Deriving Kepler's Law: Negative Sign Explained

[pureblade]

hi, I am having trouble understanding what happens to the negative sign when i try to derive keplers law.

v=d/t
where d is the distance around one orbit, so d=2rpi
and t is the time for one orbit in seconds which i will call T

so v=2rpi/T

now using Ep=-Gm1m2/r
Ep is gravitational potential energy or work,
so Ep=force x distance
where we expand the terms Ep = (mass x acceleration) x distance
further expansion i get (mass x distance /seconds^2) x distance
simplifying to get Ep= v^2 x mass

now equating v^2 x mass = -Gm1m2/r
dividing mass, v^2 = -Gm/r
rooting, v=sqrt(Gm/r) ----- (ignoring the negative)
(cant understand why the negative should dissapear)

equating v=2rpi/T and v=sqrt(Gm/r) to get:
2rpi/T=sqrt(Gm/r)
squaring both sides, 4(r^2)(pi^2)/T^2 = Gm/r
now multiplying both sides by r and dividing by 4pi^2

i end up with r^3/T^2 = Gm/4pi^2

valid method? I am not sure.

 

[Leo Klem]

Yes, the reasoning is substantially correct, for Newton's gravitation law was derived straight from Kepleran laws, basing the reasoning on the equivalence between the centrifugal force due to the planet's revolution with the supposed attraction force exerted by the Sun on the planet.

 

[D H]

Out-of-order reply:

Yes, the reasoning is substantially correct, for Newton's gravitation law was derived straight from Kepleran laws, basing the reasoning on the equivalence between the centrifugal force due to the planet's revolution with the supposed attraction force exerted by the Sun on the planet.

This is wrong, and for many reasons. Two of them are

Item 1.
Kepler's laws talk about elliptical orbits.

Here are nine elliptical orbits as viewed from the perspective of an inertial frame. All nine orbits have the same semi-major axis and orientation, but the eccentricity varies from 0.1 to 0.9 in steps of 0.1.

With this inertial perspective, the shapes of the orbits are simple and well-known (ellipses that share a common focus, and in this case, share a common semi-major axis). The descriptions of the orbits are once again simple and well-known (Kepler's laws). The equations of motion are yet again simple and well-known (Newton's second law + Newton's law of gravitation).

There is no centrifugal force in this picture.To get a centrifugal force you must be observing things from the perspective of a rotating frame. The only case for which centrifugal force exactly counterbalances gravitational force is a circular orbit (inertial point of view) viewed from the perspective of a frame rotating at the planet's orbital rate. The planet does not move from this perspective. There is no orbit from this perspective.

What if the orbit (inertial perspective) is elliptical rather than circular? The planet will move, but the behavior is, bizarre. Here are the same orbits depicted above, but this time viewed from the perspective of a frame rotating with the mean motion of these orbits:

With this rotating perspective, the shapes are complex (this family of curves may have a name, but I don't know what it is). The descriptions of the orbits are complex, as are the equations of motion (Newton's second law + Newton's law of gravitation + centrifugal force + coriolis force).Item 2.
Newton showed that Kepler's laws follow from his laws of motion and his universal law of gravitation in the very special case of one, and only one, body orbiting a much, much more massive central body. Kepler's laws are only approximately correct. Jupiter, for example, violates Kepler's laws because its mass is about 1/1000 that of the Sun's mass. The Earth, for example, violates Kepler's laws because it is also attracted to things like Jupiter. Newton's theory was able to explain these violations of Kepler's laws.

hi, I am having trouble understanding what happens to the negative sign when i try to derive keplers law.

That's because what you are doing is not correct.

now using Ep=-Gm1m2/r
Ep is gravitational potential energy or work,
so Ep=force x distance

This is the source of your error. The specific value of potential energy is arbitrary. What matters is the change in potential energy. For a circular orbit, there is no change in gravitational potential energy. It is constant.

What you need to do is to incorporate conservation of angular momentum into your work.

 

[Leo Klem]

First point: it seems that DH, beyond his long reasoning and choreographic show, doesn't know that circle lines belong to the family of ellipses.

Second point: it seems that DH hasn’t got the meaning of Newton's second Law, F = ma. This law includes Newton's first law: "Corpus omne perseverare in statu suo quiescendi vel movendi uniformiter, nisi quatenus a viribus impressis cogitur statum illum mutare". Which means that any material body changes its inertial motion path only if subjected to a force. Therefore, because of mere kinematical laws, if the motion is not uniform and linear, there is an acceleration towards the centre of the motion-line curvature, irrespective of the curve's shape. Such a "centripetal acceleration", multiplied by the body's mass, does in turn imply a "central force" applied to the body.

Third point: The Newtonian third law states that a reacting force opposes to any acting force applied to a material body. Therefore, once ascertained that a certain body undergoes a "centripetal force", there is a simultaneous opposite reaction, that is a "centrifugal force" opposed to the centripetal one. There is NO centrifugal force ONLY IF the body has no mass, because - in such a special case - there neither is a "centripetal force" associated with the centripetal acceleration.

