# A question in Kepler's second and third laws.

1. Dec 7, 2004

### pezzang

I am doing this project on kepler's law and can't figure out something.
part a (in the process of proving kepler's second law)of this is to show that h = |r|^2(d(theta)/dt) k when r = (|r|cos(theta))i + (|r|sin(theta))j and h = r x r' = r x v. I showed it by taking derivative of r and cross producting it with the original r.
I did this part, but now i don't know how to deduce that r^2(d(theta)/dt) = |h|. I actually have no understang of what h represents and stuff. If you can explain to me how to deduce this relationship, it would be helpful.

Another question that's about proving kepler's third law is that
i have to show that h^2/(GM) = ed = b^2/a.
Its previous part was to show that T = 2(pi)ab/h (where a=major axis, b=minor axis) and h = the magnitude of vector h). I proved it using the second law of kepler and showing T is equal to the total area divided by the rate of the area swept by the planet. I don't know what e and d are though. any clues? and how might i approach this problem(showing that h^2/(GM) = ed = b^2/a)?

2. Dec 7, 2004

### BobG

h is specific angular momentum per unit of mass. It is the cross product of the position vector (the radius) and the velocity vector. Angular momentum is always conserved - it stays constant.

Every coordinate system has 3 dimensions. In this case, the radius and velocity vector always lie in the orbital plane. The angular momentum vector lies along the third axis (cross products are always perpendicular to the two vectors that form the cross product).

In this case, you want a UVW coordinate system, where the U axis always points the same direction as the radius. This can be related back to the perifocal system by true anomaly (this is the theta in your equation, although you'll more often see it as the greek letter nu). True anomaly measures the angle between a vector pointing from the center of the Earth towards perigee and the position vector.

$$\hat{U}=cos \theta_i + sin \theta_j$$

V is perpendicular to U and is the derivative of U:

$$\hat{V}=-sin \theta_i + cos \theta_j$$

Since U always points the same direction as the radius, the radius vector is:

$$\vec{r}=r\hat{U}$$ or $$\vec{r}=r cos\theta_i + r sin\theta_j$$

The velocity vector is the derivative of the radius, or (using the product rule and chain rule):

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

or, substituting the relationship back to the perifocal system:

$$\dot{\vec{r}}=(\dot{r}cos\theta)_i + (\dot{r}sin\theta)_j + (-r\dot{\theta}sin\theta)_i+(-r\dot{\theta}cos\theta)_j$$

As you commented, after you rearrange the equation for rdot to group the i components and the j components, angular momentum is the cross product of the radius and velocity vector.

The angular momentum is a vector, but it lies entirely in W, perpendicular to the orbit plane:

$$\vec{h}=r^2\dot{\theta}_k$$

Since it lies entirely along one axis, you can use it as a scalar equation.

The magnitude of the cross product (the angular momentum vector) equals the area of a parallelogram that has the radius and velocity as its sides.

The velocity is normally measured in meters per second, so one 'side' of the parallelogram is the distance traveled in one second and the other is the distance from the center of the Earth. All those parallelograms don't help very much. Triangles would help more. You could string a bunch of triangles around your ellipse and get the area of the ellipse.

In fact, that's why the equation for finding the area of a segment of a circle is:

$$A=\frac{r^2\theta}{2}$$

The angular momentum remains constants and its magnitude has a relationship to area define by the cross product definition. Dividing it by 2 gives you the amount of area swept out per second. You divided a constant by a constant, so the area per second is also a constant.

In other words, the area swept out per second is:

$$A=\frac{r^2\dot{\theta}}{2}$$

Since these can be related to h, which remains constant, that means that if the radius gets bigger, the angular velocity has to get smaller.

I'm not sure if that answers what you were trying to ask.

3. Dec 7, 2004

### BobG

I don't know what you mean by e and d. If angular momentum divided by 2 is the area swept out per second, then multiplying by the number of seconds to complete one orbit gives you the total area.

The other method of determing area for an ellipse is $$\pi ab$$ where a is the semi-major axis and b is the semi-minor axis.

Normally, everything in the orbital plane is defined by the semi-major axis (a), the eccentricity (e), and true anomaly. The semi-minor axis is found by:

$$b=a\sqrt{1-e^2}$$

It takes a little doing, but you can define the magnitude of the angular momentum by:

$$h=\sqrt{\mu a (1-e^2)}$$ where
$$\mu$$ is the geocentric gravitational constant (you have to go back to why the specific energy and angular momentum remain constant in the first place).

Via substitution, you get:

$$\pi a (a\sqrt{1-e^2})=\frac{1}{2}(\sqrt{\mu a (1-e^2)})T$$

Solving this equation for T gets you the orbital period. The square root of 1-e^2 cancels out, so the orbital period depends only on the semi-major axis. A circular orbit with a radius of 10,000 km takes the same amount of time to complete one orbit as a elliptical orbit (e=.75) with a semi-major axis of 10,000 km.

