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Planetary Motion
By Brad Rodgers (P1930) on Tuesday, January 02, 2001 - 11:57 pm:

How is the result for planetary motion derived? I have been able to manipulate equations so far to get:

t=1/GM(Cr4+2GMr3)1/2+D

Where t is the amount of time taken, r is the distance from the original planet and C and D are unknown constants. I'm not sure whether this is a good way to show that a planet travels in an ellipse, or if it is, what substitutions to make after this.

Thanks,

Brad

By Michael Doré (Md285) on Wednesday, January 03, 2001 - 12:27 am:

Brad - I'm not sure how you've got that. Are you considering motion in a straight line (i.e. if a "planet" is released from rest and allowed to drop into the sun)? Otherwise there should be an angle involved somewhere; and the motion should be periodic in some cases, so there should be infinitely many solutions for t, given r.

Are you wanting to find the curve a particle takes in an inverse square law field (this is a good approximation to planetary orbits when the planet has negligible mass compared to the sun)? Are you interested in calculating the position of the particle at a certain time, or are you only interested (for the moment) in the curve?

I'll assume the latter. In this case we need to transform into polar co-ordinates. To save writing cos and sin all over the place, let's have our particle in the Argand diagram. So the complex number representing its position is:

z = reix

where the origin is the attractor (the sun). r is the distance of the particle away from the sun after time t, and x is the angle the particle has travelled through after a time t. So x and r are functions of t.

Now we need to establish two important facts. At a time t, the acceleration of the particle can be split up into two components. One component along the radius - that is in a direct line between the sun and the particle. And there is another component perpendicular to this - in the direction of increasing x. Let ' denote differentiation with respect to time, then we need to show that the radial acceleration is:

r'' - rx'2

And the perpendicular component is:

(1/r)( d/dt(r)2x')

As a hint for doing this, show that the complex number representing our position in the Argand diagram, that is z = reix satisfies:

z'' = (r'' - rx'2)eix + (2r'x' + rx'')ei(x + [pi]/2)

When you've got these two results we can begin to show that the orbit is an ellipse, parabola or hyperbola. If you get stuck let me know.

Yours,

Michael

By Brad Rodgers (P1930) on Wednesday, January 03, 2001 - 01:59 am:

I got my result by first saying that d2r/dt2=-GM/r2

I then proceeded from there in the ordinary way. I think that it could be useful if I could figure out how to put in the constants. The problem I had with putting in constants is that to solve I had to use the absract concept of m=dr/dt, which I didn't know how to deal with. I think my original idea might be flawed anyway. It could be salvaged I suppose to find the time of a falling object assuming the inverse square law.

I'm not sure I entirely understand what you mean by radial acceleration. I would think that this would be defined to be r"

I can get the result of z"=... though. All you have to do is use the product rule several times and then change ieix into ei(x+p/2) because cos(x+pi/2)=-sinx and sin(x+pi/2)=cosx. I'm not sure I know what to do for the rest of it though as it seems like the radial component would be r" and the vertical component would be x". I'm somewhat unfamiliar with the Argand diagram, I just need a little more explanation. (I've noticed that we have the radial component in z" being multiplied by eix and the x component being multiplied by ei(x+p/2)-this surely has something to do with it, I'm just not sure exactly what the x and r component are).

Thanks,

Brad

By Michael Doré (Md285) on Wednesday, January 03, 2001 - 02:08 am:

So your original question is to find r as a function of time, in the case where the particle is dropped vertically? I tried something very similar in my first message to this board here. (The only difference was I accounted for the motion of both bodies, but this is minor.) The solution which AlexB gives, while very clever, is nothing like as simple as the one you've got.

The radial acceleration means the component of the instantaneous acceleration vector, which is in the direction of the radius. At first sight you might think it's r'' but after a while you realise it's not. Think of whirling round a stone on a string in a circular path - the tension in the string causes inward (radial) "centripetal" acceleration (v2/r) yet r'' = 0. It is true that the inward velocity is r', but the tangential velocity can also contribute to the inward acceleration (when differentiated). This is because the radius is not a fixed vector in space, but depends on t.

Michael

By Michael Doré (Md285) on Wednesday, January 03, 2001 - 04:43 pm:

As you've derived the result:

z'' = (r'' - rx'2)eix + (2r'x' + rx'')ei(x + p/2)

we may as well go from there. Remember z is the complex number representing our position as a vector (so if our position is 3 + 5i, this is the same as saying we have position vector 3i + 5j). z = reix is always true (z, r and x are functions of time). r is the distance of the particle from the origin, and x is the angle it makes with some fixed arbitrary line through the origin. Are you okay with all this? It is basically a consequence of Euler's relation: eix = cos x + i sin x.

Notice that complex numbers add in the same way as vectors. It quickly follows that the complex number representing the particle's vector acceleration is z''. So the complex number representing the particle's vector acceleration is:

z'' = (r'' - rx'2)eix + (2r'x' + rx'')ei(x + p/2)

Now consider the force on the particle which causes this acceleration. It is towards the centre, and of magnitude F(r). (Later we will set F(r) = A/r2 for constant A.) So the complex number representing the vector force is:

-F(r) eix

because eix is a unit complex number pointing from the origin to the particle.