More in general: It seems that DH forgets that - as Newton was first able to prove – any "central force", whose direction is along a straight line that includes the centre of the motion together with the attracting bodies' mass-centres, implies a family of plane curves which include, in particular, ellipses, parabolas and hyperbolas, taken the motion centre as the centre of the reference frame. The variable distance "r" is from the motion centre.

At Newton's time, there was no means to detect some "irregularities" in planetary orbits, not even the most "evident" one relevant to Mercury. Newton's gravitational law is in itself the proof that Newton could not account for such "irregularities". In fact, from Newton's gravitational law one can only derive orbital curves that belong to the family of conic sections (ellipses, circles, parabolas, hyperbolas), according to given initial conditions in the body's motion. It's the mathematics of dynamics, not an opinion. That is why Newton's law is insufficient to explain Mercury's orbit (despite the consideration of other planets' influence), and why the success of General Relativity.

Referring to the laws formulated by Kepler (1571-1630), Newton (1642-1727) did only assume that the centre of each gravitational motion, like also the motion - for example - of satellites around the respective planets, had to be associated with the relative "Kepleran constant", regarding the ellipses or circles of satellite orbits. Which led him to postulate the constancy of the ratio of the planet's mass to its specific "Kepleran constant". This assumption, together with the Newtonian three basic dynamics laws and Kepleran laws, is all that is needed to obtain Newton's gravitational law. Obviously, the Kepleran laws being based on "approximate" observations, also Newton's gravitation law is an "approximate" model of gravitation.

 

[ideasrule]

Which led him to postulate the constancy of the ratio of the planet's mass to its specific "Kepleran constant". This assumption, together with the Newtonian three basic dynamics laws and Kepleran laws, is all that is needed to obtain Newton's gravitational law. Obviously, the Kepleran laws being based on "approximate" observations, also Newton's gravitation law is an "approximate" model of gravitation.

You have it the other way round. Newton's gravitation law, plus Newton's three dynamics laws, are all that is needed to obtain Kepler's laws.

It's a common exercise in introductory physics to prove Kepler's second law for circular orbits, and for circular orbits the OP's method is fine, although he did make a few mistakes. However, Kepler's second law is meant to work for elliptical orbits as well, since Kepler knew planets orbit the Sun in ellipses.

 

[D H]

Leo, the only one who doesn't get it is you. In particular, you apparently don't grok the meaning of Newton's first and third laws.

First point: it seems that DH, beyond his long reasoning and choreographic show, doesn't know that circle lines belong to the family of ellipses.

Second point: it seems that DH hasn’t got the meaning of Newton's second Law, F = ma.

I get the meaning of Newton's second law. The question is: Do you? In your previous post, you talked about the "equivalence between the centrifugal force due to the planet's revolution with the supposed attraction force exerted by the Sun on the planet." Let's go with this argument. Suppose the centrifugal force on a planet is equal but opposite to the gravitational force. That means the net force on the planet is zero. Per Newton's first law, this means the planet must be either stationary or must be following a straight line path at constant speed. Yet in this last post of yours you alluded to planets following a curved path. This of course contradicts the notion of centrifugal force counterbalancing gravitational force. There is no curved path to follow if you want to proceed with this argument.

Third point: The Newtonian third law states that a reacting force opposes to any acting force applied to a material body. Therefore, once ascertained that a certain body undergoes a "centripetal force", there is a simultaneous opposite reaction, that is a "centrifugal force" opposed to the centripetal one.

You are misconstruing Newton's third law here. Newton's third law does indeed say that there is a reaction to every action. However, in an action-reaction force pair, the action and reaction forces are always (a) the same kind of force, and (b) applied to the opposing bodies. The reaction force to the gravitational force that the Sun exerts on a planet is the gravitational force that the planet exerts on the Sun. It does not act on the planet.

At Newton's time, there was no means to detect some "irregularities" in planetary orbits, not even the most "evident" one relevant to Mercury. Newton's gravitational law is in itself the proof that Newton could not account for such "irregularities". In fact, from Newton's gravitational law one can only derive orbital curves that belong to the family of conic sections (ellipses, circles, parabolas, hyperbolas), according to given initial conditions in the body's motion. It's the mathematics of dynamics, not an opinion. That is why Newton's law is insufficient to explain Mercury's orbit (despite the consideration of other planets' influence), and why the success of General Relativity.

Nonsense. I was not talking about general relativistic precession. I was talking about precession in the context of Newtonian mechanics.

Side note: Even Mercury's precession is fairly well explained by Newtonian mechanics. Mercury's observed precession is 5700 arc seconds per century. Newtonian mechanics accounts for all but 43 arc seconds per century of that precession.

Back on topic: The axis of an orbit does not precess in the Newtonian mechanics two body problem. Newton is no precession. In Newton's terminology, the ellipse is immovable. However, that planet orbits did precess was known. Newton's proposition 45 was an initial attempt at explaining that precession in terms of how other planets perturb each others orbits.

