# Deflection angle of a comet traveling around the Sun

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1. Jan 21, 2012

### Leb

1. The problem statement, all variables and given/known data
A comet is approaching the Sun from a vast distance with velocity V. If the Sun exerted no force on the comet it would continue with uniform velocity V and its distance of closest approach to the Sun would be p. Find the path of the comet and the angle through which it is deflected.

2. Relevant equations

$\frac{d^{2}u}{d\theta^{2}}+u=\frac{\gamma}{h^{2}}$, where $\gamma=GM$

$u=\frac{\gamma}{h^{2}}+A\cos{\theta}+B\sin{\theta}$

Since $\dot{r}=-Vcos{\alpha}≈-V$ where $\alpha$ = the angle between
V and R (R is the distance from the Sun to the comet, green angle in the diagram).

Using $\alpha$≈0 → $h=RV\sin{\alpha}=pV$

At t=0, choose $\theta=0$ and u=1/R≈0, du/dθ=-$\dot{r}$/h=1/p

→$u=\frac{\gamma}{p^{2}V^{2}}-\frac{\gamma} {p^{2}V^{2}}\cos{\theta}+\frac{1}{p}\sin{\theta}$

Then u → 0 again, when θ=0, or 2∏-2δ, where $\tan{\delta}=\frac{pV^{2}}{\gamma}$ Where 2δ is shown below.

and $h=r^{2}\dot{\theta}$ In the usual polar coordinates.

3. The attempt at a solution
What I do not understand is the definition of δ. Especially when it comes to introduction of tanδ ... Can someone explain why we can write tanδ as above ?

I tried to provide as accurate drawing with my paint skills as I could. It was drawn in such a way, the deflected path almost intersected the straight line (almost like a reflection h, where a connect the shortest side with the unique angle of an isosceles triangle )

This example is actually taken from M.Lunn's A First Course in Mechanics, 1991, p33-34.

Last edited: Jan 22, 2012
2. Jan 21, 2012

### Simon Bridge

I think the notation needs a tidy up:
isn't that y in the first line supposed to be a u?
what is the relationship between R and r?

some of these letters seem to be vector quantities in one place and scalar quantities i another.

delta is defined on a rt-triangle with short-sides pV2 and GM (or is it (pV)2 ?)
the diagram says that this is also the half-angle between initial and final positions shown.
so the problem is to relate the definition with the geometry.

note: if the initial distance to the sun is r, then the diagram seems to be saying $\tan\delta = p(r^2-p^2)^{-1/2}$
however, I think tidying the notation, and thinking about what each bit means, will help.
also tidying up that diagram so it better reflects the kinds of trajectories to expect.

other notes:
the trajectory is either a parabola or a hyperbola.
the description of the problem makes more sense for a hyperbola.
would delta be the half-angle between asymptotes?

basically, tan(delta) is a tightness-of-fit parameter for the hyperbola
... call it just F, then your equation has form:

u=F-Fcos(θ)+sin(θ)

if F = 1 then the deflection angle is π/2 (from the initial path)
F < 1 the the deflection is less
F > 1 it is more.

the deflection angle is π - 2δ ...
for a deflection angle of π/2, δ=π/4, tanδ=1 = F ...

Last edited: Jan 21, 2012
3. Jan 22, 2012

### Leb

Yes, y should have been u, sorry for the mistake.

I scanned the problem with the most important parts included.

Simon, I get the tanδ epression in terms of p and r, however, I do not get the on in terms of pV^2 and GM

http://img835.imageshack.us/img835/7707/probn.gif [Broken]

Last edited by a moderator: May 5, 2017
4. Jan 22, 2012

### Simon Bridge

the delta is the half-angle between the asymptotes of the hyperbolic trajectory.
The bigger G the smaller delta, the bigger V the bigger delta... follows that delta has to be related to G and V (and p and M too - same way). This is the relationshhip that falls out.

hat I showed you at the end of last post was where this comes from - the derivation did a naughty thing and anticipated the result. fig. 2.8 is very misleading.

5. Jan 22, 2012

### Simon Bridge

I had a fiddle with octave to print a better representations for you:

The blue "o"'s represent the position of the comet at equal times. The comet enters horizontally from the right, bends around the mass, and exits lower.