4. Dec 7, 2004

### pezzang

thank you very much. ill look over and see if i understand

5. Dec 7, 2004

### pezzang

This is actually a math project. Can you prove the above in mathematical temrs? i dont know if you understand what i am saying.

6. Dec 7, 2004

Staff Emeritus
What do you mean prove? Kepler's laws say a planet moves on an ellipse with the sun at one focus, and the planet moves so that the radius vector from it to the sun sweeps out equal areas in equal times. What mathematical axioms or postulates do you suppose exist for a planet or the sun?

If you mean, can we derive Kepler's laws from Newton's laws of physics and law of gravitation, then yes we can; after all Newton did it in the Principia.

7. Dec 7, 2004

### pezzang

Ok, here are my final questions after perusing over your replies.

1.How do we deduce that (1/2)h is in fact constant? I know that angular momentum has to be constant but HOW DO WE PROVE IT?

2.in a part where I have to show that h^2/(GM) = ed = b^2/a,
I found out that e and d were predefined. e = c/ (GM) and d = h^2/c. (where c is the magnitude of constant vector c(bold c) and c is what we get when we integrate (vxh)' = GM u' and getting v x h = GMu + c (bold c = constant vector c). u is the unit vector in the direction of r (u = (1/r)(vector r)).
Now I could show that h^2/(GM) = ed by simply substituting the above values to e = c/ (GM) and d = h^2/c, but I don't know how to go from ed to b^2/a (that is, to show that ed = b^2/a). Please help me on this. I am really exhausted trying to solve this problem for the past six hours. sigh...

8. Dec 8, 2004

### pervect

Staff Emeritus
Angular momentum is conserved by any central force, for the simple reason that the torque exerted by a central force is zero.

9. Dec 8, 2004

### pezzang

pervect, how would you prove that mathematically rather than verbally saying it? Is there any equation that would justify your statement? I am not a physics person so I don't really know what you are saying until I see mathematical equations that could clarify my understanding.

10. Dec 8, 2004

### da_willem

As I haven't read the Principia I quote Robert Weinstock:
I'm sure that even as a mathematician you do know what velocity ($\vec{v}[/tex]), mass (m) and a position vector ([itex]\vec{r}[/tex]) are. Then: [itex]\vec{L}=\vec{r} \times m \vec{v}[/tex] This is called angular momentum. Now Newtons second law (sorry for the physics) for angular motion is: $$\frac{d\vec{L}}{dt}=\vec{\Gamma}$$ Where [itex]\vec{\Gamma}[/tex] is called torque wich is: $$\Gamma=\vec{r} \times \vec{F}$$ Where [itex]\vec{F}$ is force (physics again...)

Now pervects statement:
A central force has the property it is directed radially from its source (hence the name) so: $\vec{F}=|\vec{F}| \hat{r}$ with $\hat{r}$ a radial unit vector: $\hat{r}=\vec{r}/|\vec{r}|$ (we took the position of the source (in this case the sun) as our origin)

If you now calculate the torque $\Gamma=\vec{r} \times \vec{F} = |\vec{r}||\vec{F}| \hat{r} \times \hat{r}=0$ as the cross product of a vector with itself always yields zero as 'they' are parallel.

Now putting this in our equation of motion: $\frac{d\vec{L}}{dt}=\Gamma=0$ we see $\vec{L}$ is constant.

So angular momentum is indeed conserved ($\vec{L}=$constant) under a central force ($\vec{F}=|\vec{F}| \hat{r}$). And indeed for the simple reason the torque exerted by a central force is zero ($\Gamma=\vec{r} \times \vec{F} = |\vec{r}||\vec{F}| \hat{r} \times \hat{r}=0$).

11. Dec 8, 2004

Staff Emeritus
I haven't a clue who Robert Weinstock is, but this is a snotty piece of ahistoricity. Newton's theorem one does the conservation of angular momentum (not that he had the term; he proves pure central force => Kepler's equal areas) and is quite valid within its historical context. It can also be dressed up with limits to meet modern requirements. That of course is ahistorical too, limits were not truly understood until two hundred years later. Newton, however was on sounder ground with his prime and ultimate ratios than were most of his contemporaries. And Newton's subsequent arguments in Principia prove both that inverse square follows from Kepler's I and II, and conversely that they follow from central inverse square.

12. Dec 8, 2004

### pervect

Staff Emeritus
Hopefully Da Willem's post answered your question? The torque around a given point is the cross product of the radial vector and the force, and the cross product is zero when the vectors are parallel. The definition of a central force is that the force points in the direction of the radial vector.

I am assuming here that the problem you are analyzing is the problem of a body of negligible mass orbiting a very massive body which is at the origin of the coordinate system. Results are very similar if both bodies have appreciable mass, but the details are different.

If this isn't clear enough try this link:

angular momentum

You might also find the hyperphysics web page on Kepler's laws useful, if it's not too much like cheating to use them

http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c1