The motion of a planet in a central force field has been restricted to a single plane. The equations below describe its motion in the $x$ and $y$ directions.



I'm supposed to show that in order to get a circular orbit, I should choose an initial velocity $v = 2\pi/\sqrt{r}$, perpendicular to $\mathbf{r}$, where $\mathbf{r}$ is the position vector of the planet and $r = (x^2+y^2)^{1/2}$.

I know that I can just start by assuming that the orbit is circular, and use the equations of uniform circular motion to show that the velocity has the magnitude and direction required, as below:

Centripetal force on body = $F = mv^2/r$ where $m$ is the mass of the planet and $r$ the radius of its orbit.

Also, $F = GMm/r^2 = 4\pi^2m/r^2$ where $G$ is the universal gravitational constant and $M$ the mass of the central body. (Here, units are years and A.U so $GM = 4\pi^2$ (why is this?))

Then, $v^2/r = 4\pi^2/r^2 \implies v = 2\pi/\sqrt{r}$.

I'm sceptical of this solution. Is there a better way to approach this problem?

  • $\begingroup$ You're better off working in polar coordinates centred on the centre of the force field. $\endgroup$
    – Chappers
    Oct 13, 2015 at 17:05
  • $\begingroup$ What are your assumptions? If your only assumption is that the initial velocity is perpendicular to your initial position vector then your solution is wrong as you are also assuming the path is a circle. $\endgroup$ Oct 13, 2015 at 18:26
  • $\begingroup$ @H.R. The OP explains that the initial velocity is perpendicular to the initial position vector and with the specific value 2pi/sqrt(r). $\endgroup$
    – Did
    Oct 13, 2015 at 18:34
  • 1
    $\begingroup$ @H.R. How would I then show, instead, that if the initial velocity is as given, the resulting path will be a circle? $\endgroup$
    – 2good4this
    Oct 13, 2015 at 19:37
  • 1
    $\begingroup$ @H.R. Please use @user, not @ user. The space between @ and user makes that the comment is not signalled to user. $\endgroup$
    – Did
    Oct 14, 2015 at 16:33

2 Answers 2


Problem Statement

Suppose that a particle has a radial acceleration of the form

$${\bf{a}} = - {{GM} \over {{r^2}}}{{\bf{r}} \over r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$

where $G$ and $M$ are physical positive constants, $r = \left\| {\bf{r}} \right\|$ is the absolute value of the position vector, and ${\bf{r}}$ is the position vector . Prove that if the initial velocity is perpendicular to the initial position vector and has the absolute value $\sqrt {{{GM} \over {{r_0}}}} $, i.e.

$$\left\{ \matrix{ {{\bf{r}}_0}.{{\bf{v}}_0} = 0,\,\,\,\,{{\bf{r}}_0} \ne 0,\;\;\;{{\bf{v}}_0} \ne 0 \hfill \cr {v_0} = \left\| {{{\bf{v}}_0}} \right\| = \sqrt {{{GM} \over {{r_0}}}} \hfill \cr} \right. \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$

then the path of the particle will be a circle.


I tried to write an abstract solution for the problem and consequently the shortest one. Although it is not short! The strategy is to obtain the path in a general form and then apply the assumptions we have. First of all, we will show that this is a planar motion. For this purpose, consider the following

$$\left\{ \matrix{ {\bf{r}} \times {\bf{a}} = {\bf{r}} \times - {{GM} \over {{r^2}}}{{\bf{r}} \over r} = {\bf{0}} \hfill \cr {\bf{r}} \times {\bf{a}} = {\bf{r}} \times {{d{\bf{v}}} \over {dt}} = {d \over {dt}}\left( {{\bf{r}} \times {\bf{v}}} \right) \hfill \cr} \right.\,\,\,\,\, \to \,\,\,\,\,\,\,{d \over {dt}}\left( {{\bf{r}} \times {\bf{v}}} \right)\, = 0\,\,\,\, \to \,\,\,{\bf{r}} \times {\bf{v}} = {\bf{c}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)$$

where by initial conditions you can find ${\bf{c}}$ as

$${\bf{c}} = {{\bf{r}}_0} \times {{\bf{v}}_0} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)$$

and hence by assumptions ${\bf{c}} \ne {\bf{0}}$. Consequently, we have ${\bf{r}}.{\bf{c}} = 0$ which implies that ${\bf{r}}$ lies in the plane whose normal is ${\bf{c}}$. Before continuing, I just remark two identities which I will make use of it

$${\bf{r}}.{{d{\bf{r}}} \over {dt}} = {d \over {dt}}\left( {{1 \over 2}{\bf{r}}.{\bf{r}}} \right) = {d \over {dt}}\left( {{1 \over 2}{r^2}} \right) = r{{dr} \over {dt}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)$$

$${\bf{a}} \times \left( {{\bf{b}} \times {\bf{c}}} \right) = \left( {{\bf{a}}.{\bf{c}}} \right){\bf{b}} - \left( {{\bf{a}}.{\bf{b}}} \right){\bf{c}}$$