The first plot shows you three trajectories with tanδ=1.5, 1 and 0.5 from widest to tightest. Since M and p are the same for each run, these correspond to different incoming speeds.

The mass is at the black "o", the red lines are asymptotes, and the green line is the symmetry axis of the trajectory. δ is the angle between the green line and the red ones.

The second plot isolates the tanδ = 0.5 one so you can see it clearer.

Plots generated from the following code.
Code (Text):
$trajectories$ cc by sa 2012 Simon Bridge
## www.physicsforums.com

#initialize
a=0;                  # start angle
b=2*pi;               # finish angle
N=1024;               # number of points
dA=(b-a)/(N-1);       # step size

A=a:dA:b;             # angle axis

d=1.5;                # d=tan(delta)=pVV/GM
p=1;                  # impact parameter

# find radius as a function of angle
u=(1/(d*p))-(1/(d*p))*cos(A)+(1/p)*sin(A);
r=1./u;

# isolate the bits close to the mass
r(find(r>10))=nan;
r(find(r<0))=nan;

# convert to cartesian coordinates
x=r.*cos(A);
y=r.*sin(A);

# build the asymptotes:
asx=-3:0.05:10;
hz=p*ones(size(asx));

m=tan(2*atan(d));
k=p-(m*p)/d;
y1=-d*asx;
y2=-m*asx+k;

# plot:
# trajectory (blue)
# symmetry axis (green)
# asymptotes (red)
# location of mass (o)

plot(x,y,asx,y1,asx,hz,"r",asx,y2,"r",0,0,"ok",x,y,"bo")
box
grid on
axis([-3,10,-10,2])
(this code carries a copy-left creative commons license. You don't know what that is http://www.creativecommons.org.nz/licences_explained__1 [Broken] it's at the bottom. It's no big deal - it just means if you want to use it someplace else then please say where you got it. Nobody here would think of doing anything else right?)

#### Attached Files:

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• ###### traj2.png
File size:
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Last edited by a moderator: May 5, 2017
6. Jan 22, 2012

### Leb

Wow, I really appreciate that you took the time to graph and explain. When looking at graph 2.8 I had no idea how was that supposed to be the 'half-angle'.

However, I am too daft to interpret $pV^{2},\gamma$ geometrically. You said, that those are the shorter sides of a right triangle, but how did you know which one is which ? I tried thinking in terms of the gravitational force exerted by the Sun on the comet, but with no progress...

When you said:

basically, tan(delta) is a tightness-of-fit parameter for the hyperbola
... call it just F, then your equation has form:

u=F-Fcos(θ)+sin(θ)

How did you manage to express that while having u untouched, and 1/p changed to one, in front of the sin(θ) term ? That is assuming you are working with u=$\frac{\gamma}{p^{2}V^{2}}$−$\frac{\gamma}{p^{2}V^{2}}$cosθ+$\frac{1}{p}$sinθ

Again, sorry for not being able to understand.

7. Jan 22, 2012

### Simon Bridge

I kept wanting to draw diagrams on a whiteboard - and when I saw the diagram they gave you it was pretty obvious what the problem was.

Definition of the tangent. Remember SOH CAH TOA?

In order to investigate an equation, you can put any values in that you like. The trick is finding useful values.

p is what we call an "impact parameter" - it is how close the comet would have got if it was not deflected. For an arbitrary value p $$u=\frac{1}{p}(F-F\cos\theta + \sin\theta) \; : \qquad F=\frac{\gamma}{pV^2} = \frac{1}{\tan\delta}$$ So doubling p just halves u, and doubles r=1/u, so it is a scale factor. All values give me exactly the same shape, just bigger or smaller.

By putting p=1, I was basically defining the unit of distance on the graph. So if your comet would have passed at 1AU, then the x and y axis are labelled in AU and V is in AU/time. If it would have passed 1 light-second away, then the x and y labels are in light-seconds and V would be in light-seconds/time. See?

To investigate the equation, now, I just need to play with V$$F=1 \Rightarrow V=\sqrt{\frac{\gamma}{p}}$$ and, of course, p=1. So the initial speed is the square-root of an energy term. If it is exactly this square-root, then the deflection angle is exactly π/2 radiens.

