# A question on the Kepler's law

1. Sep 3, 2007

### Rainbow

According to Kepler's law of areas, the area swept by the line joining a planet(say earth) and the sun is constant with time.

i.e., dA/dt=constant
Also, while deriving we get that,
dA/dt=L/2m where L is the angular momentum of the planet, and m is its mass.

Now, in order for dA/dt to be constant, L should be constant.
I think this is possible only when the following two conditions are fulfilled:
i) the eccentricity e of the orbit is zero, and
ii) the sun is exactly at the centre of the orbit.

Now let us consider a case where these two conditions are not fulfilled-
The earth revolves around the sun in an elliptical orbit (e is not equal to zero), and the sun is not exactly at the centre of the orbit. Now, at most of the points on the orbit(almost all) the gravitational force of the sun on the earth has a tangential component. This means that there is a net non-zero torque on the earth. And this essentially means that the angular momentum of the earth is not constant, i.e. L is not constant. So,now if we have a look at the eq. above, we find that dA/dt is not constant.

Can somebody please tell what the matter is?

2. Sep 3, 2007

### cesiumfrog

It is not necessary for the eccentricity to be zero.

3. Sep 3, 2007

### mjsd

why do you think that the eccentricity must be zero in order for L to be constant? show us your proofs

4. Sep 3, 2007

### genneth

Remember that in an isolated system, angular momentum is conserved.

5. Sep 3, 2007

### Rainbow

Because, only if the eccentricity of the orbit is zero and if the sun is exactly at the centre, the force of gravity on the earth will be radial and it will have no tangential component. No tangential component means no net torque, and no torque means that the L remains constant.

6. Sep 3, 2007

### Rainbow

You are very correct in saying that the L in an isolated "system" is conserved. But here I'm talking about the L of the earth (a component of the sun-earth system). And conservation of L in a system doesn't essentially mean that the L of the individual components of the system cannot change.

7. Sep 3, 2007

### HallsofIvy

Staff Emeritus
Not only is "eccentricty 0" not necessary, it is not even necessary that the "orbit" be an ellipse.

It is, in fact, very easy to show that if a "planet" were to move by the sun in a straight line at constant velocity, it would still sweep out "equal areas in equal times". That's because the base of the triangles formed would be the same (constant velocity along the line so the same distance in the same time) while the altitudes of the triangles are simply the shortest distance from the sun to the line.

8. Sep 3, 2007

### genneth

However, you are only interested in the orbital angular momentum. Your model doesn't take into account anything else. The only angular momentum is due to the orbit of one body about another (massive one, so it doesn't even move). It is conserved.

9. Sep 3, 2007

### D H

Staff Emeritus
Rainbow, Kepler's second law is approximately true. It obviously is not true in the multibody problem or in the case of two bodies with comparable mass orbiting around each other. (It is also not true in the sense that Newton's law of gravitional is not exactly true because of relativistic effects; I'll ignore that.)

In the case of two comparably sized bodies, each body moves along a conic section (circle, ellipse, parabola, or hyperbola) around the common center of mass. The equal area law is true in the center of mass frame, regardless of eccentricity.

Now, what about the solar system? Since the Sun is much more massive than any of the planets, the solar system center of mass is very close to the center of the sun. The error induced by ignoring the motion of the Sun is small, and thus Kepler's laws are very close to true.

10. Sep 3, 2007

### Rainbow

As you said, that the planet will be "moving in a straight with a constant velocity", which essentially means that no net non-zero force is acting on the planet. This is the reason why it sweeps equal areas in equal times. This is a similar argument which I gave in my second post in this thread, where I justified my point that in order for the L to be constant the eccentricity should be zero, and that the sun should be at the centre of the orbit. In other words, the orbit should be circular and the sun should be at the centre of the circular orbit. Now, if a net non-zero force acts on your straight moving planet, it will not sweep equal areas in equal times, as its speed will now change. Similarly, as I mentioned in my question, if the conditions I have given are not fulfilled, there will be a net non-zero torque on the planet which will change its L and hence the rate at which the area is swept.

11. Sep 3, 2007

### Rainbow

D H, I'm sorry but, I'm not clearly getting what you said. Could you kindly elaborate what you said.