Now we go on. Consider the followings

$$\eqalign{ & \left\{ \matrix{ {\bf{a}} \times {\bf{c}} = {{d{\bf{v}}} \over {dt}} \times {\bf{c}} = {d \over {dt}}\left( {{\bf{v}} \times {\bf{c}}} \right) \hfill \cr {\bf{a}} \times {\bf{c}} = - {{GM} \over {{r^2}}}{{\bf{r}} \over r} \times \left( {{\bf{r}} \times {\bf{v}}} \right) = - {{GM} \over {{r^3}}}{\bf{r}} \times \left( {{\bf{r}} \times {{d{\bf{r}}} \over {dt}}} \right) \hfill \cr \,\,\,\,\,\,\,\,\,\, = - {{GM} \over {{r^3}}}\left[ {\left( {{\bf{r}}.{{d{\bf{r}}} \over {dt}}} \right){\bf{r}} - \left( {{\bf{r}}.{\bf{r}}} \right){{d{\bf{r}}} \over {dt}}} \right] \hfill \cr \,\,\,\,\,\,\,\,\,\, = - {{GM} \over {{r^3}}}\left( {r{{dr} \over {dt}}{\bf{r}} - {r^2}{{d{\bf{r}}} \over {dt}}} \right) \hfill \cr \,\,\,\,\,\,\,\,\,\, = GM\left( {{1 \over r}{{d{\bf{r}}} \over {dt}} - {1 \over {{r^2}}}{{dr} \over {dt}}{\bf{r}}} \right) = GM{d \over {dt}}\left( {{{\bf{r}} \over r}} \right) \hfill \cr} \right.\,\, \cr & \to \,\,\,{d \over {dt}}\left( {{\bf{v}} \times {\bf{c}}} \right) = {d \over {dt}}\left( {GM{{\bf{r}} \over r}} \right)\,\, \cr & \to \,\,{\bf{v}} \times {\bf{c}} = \,GM{{\bf{r}} \over r} + {\bf{b}} \cr} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(6)$$

Again use the initial conditions to determine ${\bf{b}}$ from the result of $(6)$ as follows

$$\eqalign{ & {\bf{b}} = {{\bf{v}}_0} \times {\bf{c}} - \,GM{{{{\bf{r}}_0}} \over {{r_0}}} = {{\bf{v}}_0} \times \left( {{{\bf{r}}_0} \times {{\bf{v}}_0}} \right) - \,GM{{{{\bf{r}}_0}} \over {{r_0}}} \cr & \,\,\,\, = \left( {{{\bf{v}}_0}.{{\bf{v}}_0}} \right){{\bf{r}}_0} - \left( {{{\bf{r}}_0}.{{\bf{v}}_0}} \right){{\bf{v}}_0} - \,GM{{{{\bf{r}}_0}} \over {{r_0}}} \cr & \,\,\,\, = \left( {v_0^2 - {{GM} \over {{r_0}}}} \right){{\bf{r}}_0} - \left( {{{\bf{r}}_0}.{{\bf{v}}_0}} \right){{\bf{v}}_0} \cr} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(7)$$

Now, dot product the last equation in $(6)$ by ${\bf{r}}$ to eliminate ${\bf{v}}$ and obtain the equation of the path

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,{\bf{r}}.\left( {{\bf{v}} \times {\bf{c}}} \right) = \,GM{{{\bf{r}}.{\bf{r}}} \over r} + {\bf{r}}.{\bf{b}} \cr & \to \,\,\,\,\,\,\,{\bf{c}}.\left( {{\bf{r}} \times {\bf{v}}} \right) = GM{{{r^2}} \over r} + {\bf{r}}.{\bf{b}} \cr & \to \,\,\,\,\,\,\,{\bf{c}}.{\bf{c}} = GMr + {\bf{r}}.{\bf{b}} \cr & \to \,\,\,\,\,\,\,\,{c^2} = GMr + {\bf{r}}.{\bf{b}} \cr} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(8)$$

and finally the equation of path is

$${c^2} = GMr + {\bf{r}}.{\bf{b}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(9)$$

where by $(4)$ and $(7)$ we have $c$ and ${\bf{b}}$ in terms of initial position vector and velocity, and hence we have the general form of the path in terms of initial conditions. When does equation $(9)$ correspond to a circle? Consider the case ${\bf{r}}.{\bf{b}}=0$. But according to ${\bf{b}}$ being a constant vector and ${\bf{r}} \ne {\bf{0}}$ which varies with time, this can happen only when ${\bf{b}} = {\bf{0}}$ which is equivalent to