To put this in terms of energy, just multiply both sides by m/2 so that this means:[tex]\frac{1}{2}mV^2 = \frac{1}{2}\frac{GMm}{p}[/itex]... the LHS is the initial kinetic energy and the RHS is half the potential energy at distance p.

That should put you well on your way.

8. Jan 23, 2012

### Leb

Simon Bridge, thank you very much for your time, you gave some valuable insights.

However, I was only able to work out tanδ using the following:

We look for $\cos{\theta}-m\sin{\theta}=1, \qquad \tan{\delta}=m$

If $t=\tan{\theta/2}$ then $\sin{\theta/2}=\frac{2t}{1+t^{2}}, \cos{\theta/2}=\frac{1-t^{2}}{1+t^{2}}$

Sub these values in the first equation to get $2t^{2}+2mt=0$
So either t=0 or t=-m. For t=0, we get θ=2nπ, if t=-m, we get θ=2nπ-2δ, n in Z.

When I said that I do not understand how to interpret $pV^{2},\gamma$, I meant that I have no Idea how could these create sides of a right triangle (I know the SOC CAH TOA ) and furthermore, how to know which side is which. That is, how to view it physically. I have never seen something like this before.

This mechanics course is actually taught in mathematics.

9. Jan 23, 2012

### Simon Bridge

sure:- using my example: $u=F(1-\cos\theta+(1/F)\sin\theta)$
we want the angle between the asymptotes;
the asymptotes are at θ: r→∞ i.e. u=0, so ...
$\cos\theta=1+m\sin\theta : 1/F = \tan\delta = m$ ...(1)

This relation will work for two angles - the first being 0 (comet starts a log way away).
The other is the critical angle C. Which is the angle to the other asymptote measured anticlockwise from the the first one (see the plots - C goes around the outside.)

(If you run that octave code, but plot r against angle, you'd see this effect graphically.)

Fortunately, this means C and delta are related by $C+2\delta=2\pi$ ... (2)

Now you have two equations and two unknowns.

This is what you did but more direct.

The deflection-angle D is usually defined to be measured anti-clockwise from the straight-through direction ... so D = C - π = π - 2δ.

I was able to go into so much detail because it did not involve actually doing the problem for you :) however, getting to those insights is very likely the whole point of giving you the exercise.

As for the triangles - you can make a triangle out of anything. It just describes a relationship. It used to be that all math was done geometrically like that. Any fraction can be represented as a tangent ... the second you say "a/b=tanA" you have just put a and b on a triangle. This is also what you are doing when you make a trig-substitution in calculus.

In physics you want a and b to have the same units... which suggests to me that M is measured in cometary masses or something. (GM has an extra unit of inverse-mass otherwise.) The triangle describes the relationship between kinetic and potential energies at particular places in the trajectory and is a direct consequence of the conservation of energy.

What I considered "naughty" was that this relationship was just plucked out of the air and inserted - which you noticed. Engineering people do that all the time when they are teaching - it is supposed to be a shortcut. If it were me, I'd have not mentioned it directly like that - but maybe I'd have given a hint to find the half-angle between the asymptotes. I would also have given you a better diagram... or got you to sketch one as part of the exercise.

The way to look at this is to see the green-line in my plots as the x-axis, with x=0 at the place where the two red lines cross. From there you can think of the usual equation for the hyperbola you learned when you did conic sections. Remember how to work out the angle of the asymptotes? Your tanδ will drop out of that automagically.

[short speech:]
When I was in your place, I found it very useful to have a computer plotting the actual functions for me ... this would tell me what would be a useful line of inquiry. Actually, it's still useful. I had to write my own program to do this way back then - in between dodging T-rexs and eating caterpillars that is. Hmmmm .... caterpillars....

You have this great advantage that there are a range of math scripting languages available to you for a variety of prices - octave is free/gratis and free/libre and you can own it, mathematica is available for \$ but iirc wolfram will let you use an online engine for free, and matlab is very widely used at colleges. There's no end of others.

I personally consider a basic knowledge of octave, as well as LaTeX, to be about equally important - I predict you will find they become more important as your studies progress too.