1. Nov 19, 2004

### Machinus

This problem is pretty complicated, and I have fooled around with it in Maple but I don't understand the concepts well enough to work it out yet.

Suppose we have a simple hyperbolic orbit where there is a central mass M and a smaller mass m, and there is more than enough energy in this system for the smaller mass to whiz by and fly back out into space, reasonably deflected. For this problem we have at least two conservation principles to work with, one being total energy

$$E=\frac{1}{2}m\vec{v}^2-\frac{GMm}{r}+\frac{ml^2}{2r^2}$$

$$\frac{d}{dt}E=0$$

and total angular momentum

$$L=mr^2\dot\theta$$

$$\frac{d}{dt}L=0$$

I am attempting to solve for maximum $$\dot{r}$$, with as few variables remaining in the equation as possible.

I have attempted to solve for $$\ddot{r}=0$$ and simplify that, but it is really complex, due in part to the transformation to polar coordinates:

$$\vec{v}^2=\dot{r}^2+r\dot{\theta}^2$$

If anyone knows of a solution or a way to work this out that they think I will be able to do, I would really appreciate the help.

Last edited by a moderator: Nov 19, 2004
2. Nov 20, 2004

### Jenab

It has been a while since I had a look at a celestial mechanics textbook other than the one by Dubyago. What does the variable (lower case l) represent? It seems to have units of m^2 sec^-1. What form of energy is that whole third term in that equation supposed to be?

Jerry Abbott

3. Nov 20, 2004

### BobG

If you're only concerned about the motion of the object, not it's actual energy. You can toss the mass out of the equation and use specific energy per unit of mass. You can also toss out rotational kinetic energy.

$$E=\frac{1}{2}v^2-\frac{GM}{r}$$

You can do the same with the angular momentum. You want specific angular momentum per unit of mass. For trajectories, your specific angular momentum is the cross product of the radius and velocity vectors.

$$H=\vec{r}\times\vec{v}$$

Something has to be given. Since you're talking about hyperbolic orbits, you want E to be some value greater than 0 (If E=0, you have a parabolic orbit; if E<0, you have a closed elliptical or circular orbit).

Your max velocity is going to depend on the lower limit you set for r (there's some point where r is smaller than the central mass's radius).

4. Nov 20, 2004

### Machinus

This doesn't make any sense to me. The lower limit of r is simply where $$\dot{r}=0$$, and it is not smaller than the radius of the central mass. The given is $$E_{0}$$, which is the maximum energy. Can you explain what you are talking about?

5. Nov 20, 2004

### marcus

I think the maximum radial velocity is c and the maximum is achieved
whenever some light goes whizzing past any massive object and is slightly deflected.

Eventually the speed of the light is almost all radial---when it is again far from the object.

But let's be definite and say that the object is a Schwarzschild black hole and that the light comes just to within 2 Radii of the center of the hole.

Someone may wish to calculate this more accurately but as a rough estimate, at least, the light is deflected by an angle of one radian.

One radian is, I hope, what you would call a "reasonable" deflection.

at times minus infinity and plus infinity the radial component of the light's speed is c

Most of the rest of the time it is approximately c.

Now, if you do not like light, then you can take a small rocket ship that is traveling nearly c, say at 0.999 c.
then it will follow essentially the same geodesic and do approximately the same thing as the light.

and so the maximum is c, although a material object like a rocket ship can only approximate the maximum arbitrarily closely but cannot achieve it.

6. Nov 20, 2004

### Machinus

This is incorrect. The maximum radial velocity does not occur at infinity. The gravitational potential well increases the kinetic energy of the object and the radial velocity is highest between time=0 and the turning point of the object. I am attempting to solve for this energy.

This is easily seen since the maximum total velocity occurs when r reaches its minimum point. The maximum radial velocity occurs some time shortly before and after this point, however I am unable to determine precisely when.

7. Nov 20, 2004

### marcus

to me, since the light is always traveling at speed c, it is not too important to ask "at what point" does it have maximum speed

it has the same speed, namely about 300,000 km/s, all along its trajectory.

But it is only at minus and plus infinity that the velocity is purely radial.

at other points it has a (possibly small) transverse component

but none of this matters because your main question is what is the
maximum radial velocity, and I believe the answer is c, or approx
300,000 km/s

This would also be the case for a small object like a rocket ship which is launched towards the massive object at some speed very close to c,
like 0.999 c,
the maximum radial velocity (which could be achieve arbitrarily closely) is again c.

In this case, of a material object, I am not telling you where on the trajectory it would reach top radial speed, but simply that the least upper bound is c.