12. Sep 3, 2007

### Cthugha

No, there is just a tangential component, if the sun is moving significantly due to gravity or if you do not take the sun as the origin of your coordinate system. But as Kepler's second law is about the area covered by the vector sun->earth, this is not interesting here anyway.

$$\vec{L}=m \vec{r} \times \vec{v}$$

Do you agree?

Then

$$\frac{d \vec{L}}{dt}=m \vec{v} \times \vec{v} + m \vec{r} \times \vec{a}$$

So $$\vec{v} \times \vec{v}$$ is 0 anyway and $$\vec{r} \times \vec{a}$$ is 0 too, because the gravitational force always points in the direction of the line sun-planet.

As said before, this is not true, if you consider some other point as being the origin for calculating the area, but these cases are not covered by Kepler's second law anyway.

Last edited: Sep 3, 2007
13. Sep 3, 2007

### Parlyne

I'm sorry, but your rationale for thinking that gravity causes a torque in non-circular orbits is simply incorrect. Torques only occur when there is a force (or component of a force) that acts perpendicularly to the radial direction. This is implicit in $$\vac{\tau} = \vec{r} \times \vec{F}$$. In a circular orbit, this perpendicular direction is tangent to the orbit, so we can just refer to it as a tangential force. However, in an elliptical orbit, the tangential direction is not perpendicular to the radial direction. This means that a totally radial force can, at some points in the orbit, have a tangential component. However, since the tangential direction is not perpendicular to the radial, this cannot be taken as an indication that there is a net torque, which still requires a force perpendicular to the radial direction.

To be explicit, taking the larger object as the origin of a coordinate system, the force of gravity on the smaller object is:

$$\vec{F} = -\frac{GMm}{r^2}\hat{r}$$.

The torque on the smaller object will, then, be:

$$\vec{\tau} = \vec{r} \times \vec{F} = -\frac{GMm}{r^2}\vec{r} \times \hat{r}$$.

But, since $$\hat{r}$$ is just the unit vector in the direction of $$\vec{r}$$, it is clear that the two vectors are antiparallel, meaning that $$\vec{r} \times \hat{r} = 0$$, which means that we must conclude that $$\vec{\tau} = 0$$.

Notice that the direction of motion does not enter into these considerations, meaning that the shape of the orbit is not relevant. All that matters is the direction of the force as compared to the radial.

14. Sep 3, 2007

### cesiumfrog

Isn't it equivalent to conservation of angular momentum? So for two bodies of comparable mass, you just need to consider the area swept around the centre of mass of the system by either body. (And if that's the case then for more bodies, which can exchange momentum, you'd have to rephrase the law as the sum of rates of area sweep by each body, around the centre of mass, that is constant at different times?)

15. Sep 6, 2007

### Rainbow

But when I draw a diagram of the situation I mentioned in my question, I get something like this(view attachment).

#### Attached Files:

• ###### Orbit2.bmp
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16. Sep 7, 2007

### cesiumfrog

It's pulled so as to go faster when it's closer, like a skater that draws in their arms. Hence the equal areas in equal times. What's the problem?

17. Sep 7, 2007

### Shooting Star

Your diagram is correct. The tangential component of the accn does change the linear speed of the planet, speeding it up as the planet approaches the perhelion. The normal component cannot change the speed, only the direction. (Note that the normal component does not point toward any fixed point, as would happen in a circular orbit.) The net result is that the planet speeds up and comes nearer in such a way as to keep the ang. mom. constant.

The reason that you thought that there must be a torque was beacuse you were considering only the tangential component. The normal component also contributes, beacuse the normal, in this case, does not lie along the st line joining the planet and the sun.

Last edited: Sep 7, 2007
18. Sep 7, 2007

### Rainbow

Oh now I get it. In case of the sun-earth system, the gravitaional force is an internal force, which cannot cause any change in the angular momentum of the system. And here, though the speed of the planet changes due to the force, the distance of the planet from the sun also changes by just the right amount to keep the L constant. right?

19. Sep 13, 2007

### Gaenn

dA/dt=L/2m is not correct.

should be dA/dt = 0.5L

20. Sep 13, 2007

### rcgldr

Last edited: Sep 13, 2007