If a body is acted on by a force which obeys the inverse square law, will it mean the body follows an ellipse, if so, equating Newton's law of universal gravitation $(f=(GMm/(r^2)))$ to centripetal force $(mv^2)/r$ is erroneous.

  • $\begingroup$ Equating the gravitational and centripetal accelerations only works for circular orbits i.e. $\ddot{r}=0$. Is that the point you are trying to make? $\endgroup$ Nov 22, 2016 at 17:11
  • $\begingroup$ yes , in my class however I have been told to equate these two. $\endgroup$
    – Think
    Nov 22, 2016 at 17:16

2 Answers 2


Let the interaction potential be of the form $V=kr^n$.

The eccentricity of the orbit is given by the expression (Refer "Goldstein, Classical Mechanics, 2nd edition, chapter 3)


If $e>1$, the shape of the orbit is hyperbola,

If $e=1$, the shape of the orbit is parabola,

If $e<1$, the shape is ellipse,

If $e=0$, the shape is circle,

So the shape of the orbit depends on the total energy of the system, not on the form of Force. The shape of the orbit can be ellipse or hyperbola or parabola or circle depends on the total energy of the system.

The total energy of the system is given by $$E=\frac{1}{2}mv^2+V_{eff}$$

Where $V_{eff}=V+\frac{l^2}{2mr^2}$, $l$ is the angular momentum which is a constant for a central force problem.

From Virial's theorem, it is known that, $$<T>=\frac{-1}{2}<V>$$

So $E=\frac{-k}{2r_{0}}$, if $r_{0}=\frac{l^2}{mk}$, the orbit is circular, For Elliptical Orbit, $E=\frac{-k}{2a}$, where $a$ is the semimajor axis.

Finally only for uniform cirular motion you can equate Centripetal force with Gravitational force.


Yes, if you are considering uniform circular motion, then you can use $$ mv^2\,r^{-1}=GMm\,r^{-2}. $$ And I imagine most textbooks try to make it clear that this is valid only in that case.

However, what you want to solve is the sum of the forces: $$ \sum F=F_g $$ That is, $$ m\frac{d^2r}{dt^2}-mr\omega^2=-\frac{GMm}{r^2}\tag{1} $$ where the $r\omega^2=v^2/r$ term is the rotational acceleration and $M$ the mass of the larger body (i.e., the star). If you let $L=mr^2\omega$ be the angular momentum (which is conserved here, so it is a constant in time), then (1) can become $$ \frac{d^2r}{dt^2}=-\frac{GM}{r^2}+r\left(\frac{L}{mr}\right)^2 $$ which, with some substitutions and work (see the section labeled "Kepler's First Law"), can be reduced to, $$ \frac{a\left(1-e^2\right)}{r}=1+e\cos\theta $$ where $a$ is the semi-major axis of the ellipse, $e$ the eccentricity and $r,\,\theta$ are the typical polar coordinates. And this equation is the standard equation for an ellipse


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