Last edited: Nov 20, 2004
8. Nov 20, 2004

### Machinus

I stated in my post that the object has a mass m. Photons do not have mass. Light is not an appropriate example for this orbit.

The radial velocity reaches a maximum at some point between infinity and the minimum r. The fact that the velocity is all radial at infinity is irrelevant to finding the maximum $$\dot{r}$$.

9. Nov 20, 2004

### marcus

the massive particle that is sent at the black hole at speed 0.999 c
will follow approximately the same trajectory as the photon

if you want to make the approximation better, send a new particle at
0.9999 c

and then one at 0.99999 c

the particles follow geodesics in spacetime---the spacetime created by the massive central body, the black hole. approximately the same paths as photons of light

if you think about it in an open-minded fashion, I believe you will come to the conclusion that the maximum radial velocity (what you asked about in your question) is c.

10. Nov 20, 2004

### Machinus

What the balls are you talking about? There is no black hole here and this problem has nothing to do with relativity AT ALL. There are no speeds even close to c anywhere in this problem. Please stop saying retarded things in my thread.

11. Nov 21, 2004

### Jenab

Q = true anomaly (perihelion at Q=0)
u = hyperbolic eccentric anomaly

You choose Q, then...

u' = ArcCosh {(e + cos Q) / (1 + e cos Q)}

Or, equivalently,

B = (e + cos Q) / (1 + e cos Q)

u' = ln { B + ( B^2 - 1 )^0.5 }

Then...

if sin Q < 0 then u = -u' else u = u'

a = semimajor axis
GM = solar gravitational parameter = 1.32712440018E+20 m^3 sec^-2

The heliocentric distance is found from

r = a (e^2 - 1) / (1 + e cos Q)

The canonical state vector is found as follows:

x = r cos Q
y = r sin Q

Vx = -(a/r) { GM / a }^0.5 sinh u
Vy = +(a/r) { GM / a }^0.5 (e^2 - 1)^0.5 cosh u

V = [ (Vx)^2 + (Vy)^2 ]^0.5

Getting the radial velocity, Method 1

Vr = (x Vx + y Vy) / r

Getting the radial velocity, Method 2

The angular momentum per unit mass is

h = x Vy - y Vx

The transverse velocity has magnitude

Vt = h/r

The radial velocity is found from the Pythagorean theorem.

Vr = [V^2 - Vt^2]^0.5

You could set up a program to find the value of Q for which Vr is a maximum. Or you might do it analytically by combining the equations and differentiating the result.

I just now wrote such a program, and here's what things look like on the outbound hyperbolic branch.

Fields:
Semimajor axis (AU)
eccentricity
asymptotic true anomaly (deg)
true anomaly for maximum radial speed (deg)
eccentric anomaly for maximum radial speed (deg)
heliocentric distance (AU)
canonical x coordinate (AU)
canonical y coordinate (AU)

1, 2, 120, 90, 75.456, 34.392, 3, 0, 3
2, 2, 120, 90, 75.456, 24.319, 6, 0, 6
1, 4, 104.378, 90, 118.214, 30.761, 15, 0, 15

Apparently, the maximum outward radial velocity for a hyperbolic orbit always occurs at a true anomaly of 90 degrees, and the maximum inward radial velocity always occurs at a true anomaly of -90 degrees.

Jerry Abbott

Last edited: Nov 21, 2004
12. Nov 21, 2004

### BobG

I think I forgot the title by time I finished reading the text.

I take it you only want to know the radial velocity. That is a coordinate transformation. For a closed orbit, the principal direction of a perifocal coordinate system is perigee. That should apply to hyperbolic orbits, as well.

To break the velocity into radial and tangential coordinates, you want the radius as the principal direction (a non-inertial system). You have to know the angle between the radius and the direction of perigee to relate the two.

The transformation is kind of tedious, but it winds up that for an elliptical orbit, radial velocity is (I don't have a reference availabe, so I hate to guranatee this):

$$v_r=\frac{eh}{p}sin \nu$$
h is angular momentum and stays constant.
e is eccentricity and statys constant.
p is the semi-latus rectum.
nu is the true anomaly.

Since the sine of an angle is at its max at 90 degrees, the maximum radial velocity always occurs at the semi-latus rectum. I'm pretty sure that's true for hyperbolic orbits as well as closed orbits.

