I was doing Kepler's first law (basically that is the only law in which I have a problem) other two laws are easy to understand). During the explanation of the first law my instructor said that we need to establish the fact that angular momentum does not change per unit time i.e. stays constant.

Problem 1.) Instructor says: "First of all, it can be shown that the angular momentum $L$ is constant. Since in the first term, the centripetal force $F$ is collinear with $r$, and in the second term $v$ is obviously collinear with $v$, both terms are zero, which implies that $L$ is constant and that the orbit lies in a plane!" What does this mean (specifically what does "lies in a plane mean"?)

Problem 2.) I learned Kepler's second law which was about equal areas and we PROVED that law of areas mathematically and also in his third law we PROVED, mathematically, that the period squared is equal to semi-major axis cubed, but in his first law which is:"All planets move about the Sun in elliptical orbits, having the Sun as one of the foci" we did not PROVE this law. We just said that the orbit lies in a plane, not that it is elliptical which is actually the law. Should not we be proving that the orbit is elliptical with Sun as on of the foci? Or am I missing something?

Problem 3.) Here is a screenshot from the website:


It says dr/dt=v (vector) and it is in the same direction as p (vector), but the change in r (vector) with respect to time is not v, it is the one component of v which is directed inwards. Obviously v is collinear with v but it is not "v" which is causing change in r, it is the component of v which causes the change in r and it is directed inwards, not collinear.

  • $\begingroup$ Please at least reformat this to avoid the Wall Of Text which makes it hard to try and read. $\endgroup$ – StephenG Nov 1 '17 at 7:07

Solution 1) "Lies in a plane" means that the orbit forms a 2-dimensional shape. It means that you could define some x, y, and z axes and the paths for both the Sun and the planet would always have a z coordinate of $0$.

Solution 2) It's true that saying the orbit lies in a plane is not the same thing as proving Kepler's Law of Orbits mathematically. However, if you go back to the website your screenshot comes from, you'll find that they continue on to show the complete and rigorous derivation for the orbital path and show how that is a conic section, which, for a gravitationally bound orbit, is the definition of an ellipse.

Solution 3) it is important that we distinguish between $\vec r$ and $r$. $r$ is the magnitude of the radius. You are correct in saying that $\frac{dr}{dt}$ is the component of $\vec v$ that is directed along the radius, because it corresponds to a change in distance from the center. However, $\vec r$ is the position vector. It represents the coordinates of the orbiting body in the orbital plane. As a vector, it would be written as $\vec r=(r,\theta)$; that is, it has components that represent the distance from the origin, $r$, and the angle from some pre-defined line of reference, $\theta$. So you can see that $\frac{d\vec r}{dt}=\frac{dr}{dt}\hat r+r\frac{d\theta}{dt}\hat\theta$ (the hat notation indicates a direction unit vector). What was used in the equation you showed was $\frac{d\vec r}{dt}$, which doesn't necessarily have to point in the radial direction because the position vector might only be changing in $\theta$. So you see that $\frac{d\vec r}{dt}=\vec v$, which is a definition.

  • $\begingroup$ Okay, so the tangential velocity is causing two changes? The radial component and the tangential component? Also the gravitational attraction is also playing its role to decrease its radius right? In summary the radial velocity and the gravitational attraction BOTH are decreasing its radius? Also you said:"...position vector might only be changing in θ" is it because the position vector is changing ONLY because of the tangential component and not radial one? $\endgroup$ – FlightMuj Nov 1 '17 at 15:46
  • $\begingroup$ Basically saying that the orbit lies in a plane is the beginning of understanding an ellipse, right? Also calculus was not invented back then, it would have been hard for Kepler to figure this out. How did he find that the orbits are elliptical? $\endgroup$ – FlightMuj Nov 1 '17 at 15:54
  • $\begingroup$ @FlightMuj the tangential component of velocity doesn't change the radius, only $\theta$. It is multiplied by the radius to make the units agree and turn the angular velocity into a linear, tangential velocity. $\endgroup$ – Jim Nov 1 '17 at 17:27
  • $\begingroup$ By the way, just to confirm, can you draw a diagram and show where θ is? $\endgroup$ – FlightMuj Nov 1 '17 at 17:30
  • $\begingroup$ @FlightMuj draw a line from the focus to a point on the ellipse. $\theta$ is the angle between that line and a line from the focus to the orbiting body $\endgroup$ – Jim Nov 1 '17 at 19:33

I am only going to answer the first part of your question as the second part, that is why planets move in elliptical (more generally in a conic section) orbit is just solving the differential equation for Newton's Law and if you are not an undergraduate student, I doubt that it is relevant for you.

First of all note that the definition of $\vec v$ is in fact $\dot{\vec r}$. I don't know how else you would define it. You said the change of $\vec r$ is not $\vec v$ but "it is the one component of $v$ which is directed inwards". Is it possible that you are confusing the change of $\vec v$, which is the acceleration and is directed inwards due to Newton's law, with the change of $\vec r$, which is just $\vec v$ ?

Let's take an example. Assume that the orbit is a circle with radius $r$ and the planet is in the position $(r,0)$ now the radius has to stay constant, so the planet is going inwards as you guessed but it also has to go sideways because otherwise it would be at a point $(r-\Delta x, 0)$, which would not be on the circle, so the change in $\vec r$ is not just downwards but also sideways too!

I don't know if this helps but if you can provide your definition of $\vec v$ and why you think think that only the radial part of $\vec v$ is changing the position, I can help more.

  • $\begingroup$ See, if the radius is changing isn't it going to be the velocity directed towards the Sun causing the change? Or is it something like this as I think it might be: Radius is a position vector (vector has a magnitude and a direction), definitely the position vector's direction is changing, because the planet is rotating, and it is only the velocity vector which can cause radius vector's direction to change and definitely the sideways component itself causes the direction of r to change, and so as the direction of r changes then simply dr/dt can be said to be caused by v? Right or wrong? $\endgroup$ – FlightMuj Nov 1 '17 at 12:17
  • $\begingroup$ Well $dr/dt$ is $v$! They are not two separate things. I corrected my answer above slightly. The velocity vector has even no radial component and is perpendicular to the radius vector. $\endgroup$ – Gonenc Nov 1 '17 at 12:43
  • $\begingroup$ So, how is dr/dt= v? $\endgroup$ – FlightMuj Nov 1 '17 at 12:46
  • $\begingroup$ Because it is the definition of $v$. How else would you define $v$?? $\endgroup$ – Gonenc Nov 1 '17 at 12:47
  • $\begingroup$ The change in radius length is caused by the v component directed inwards. No? $\endgroup$ – FlightMuj Nov 1 '17 at 12:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.