# Keppler's 2nd & 3rd

1. Apr 1, 2003

### mich

[SOLVED] Keppler's 2nd & 3rd

While I have the correct topic, I'm not sure if this belongs to
the crackpot theory or not. I appologize if I haven't placed my post in the correct category.

I'm having a bit of a problem concerning Keppler's 2nd.law.
" ...moving planets sweeps out equal areas at equal time."
This clearly states that a planet which approaches or get's farther
from the sun will experience an increase or a decrease of velocity
by a specific amount.
Let a planet revolve around the sun at R=1, if the orbit is circular, then the distance traveled would be 2*pi*R= 2*pi. Let us claim the period to be 1 unit of time.It's velocity would then be 2*pi/1.
According to the 2nd law, it would take a radius of sqrt2 in order
for the period to be equal to 2. the velocity would then be
2*sqrt2*pi/2; or simply sqrt2*pi.
If we continue to increase the radius, we get:
for R=sqrt3 T=3; v=2*sqrt3*pi/3
for R=sqrt4 T=4; v=2*sqrt4*pi/4 = 4*pi/4 = pi (notice that when the radius is doubled, the velocity, being directly affected by the gravitational force,is halved. This "could" seem to claim as you did,that gravitational force is inversally proportional to R...not R^2.
Nevertheless,My question would be: could Newton's 3rd law be centered more on the kenetic energy of a planet relative to it's distance from the gravitational force? Force=Kenetic Energy/R?
F=mv^2/2R

R=1 T=1 V=2 F= m*4/2*1 = 2 (let m=1)
R=2 T=4 v=1 F= m*1/2*2 = 1/4
R=3 T=9 v=.66 F= m*.4356/2*3 = .0726
R=4 T=16 v=.5 F= m*.25/2*4 = .031
R=8 T=64 v=.25 F= m* .0625/2*8 = .0039

This would show that when the radial distance is doubled, the kenetic energy is halved, and the Force is reduced by a factor of 8.

2. Apr 2, 2003

### mich

Re: Re: Re: Gravity decreases as 1/r

Thank you for replying, Janus, and I do agree that Keppler's 2nd law was meant to speak of one planet travelling on a elliptical orbit.
On the other hand, wouldn't you agree that the "covering of areas at equal time" law must indeed state that the planet having a specific velocity at R1 will change to a different "specific" velocity when the planet is at R2?Let us say that at R1 the velocity is 1 unit of distance/unit of time. According to the 2nd law, one ought to be able to calculate the velocity of the planet when the distance changes to R2 from the gravitational source.According to your calculation,what would be R2 when the velocity is halved, if R1= 1 unit of distance?

3. Apr 2, 2003

### Janus

Staff Emeritus
The formula for velocities at perigee and apogee repectively are:

Vp = [squ]((2GM/(Rp+Ra)) Ra/Rp)

and

Va =[squ]((2GM/(Rp+Ra)) Rp/Ra)

In this case, Rp = 1/2 and Ra =1, so

Vp = [squ]((2GM/(1/2+1)) 1/(1/2))

= [squ](8GM/3)

and

Va =[squ]((2GM/(1/2+1)) (1/2)/1)

= [squ](2GM/3)

The ratio of Vp to Va is

Vp/Va =[squ](8GM/3)/[squ](2GM/3)

=[squ]((8GM/3)/(2GM/3))

GM/3 cancels out, leaving

=[squ](8/2) =[squ]4 = 2.

The area traced out by the orbit at perigee in one sec is;

Ap/sec = (2*1/2)/2 = 1/2 square units.

at apogee:

Aa/sec = (1*1)/2 = 1/2 square units, which satisfies Keppler's second law.

One thing that should be pointed out however, is that Keppler's Second law by itself cannot tell us anything about how the gravitational field behaves other than that it acts along a certain line.
It is true regardless whether the field is attractive, repulsive, constant, falls off lineary or by the square of the distance. All it requres is that the force always acts along a line towards a central point.

This has been proven through Newton's "Theorem of Areas".

Last edited: Apr 2, 2003
4. Apr 3, 2003

### mich

O.K. this would seem to come from Newton's force of gravitation formula where [sqrt 2GM/R] I believe would be considered the velocity of one of the bodies;. would that be correct?