The semi-latus rectum for closed orbits is:

$$p= a(1-e^2)$$

The semi-latus rectum for hyperbolic orbits is:

$$p= a(e^2-1)$$

Edit: a is the semi-major axis.

$$r=\frac{p}{1+e cos \nu}$$

For hyperbolic orbits, multiply by negative 1 (you don't want a negative radius)

The transformation is based on the idea that, if the radius is the principal direction, then the object's position is:

$$\vec{r}=r\hat{U}$$

Velocity, being the derivative of position, is found using the product rule.

$$\dot{\vec{r}}=\dot{r}\hat{U}+r\dot{\nu}\hat{V}$$

$$\hat{U}=cos\nu_{e1}+sin\nu_{e2}$$
$$\hat{V}=-sin\nu_{e1}+cos\nu_{e2}$$

Through some cross product differentiation:

$$\dot{\nu}=\frac{h}{r^2}$$

When I think about it, a person might get wrapped up over how a hyperbola could have a semi-major axis. It still works, even if it doesn't seem quite right:

$$a=-\frac{\mu}{2E}$$
E is specific energy
mu is GM (universal gravitational constant times mass of attracting object)

Try this link for a little more detailed description of how hyperbolic orbits are a little different than closed orbits.

http://scienceworld.wolfram.com/physics/HyperbolicOrbit.html

Last edited: Nov 21, 2004
13. Nov 21, 2004

### Jenab

For hyperbolas, the semimajor axis is the distance from the intersection of the asymptotes to the perihelion. The product ae is the distance from the intersection of the asymptotes to the sun (i.e., the focus of the hyperbola).

14. Nov 21, 2004

### pervect

Staff Emeritus
Suppose we have a simple hyperbolic orbit where there is a central mass M and a smaller mass m, and there is more than enough energy in this system for the smaller mass to whiz by and fly back out into space, reasonably deflected. For this problem we have at least two conservation principles to work with, one being total energy

$$E=\frac{1}{2}m\vec{v}^2-\frac{GMm}{r}+\frac{ml^2}{2r^2}$$

I'm guessing that the problem in your understanding is in this equation. (I could be wrong, it could just be the usual problem of me not understanding someone elses notation)

The key point is that we can re-write

$$\frac{1}{2} m \vec{v}^2$$ as
$$\frac{1}{2} m (\dot{r}^2 + (r \dot{\theta})^2)$$

This is because the radial velocity is perpendicular to the angular velocity. Another way of saying this is that a unit vector in the radial direction $$\hat{r}$$ is prependicular to a unit vector in the theta direction $$\hat{\theta}$$

Now, given that $$L = m r^2 \dot{\theta}$$, we can re-write the expression for E as

$$\frac{1}{2} m (\dot{r}^2 + \frac{L^2}{m^2 r^2}) - \frac{GmM}{r}$$

Substituting l = L/m gives

$$E = \frac{1}{2} m (\dot{r}^2 + \frac{l^2}{r^2}) + GmM/r$$

Writting EE = E/m, we can eliminate m from the equation

$$EE = \frac{1}{2} (\dot{r}^2 + \frac{l^2}{r^2}) + GM/r$$

Solving for the minimum distance r is just arithmetic. Let $$\dot{r}$$ = 0, which it must be at the minimum, and solve for r as a function of EE, l, and M.

Some physical insight can be gained by noting that the equation of motion of the particle for r can be expressed in terms of a one dimensional potential. But I won't get into that, I think I've written about it before on this board (perhaps not too clearly). The whole issue is discussed in texts such as Goldstein's "Classical Mechanics", though, if you want a textbook reference.

Last edited: Nov 21, 2004
15. Nov 21, 2004

### Machinus

pervect:

I tried all those calculations already, and I thought I had made that clear by posting the same ones in my post that you have used in yours. I transformed to polar and I have already tried to utilize the two conservation principles.

You have misread my post. I want to solve for $$\ddot{r}=0$$. This is obviously not the same problem as solving for $$\dot{r}=0$$, and it certainly much more difficult than just arithmetic.

I have already realized that there are five zeroes (two infinite, one zero, two finite) of $$\ddot{r}=0$$, and that the two I am looking for occur near the central turning point of the orbit. However I have no function for these zeroes, and I haven't been able to derive an equation that expresses them in reasonably simple terms.

16. Nov 22, 2004

### pervect

Staff Emeritus

The expresion for $$\ddot{r}$$ is fairly simple

$$m \ddot{r} = m r \dot{\theta}^2 - GmM/r^2$$

The easiest way to derive this is from the Lagrangian, there's probably other ways of doing it. As a matter of fact, you say that you already know the formula

$$\dot{r} = \sqrt{2 e + 2 G M / r - l^2/r^2}$$

here e = E/m, and l = L/m, the energy and momentum per unit mass
(I got this from my last post, I fixed up a sign error too :-().

