\frac{\dot{x}_{0}}{1+\epsilon},\dot{x}=\frac{\dot{x}_{0}}{(1+\epsilon)^{2}}$$

4) Expressions used in the radial equation:
$$\frac{d^{2}\vec{r}}{dt^{2}}\cdot\vec{i}_{r}=(\ddot{y}-\frac{(1+\epsilon)}{R}\dot{x}^{2})=(\ddot{y}-\frac{\dot{x}^{2}_{0}}{R(1+\epsilon)^{3}})$$

Furthermore:
$$\dot{y}=\frac{dY}{dx}\dot{x}=\frac{dY}{dx}\frac{\dot{x}_{0}}{(1+\epsilon)^{2}}$$
And:
$$\ddot{y}=\frac{d^{2}Y}{dx^{2}}\frac{\dot{x}_{0}^{2}}{(1+\epsilon)^{4}}-(\frac{dY}{dx})^{2}\frac{2\dot{x}_{0}^{2}}{R(1+\epsilon)^{5}}$$

5) Radial component of Newton's 2.law pr. unit mass
Thus, we gain:
$$\frac{d^{2}Y}{dx^{2}}=-\frac{g}{\dot{x}_{0}^{2}}(1+\epsilon)^{2}(1-\frac{\dot{x}_{0}^{2}}{gR}(\frac{2(\frac{dY}{dx})^{2}}{(1+\epsilon)^{3}}+\frac{1}{(1+\epsilon)}))$$

6) Speed and curve length s:
We have the relation for the curve-length s:
$$\frac{ds}{dx}=\sqrt{1+(\frac{dY}{dx})^{2}}$$
and also:
$$\frac{ds}{dx}=\frac{\frac{ds}{dt}}{\frac{dx}{dt}}=\frac{V}{V_{x}}=\frac{V}{\dot{x}_{0}}(1+\epsilon)$$
where V is the speed of the particle.

Thus, we may rewrite Newton's law as follows:
$$\frac{d^{2}Y}{dx^{2}}=-\frac{g}{\dot{x}_{0}^{2}}(1+\epsilon)^{2}(1-\frac{V^{2}}{gR}(\frac{2}{(1+\epsilon)}+(\frac{\dot{x}_{0}}{V})^{2}(\frac{1}{(1+\epsilon)}-\frac{2}{(1+\epsilon)^{3}}))$$

7) Order of magnitude estimates
a) The bracket $$(\frac{2}{(1+\epsilon)}+(\frac{\dot{x}_{0}}{V})^{2}(\frac{1}{(1+\epsilon)}-\frac{2}{(1+\epsilon)^{3}}))$$ is positive, and if the internal bracket $$}(\frac{1}{(1+\epsilon)}-\frac{2}{(1+\epsilon)^{3}})$$ is negative, then we have:
$$0\leq(\frac{2}{(1+\epsilon)}+(\frac{\dot{x}_{0}}{V})^{2}(\frac{1}{(1+\epsilon)}-\frac{2}{(1+\epsilon)^{3}}))\leq\frac{2}{1+\epsilon}\leq{2}$$
whereas if the internal bracket is positive, then the whole bracket is maximized when V is minimized, i.e, at the apogee where $$\frac{dY}{dx}=0$$

Thus, in order to neglect the bracket expression with respect to unity, we must either have
$$\frac{2V^{2}}{gR}\leq\frac{2V_{0}^{2}}{gR}=2\gamma<<1$$
or:
$$\frac{\dot{x}_{0}^{2}}{gR(1+\epsilon)}\leq\frac{V_{0}^{2}}{gR}=\gamma<<1$$

It remains to be shown that $$\gamma<<1\to\epsilon<<1$$
This is easiest done by energy conservation, noting that
$$\gamma<<1\to\frac{V_{a}^{2}}{gR}<<1$$
where $$V_{a}$$ is the speed at the apogee.
It follows, that the allowable value of $$\epsilon$$:
$$\gamma<<1\to\epsilon\approx\frac{\dot{y}_{0}^{2}}{2gR}\frac{1}{1-\frac{\dot{y}_{0}^{2}}{2gR}}<<1$$