The other day we derived Kepler's third law.

$$ \left( \frac{T_1}{T_2} \right)^2 = \left( \frac{r_1}{r_2} \right) ^3 $$

In order to derive this, you can look at a given planet that revolves around the sun. If you assume that the sun is way heavier than the planet and that the planet moves on a circle, you can simply state the force on the Planet by the sun:

$$ F_G = G \frac{mM}{r^2} $$

Where $F_G$ is the force on the planet, $G$ is the gravitational constant, $m$ is the mass of the planet, $M$ the mass of the sun and $r$ the distance of the planet to the sun.

Our tutor said that you now have to have a centrifugal force $F_c$ that is equal in absolute value to $F_G$, but opposite in direction so that the sum of the forces would be zero. The absolute value of this $F_c$ would then be equal to $F_G$:


Then you plug in the formulas for the respective forces:

$$ |\vec{F_c}| = |\vec{F_G}| $$ $$ m \frac{v^2}{r} = G \frac{mM}{r^2} $$ $$ \frac{v^2}{r} = G \frac{M}{r^2} $$ $$ v^2 = G \frac{M}{r} $$ $$ v = \sqrt{G \frac{M}{r}} $$

Where $v$ is the tangential velocity of the planet.

Together with $v=\frac{2\pi r}{T}$, where $T$ is the revolution period, you get:

$$ \frac{2\pi r}{T} = \sqrt{G \frac{M}{r}} $$ $$ \frac{4\pi^2 r^2}{T^2} = G \frac{M}{r} $$ $$ \frac{r^3}{T^2} = G \frac{M}{4\pi^2} $$

Since the right part is constant for each solar system, you can form a ratio and optain Kepler's law. So this approach does work.

I think that this view is only valid if you construct it from a rotating observation point, i. e. standing on the planet and facing the sun at all times. Our tutor said that this is from a static (with respect to the sun) observation point.

In the rotating system, you do have the centrifugal force, but it is not a real force in this sense. So by saying that you have a centrifugal force, I think that you are in this rotating reference frame. And in the rotating frame, the planet is not moving at all. And by not moving in this rotating frame, it does revolve around the sun in the static frame. Since the planet is not moving, there must not be any force on it. Constructing a force equilibrium is the way to achieve this.

But our tutor told me that there is a force equilibrium in the static reference frame as well. I say that if you have an equilibrium there, there is no net force, therefore the planet does only move along a straight line in a static frame. I was then told that although there is no net force, the forces are still there.

At that point, I think that this is converting dynamics into statics by adding opposing forces so that everything cancels out.

Say we do have a net force (gravity) on the planet.


This force is something like this, if the planet has rotated $\theta$ around the sun:

$$ \vec{F_G} \propto \left( \begin{matrix} -\cos(\theta) \\ -\sin(\theta) \end{matrix} \right) $$

If you integrate this, you will get the velocity and the position of the planet.

$$ \int \vec{F_G} \, \mathrm d\theta \propto \left( \begin{matrix} -\sin(\theta) \\ \cos(\theta) \end{matrix} \right) $$

This seems very good, since it goes along the circle, so this is the tangential velocity.

$$ \iint \vec{F_G} \, \mathrm d\theta \propto \left( \begin{matrix} \cos(\theta) \\ \sin(\theta) \end{matrix} \right) $$

And that is simply the position around the unit circle.

You can substitute $\theta$ with $\omega t$ if you wish to express the angle with time.

So I claim that you need a net force in order for the circular trajectory to appear at all. My tutor told me that if the force is too high (too weak), the planet will spiral into (away from) the sun, and therefore the equilibrium has to be there. That makes no sense to me. What makes sense to me is that if the trajectory is deflected too much (not enough) to form a circle, there is no circle.

If you have no force, integration turns out quite different:

$$ F = 0 $$

$$ \int F \, \mathrm dt = \underbrace{0t}_{a=0} + \underbrace{c_1}_{v_0} $$

$$ \iint F \, \mathrm dt = \underbrace{0t^2}_{a=0} + \underbrace{c_1t}_{v_0} + \underbrace{c_2}_{d_0} $$

The latter is your $d = v_0 t + d_0$, which models a linear, unaccelerated movement. In the rotating system, $v_0$ is just zero, and it works out. In the static system, that would mean that the planet either remains still or moves on a straight line -- which is not what is really happening.

So I claim that you must not have a force equilibrium for the earth to rotate -- that is the basic disagreement we have.

Who is right? Or are we both right, just looking differently at the same problem?


1 Answer 1


In an inertial reference frame, there most certainly is NOT zero net force on a planet. There is a force pulling the planet towards the sun, and that's it. There are no other forces on the planet. You are 100% correct: If there really were no net force on the planet, then the planet would be stationary or travel in a straight line at constant velocity. It would not travel in a circle or ellipse. This is Newton's First Law. If your tutor disagrees with Newton's First Law, I suggest you find a better tutor!

  • $\begingroup$ My tutor told me that it would come obvious once I would see the derivation of the centrifugal force in theoretical physics lecture (I am a freshman). Is there anything that will come up eventually, or is there nothing else that could explain my tutor's position? $\endgroup$ Nov 15, 2011 at 15:56
  • $\begingroup$ No matter how much physics you learn, Newton's First Law will still be true. Well, I suppose general relativity changes the story, but it sounds like that's not what the tutor was talking about. $\endgroup$ Nov 17, 2011 at 13:46
  • $\begingroup$ @Steve Byrnes is correct. The centrifugal force (and the other fictitious forces: coriolis, transverse, ...) all only appear in a non-inertial (accelerating) reference frame. They never appear in an inertial frame. $\endgroup$
    – John Darby
    Jan 16, 2021 at 23:01

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