So now you take

d(rdot)/dt = d(rdot)/dr * dr/dt

or

$$f(r) = \sqrt{2 e + 2 G M / r - l^2/r^2}$$

and $$\ddot{r} = f(r) \frac{df}{dr}$$;

It's annoying to work this out by hand, but you say you have maple, so that makes it a breeze.

The result is l^2/r^3 - G M / r^2, equivalent to the result in my first line.

This isn't hard to solve for r = l^2/GM then you'll have to back substitute to find the appropriate value of rdot.

You can do a quick sanity check on the answer for r-double-dot by seeing that it correctly predicts the radius of a circular orbit.

I hope I didn't make any more typos/sign errors, the general approach should be right though.

17. Nov 23, 2004

### pervect

Staff Emeritus
So did that help? I don't know anything about your background, so I don't know exactly at what level to reply, I hope I didn't come across as condescending, I also hope I didn't zip by too tersely. This happened to be rather similar to another problem I was just happening to work on BTW.

18. Nov 23, 2004

### Machinus

I am very familiar with the equations you are using. I am a 3rd year undergraduate but I have already taken all the physics courses I need to save E&M theory. What I am trying to get out of this problem is a way to express maximim $$\dot{r}$$ as simply as possible. I am having some trouble accepting L and l to be the same quantity - but keeping mass in the solution is irrelvant to me since v << c and it will be a constant.

The lagrangian does provide a short formula which contains $$\dot{r}$$, but it gets complicated when you take the derivative.

Someone suggested that the quantity I am looking for occurs at $$\pm90^o$$ relative to $$\ddot{\phi}=0$$. Is there any way to check this in your solution?

19. Nov 23, 2004

### pervect

Staff Emeritus
Huh?

What's complicated about mr'' = -GmM/r^2 + mrw^2 where w = d(theta)/dt?

In addition, L and l are not the same quantity. l is L/m

You might try re-reading my post - your response indicates that you clearly don't understand what I've just said, but it also that it's not because you lack the necessary background.

If you ask for help, you might actually want to make an effort to look at it when it's offered.

Last edited: Nov 23, 2004
20. Nov 23, 2004

### BobG

It's hard to figure out what you're asking.

Max speed occurs at the vertex.
Max tangential velocity occurs at the vertex.
Tangential acceleration is zero at the vertex.

Max radial velocity occurs at a true anomaly of 90 degrees.
Radial acceleration is zero at a true anomaly of 90 degrees.

If you're looking for a way to find the radial velocity, you take the derivative of the radius.

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

All of the elements are constant except true anomaly.

$$\dot{r}=\frac{(1 + e cos \nu)*0-[a(e^2 - 1)*(0 - e sin \nu)*\dot{\nu}]}{(1 + e cos \nu)^2}$$
$$\dot{r}=\frac{ae(e^2 - 1)\dot{\nu}sin\nu}{(1 + e cos \nu)^2}$$

The derivative of true anomaly is:

$$\dot{\nu}=\frac{h}{r^2}=\frac{h}{\left(\frac{a(e^2 - 1)}{1 + e cos \nu}\right)^2}$$

Substituting into the derivative of the radius (convering to simple fraction along the way)

$$\dot{r}=\frac{\left(ae(e^2-1)\frac{h(1 + e cos \nu)^2}{a(e^2-1)^2}sin \nu\right)}{(1 + e cos \nu)^2}$$
$$\dot{r}=\frac{ea(e^2-1)h sin\nu}{(a(e^2-1))^2}$$

Eventually, simplifying to:
$$\dot{r}=\frac{eh sin \nu}{a(e^2-1)}$$
$$\dot{r}= \frac{eh}{p}sin \nu$$

In terms of your orbital elements, h is equal to:

$$h=\sqrt{\mu a(e^2-1)}=\sqrt{\mu p}$$

Angular momentum stays constant, so there's not much advantage to substituting this into your equation (it makes it look ugly).

$$\ddot{r}=\frac{eh}{p}cos \nu$$

You get radial acceleration. (This won't give you a big picture of your velocity, since we're not looking at the tangential velocity).

Edit: I take it back. You should make the substitution.

$$\dot{r}=\frac{e\mu^{1/2}}{\sqrt{p}} sin \nu$$

That way, you have emu to the 1/2 over the square root of your semi-latus rectum times the sine of true anomaly. That should get some strange looks.

Last edited: Nov 23, 2004