I believe that this is what I had, Janus. when the Radius increases
twofold the orbital velocity is reduced to half it's original speed.
I think I do understand what you are saying though. If a body has a
circular orbit, it's velocity needs to be inferior or superior (depending on it's position relative to the gravitational force), to the velocity of the body when it's orbit is an elliptical one.
If that's the case, then I "personally" disagree. The orbital velocity has two vectors; one in the direction towards the point of gravitation and the second is lets say 90 degrees to it.
The sideral velocity displaces the path of the orbit making the body having either a circular orbit, or elliptical or whatever. It is responsible for leaving the body at a constant distance from the point
of gravity or increasing/decreasing it's distance to it.It is in itself independant to the gravitational force.Nevertheless,
when the body is at a specific distance to the gravitational point,
it seems that it's velocity can be calculated the same way that I have done so; that is the way I see it anyway. Your calculation seems to comes indirectly from Newton and not Keppler, although, as you showed me, it does indeed agree with Keppler's 2nd law.
The reason why I'm circling around this, is because, if my calculations are correct,then it seems that there's a discrepancy between Keppler's 2nd and 3rd law;and yes of course I know that I am in all probabilities wrong.Nevertheless, this is what bother's me right now.
You see, the idea stems from the fact that Kepler's calculated values of Tycho Brahe's observations could not have taken into account the red and blue shifts since during this period, light was thought to have possibly an infinite velocity.

Thanks for your time; I greatly appreciate it.
Mich

Last edited by a moderator: Apr 3, 2003
5. Apr 3, 2003

### Janus

Staff Emeritus
[squ](GM/R) gives the orbital velocity for a circular orbit of radius R . (The formula you gave is for escape velocity.)

Now it is important to note that a body in an eliptical orbit with a perigee of R will have a velocity greater than [squ](GM/R) (yet less than [squ](2GM/R) ) at perigee. So you can't use Keppler's second law to determine the period of a circular orbit at that distance.

No discrepancy. One talks about the velocity of a eliptical orbit at different points of the orbit, the Other deals with the period due to the Average radius of an orbit. One works within a given orbit, the other works between different orbits.

There can't be a discrepancy between them because they are not inter-related.

Just as it can be shown that Keppler's second law is compatable with Newtonian physics, Keppler's third law can be shown to be so.

V = [squ](GM/R)

the distance travled in one orbit is

D = 2[pi]R

The period of the orbit is therefore

P = D/V =2[pi]R / [squ](GM/R)

= 2[pi]R[squ](R/GM) = 2[pi][squ](R³/GM)

If we compare the period of two circular orbits we get

P1/P2 = 2[pi][squ](R1³/GM)/2[pi][squ](R2³/GM)

2[pi] cancels out.

=[squ](R1³/GM)/[squ](R2³/GM)

square both sides.

P1²/P2² =(R1³/GM)/(R2³/GM)

GM cancels out;

P1²/P2² =R1³/R2³

Leaving us with Keppler's Third Law.

The problem here is that the Doppler shifts in light are so small at orbital velocities that they fall far below the accuracy limits of the measurements made at that time. They would have had negligible effect on Keppler's measurements.

6. Apr 3, 2003

### mich

Thanks alot, Janus; you clarified the situation for me.

7. Apr 4, 2003

### mich

Excuse me Janus,I came back because something was bothering me..
I sound like Columbo

What you wrote makes sense but I may have a small problem still.
When the planet's orbit is at it's perigee, then the velocity
is greater than if it was in a circular orbit at that radius, which is the reason why it gets further from the sun. Also when the planet is at it's apogee , then the planet's velocity is slower than if it had a circular orbit at that particular radius, which is the reason why the planet moves closer to the sun. Therefore, the ratio between
those two velocities (apogee/perigee) could never be equal to the ratio of two different circular orbits having two different radii, one equal to the apogee of the elliptical orbit, and the other equal to the perigee of that same elliptical orbit...so how is it that in your calculations, the ratio between the two velocities (apogee/perigee) was the same as the calculations I've made using circular orbits?

8. Apr 4, 2003

### Janus

Staff Emeritus
First off, I split this discussion off into a new thread, as it was off the original topic.

Going back over your posts, it look's like you tried to use Keppler's 2nd Law to calculate those circular orbits. That won't work, because you are dealing with two separate orbits, not just one.

What you ended up getting is the velocity difference between two points with different distances in the same orbit. You used the wrong law.

To get the velocity difference between two different orbits, you have to use the 3rd law.

9. Apr 4, 2003

### mich

Yes, of coarse I understand. I thought I had something in common with their thoughts on g being proportional to R and not R^2, but
I guess I didn't.

Actually, I only used those calculations to figure out the velocity of a body at different distances from the gravitational force (R).
It seems that Keppler did not write this law for nothing, it had to be in order to calculate the ratio of different velocities when the planet changes it's distance from the sun... but how does one calculate it?
You used a version of Newton's law, something that didn't exist yet.
I took the only tool that existed then,(the covering of areas at equal time law) and tried to figure out different velocity ratios for different distances of the planet's path. The only way I could think of doing this was to calculate the respected velocities
"as if they were two different circular orbits of different radii".

But if I used the method of circular orbits, how could I have ended up
getting the velocity difference between two points with different distances "in the same orbit"? I'm a bit confused.

I will leave the 3rd law aside for now so as not to get too confused.