The acceleration a = F/m, we therefore get:

(r'' - rx'2)eix + (2r'x' + rx'')ei(x + p/2) = -F(r) eix/m

Divide through by eix and write eip/2 = i:

(r'' - rx'2) + i(2r'x' + rx'') = -F(r)/m

Now we can equate real and imaginary parts:

r'' - rx'2 = -F(r)/m
2r'x' + rx'' = 0

Let's have a look at the second equation. If we multiply it by r we get:

2rr'x' + r2x'' = 0

Notice that this can be written:

d/dt(r2x') = 0

So r2x' is constant. Therefore the particle sweeps out area at a constant rate. This is known as Kepler's Third Law. Notice that we haven't even assumed the inverse square law to derive this! All we needed to know was that the force on the particle was central. Newton was the first to realise this, although he did not derive the result in this way.

So anyway, we can write r2x' = k. Now the first equation is:

r'' - rx'2 = -F(r)/m

And we can write -F(r)/m as -A/r2 for convenience, where A (>0) depends on G and M. So:

r'' - rx'2 = -A/r2

where r2x' = k. See if you can get anywhere with these... (Hint: it is useful to work with u = 1/r.)

Yours,

Michael

By Michael Doré (Md285) on Wednesday, January 03, 2001 - 10:48 pm:

If you want to show the curve is an ellipse/parabola/hyperbola then the trick is to get rid of the time variable as quickly as possible.

Let's start off with the more general equation:

r'' - rx'2 = F(r) (*)

with r2x' = k

Now the best thing to do is to change variables u = 1/r. I can't really give you a good explanation of how you would think to do this I'm afraid; it just simplifies the equations.

Our aim is to write (*) in terms of u and x only (so we need to eliminate time).

Firstly x' = ku2. Also:

r = 1/u

r' = (-1/u2) du/dt = -1/u2 du/dx x' = (-1/u2) du/dx (ku2) = -k du/dx

Differentiate again:

r'' = -k d2u/dx2 dx/dt = -k2u2 d2u/dx2

Substitute these into (*):

-k2u2 [d2u/dx2 + u] = F(r)

If you substitute F(r) = -A/r2 = -Au2 and divide through by u2:

d2u/dx2 + u = c

where c is some new constant. I'm sure you can solve this for u. Then it is simply a matter of writing r = 1/u and showing that the relationship between r and x is indeed that of an ellipse, hyperbola or parabola. If you want to work out the orbital period, work out the area of the ellipse and the (constant) value of r2x', and find their ratio (can you see why this works?) Tell me if you get stuck.

Michael

By Michael Doré (Md285) on Thursday, January 04, 2001 - 11:49 pm:

First of all complete the square in the denominator, and then set a suitable expression as cos q. Try to utilise the identity cos2 + sin2 = 1.

There is a neater (and very standard) way of going about it. First of all let w = u - c. Then:

d2w/dx2 + w = 0.

This is the same DE as a particle on a spring that obeys Hooke's law (except w would be the extension and x would be time).

As a clue for solving this, write it as:

d2w/dx2 + i dw/dx - i dw/dx + w = 0

So if you let v = dw/dx + iw then:

dv/dx - iv = 0

If you solve these for v then w and u then you should arrive at:

u = A cos x + B sin x + c

where A and B are arbitrary.

By Brad Rodgers (P1930) on Friday, January 05, 2001 - 10:52 pm:

How do you find the exact orbit given vo and ro?

I've had some thoughts on that question, here are my ideas so far. First of all, I'm going to simplify the equation to that with only a sine factor e.g. u=sin(x+D)+C. From this, I know that the D term only tells at what angle the ellipse/hyperbola axis of symmetry ends up being tilted at. So we can ignore it because the angle of view is relative. We also know that C=GM/-k2. I know that we can let the r term in k be ro, and I think that vo equals x' at t=0, but I'm not too sure about the latter though. Once we know this, we could use my earlier attempt (or possibly an easier way) to find r and x in terms of t and use our defintion of the position of the particle to find the particles exact position. But, I still can't find concrete evidence that vo=x'.

By the way-I've done some calculations on my own and have found (I think) that double the ratio of the orbibital area and k is the time. Is this right?

Thanks,

Brad

By Michael Doré (Md285) on Monday, January 08, 2001 - 12:53 am:

Hi,

Okay, the expression for the transverse speed (that is the component of the velocity that is perpendicular to the radius) is rx'. The radial speed is r'.

So for simplicity let's say that at t = 0:

r = r0
rx' = v0
r' = 0

In other words the velocity is entirely transverse. We are not actually losing generality by saying this. We're simply assuming that at t = 0, the motion is in the phase at which there is no radial component to the velocity.

Okay, so r2x' = const = r02 (v0/r0) = r0v0

So you are correct that the expression for u is:

u = A sin(x + D) + c

for some A,D,c to be determined (hopefully!)

You should find that as r' = 0 when t = x = 0 then D = p/2, so the expression becomes:

u = A cos x + c (*)

Then when t = 0, u = 1/r0 and x = 0 so:

1/r0 = A + c

We just need one more condition. Here's where energy conservation can help us.

The kinetic energy + gravitational potential energy is always the same, at any point during the orbit because of energy conservation. The expression for KE is 1/2 mv2 where v is speed and the expression for GPE is -GMm/r. So:

1/2 mv2 - GMm/r = const

But v2 = (rx')2 + r'2:

1/2 m(r2x'2 + r'2) - GMm/r = const (**)

If you want to check this mathematically differentiate the left-hand side and try to show the derivative is zero.

From (**) we can get another condition between A and c and thus obtain an exact equation of orbit.