 

[Leo Klem]

Misunderstandings are the normality, especially if one strives to be concise in dealing with such arguments.
My first reaction was against your statement, according to which centrifugal forces appear only in circular orbits (didn’t you say that?). Yesterday, I prepared one page in WORD to recall the basics of kinematics about plane motions, to show, by use of the appropriate equations, that a point in any non-linear plane motion undergoes centripetal accelerations; therefore, this is true also of any point moving along orbital paths, with a particular characteristic if the motion obeys Kepler’s laws. This characteristic consists of the existence of a centripetal acceleration that is constantly directed toward the center of motion (i.e., the Sun for planets), this center being also the origin of the fixed reference frame. Unfortunately, I wasn’t able to transfer my memo and relevant formulas in this reply-window of the Forum, but any good manual of analytical mechanics can work much better.

My second point (referring to Newton’s second law), was the obvious remark that if the moving point has a mass and undergoes an acceleration, the material point is subjected to a force. Do you disagree? If so, I cannot comment further.

Surprisingly, you now state that the combination of centripetal and centrifugal forces should have nil effect. Instead, just to begin, I think you should know about the existence of “tension” and “pressure” effects. By the way, in the case of planetary orbital motions it regards “tension” on the planets, which, according to scientists, should also be viewed as the best explanation for tides, whose “strange” characteristic is that tide effects (on the Earth it's a well-known experience) appear symmetrically and simultaneously at opposite geographical areas of the Globe.

The main point, however, is that gravitational attraction between the Sun and the planet (let’s limit to the two-body attraction, for the sake of Newtonian simplicity) establishes a mutual dynamical constraint, quite analogous to the force transmitted by a band to the stone in an ancient sling weapon. In the sling’s case, the stone orbits the hand that holds the band, thanks to the dynamical equilibrium established by two opposite forces, the one transmitted by the band (which is a “constraint” analogous to the effect of the mutual attraction between two celestial bodies) and the centrifugal force the stone undergoes because of its constrained “revolution”; (in this connection, among other things, note that the stone’s orbit cannot be considered as “circular”, since also the holding hand rotates. In such devices, as you know, the band may lacerate if the stone's revolution speed exceeds a certain limit: is it not the tension due a centrifugal force?). Strange indeed that I’ve to ask this question.

Therefore, once ascertained the existence of a centripetal force acting upon a MASS IN MOTION, an opposite centrifugal force must also affect the same mass: it’s an every day experience for every one in the world, especially when the vehicle in which we travel is compelled to swerve. Is it not so? Your interpretation of Newton’s third law seems peculiar.

Yet; the initial topic was about the derivation of Newton’s laws from Kepler’s, and/or vice-versa.
Newton was able to prove that the planetary orbits, because of the Kepleran laws, are characterized by a centripetal acceleration that is constantly directed as the straight line connecting the planet to the Sun. This was the key point proved by Newton to obtain his gravitational law.
(In my view, your mention of polar/equinox precession, or Coriolis forces, or rotating reference frames etc. has nothing to do with the point in question, and does only contribute to create confusion).
By a very quick summary, and sticking to kinematics only: in any plane motion, as possibly described by a fixed polar reference frame, the point in motion undergoes in general an acceleration A, which can be thought of as resulting from two component accelerations AR and AT, dubbed “radial acceleration” and “transverse acceleration”, respectively.
Radial acceleration AR points constantly to the center of the motion (i.e., toward the origin of the reference frame along the vector radius), whereas the direction of transverse acceleration AT is orthogonal to the former.
It can easily be proved that transverse acceleration AT is nil (i.e, AT = 0), if the point moves according to any orbit that obeys Kepler’s laws. Thus, the only acting acceleration in Kepleran kinematical orbits is the so-called “radial” or “central acceleration” AR . From which all subsequent considerations.

As to precession, I didn’t misunderstand you. I have deliberately mentioned the perihelion precession to mean only that Newton’s gravitational law is an excellent though approximate model, which is not sufficient for an adequate explanation of a number of astronomical “anomalies”. The reason is just that Newton’s law is a direct logical implication of Brahe’s observations and Kepler’s laws.

 

[D H]

Misunderstandings are the normality, especially if one strives to be concise in dealing with such arguments.
My first reaction was against your statement, according to which centrifugal forces appear only in circular orbits (didn’t you say that?).

No, I did not say that. I said

• That in an inertial frame there is no centrifugal force, period.
• That explaining orbits as a balance between centrifugal force and gravitational force only works for circular orbits, and then only in a frame in which the orbiting object isn't moving.

Centrifugal force is an observer effect and only appears when the observer is rotating.

 

[Ich]

Therefore, once ascertained the existence of a centripetal force acting upon a MASS IN MOTION, an opposite centrifugal force must also affect the same mass

No. Actio=reactio means that the sun is pulled towards the planet, not that there is centrifugal force. D H is right in his statements about centrifugal force. It is a fictitious force.