$$\left( {v_0^2 - {{GM} \over {{r_0}}}} \right){{\bf{r}}_0} - \left( {{{\bf{r}}_0}.{{\bf{v}}_0}} \right){{\bf{v}}_0} = {\bf{0}}\,\,\,\,\, \to \,\,\,\,\,\,\left\{ \matrix{ v_0^2 = {{GM} \over {{r_0}}} \hfill \cr {{\bf{r}}_0}.{{\bf{v}}_0} = 0 \hfill \cr} \right.\,\,\,\,\, \to \,\,\,\,\,\,\,\,\left\{ \matrix{ {v_0} = \sqrt {{{GM} \over {{r_0}}}} \hfill \cr {{\bf{r}}_0}.{{\bf{v}}_0} = 0 \hfill \cr} \right. \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(10)$$

where I used the linear in-dependency of ${{{\bf{r}}_0}}$ and ${{\bf{v}}_0}$. In this case, equation $(9)$ becomes

$${c^2} = GMr + 0\,\,\,\,\, \to \,\,\,\,\,\,r = {{{c^2}} \over {GM}} = {{r_0^2v_0^2} \over {GM}} = {{r_0^2{{GM} \over {{r_0}}}} \over {GM}} = {r_0} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(11)$$

which indeed is the equation of the circle. Finally, we are done! :)

  • $\begingroup$ Nicely done! After this can we now derive eccentricity of orbit as a function of arbitrary initial velocity perpendicular to position vector? $\endgroup$
    – Narasimham
    Oct 14, 2015 at 9:09
  • $\begingroup$ Thanks. I didn't get you. In general, we can obtain the eccentricity in terms of initial conditions, i.e., the initial position vector and velocity. In this case which the path is a circle, it turns out to be zero. :) $\endgroup$ Oct 14, 2015 at 9:13
  • $\begingroup$ Yes I was hinting about including the general part. The result is classical Newtionian. Escape velocity $ = \sqrt {\frac {2GM}{r_o}}$... $\endgroup$
    – Narasimham
    Oct 14, 2015 at 9:29
  • $\begingroup$ Yeah, I think you are asking for initial conditions which the path is a hyperbola. :) That's interesting too! You can derive eccentricity from Eq. (9) and investigate this case too. $\endgroup$ Oct 14, 2015 at 9:34
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    $\begingroup$ Yes. Remember, Newton drew that famous picture.. as the velocity is increased the projectile trajectory goes from parabola to circle to ellipse to hyperbola.. $\endgroup$
    – Narasimham
    Oct 14, 2015 at 10:10

Problem Statement:

$${\ddot {\bf{r}}} = - {{k} \over {{{\left\| {\bf{r}} \right\|}^3}}}{\bf r}\qquad k\gt 0$$ $$\left\{ \matrix{ {{\bf{r}}(0)}.{\dot{\bf{r}}(0)} = 0\hfill \cr \left\| \dot{\bf{r}}(0) \right\|^2 = {k \over {\bf{r}}(0)} \hfill \cr} \right.$$ we want to prove: $${\left\| {\bf{r}} \right\|}=constant$$ Lemma: $${\left\| {\bf{r}} \right\|}=constant \quad \iff \quad {{\bf{r}}}.{\dot{\bf{r}}} \equiv 0 $$ So I prove in this case we have: $$u:={{\bf{r}}}.{\dot{\bf{r}}} \equiv 0$$ Suppose $${\bf A}:={\dot u}\,{\bf r}-u\,{\dot{\bf{r}}}$$ It is easy to prove that ${\dot{\bf{A}}}\equiv 0$ and so we have ${\bf{A}}=constant$.
But $u(0)={\dot u}(0)=0$ (why?) So: $${\bf{A}}\equiv {\bf{A}}(0)=0 $$ So: $${\bf{A}}.{\dot{\bf{r}}}\equiv 0$$ Or: $${\dot u}u-u\,{\dot{\bf{r}}}.{\dot{\bf{r}}}\equiv 0$$ Or: $$u({\dot u}-{\dot{\bf{r}}}.{\dot{\bf{r}}})\equiv 0$$ But ${\dot u}-{\dot{\bf{r}}}.{\dot{\bf{r}}}$ is nowhere zero because: $${\dot u}-{\dot{\bf{r}}}.{\dot{\bf{r}}}=\bf{r}.{\ddot{\bf{r}}}=- {{k} \over {{{\left\| {\bf{r}} \right\|}}}}$$ So: $$u\equiv 0$$ That's it :) I love also Solution of my friend H. R.

  • $\begingroup$ It's a good proof. Please write down that part of proof to show $u(0) = u'(0) = 0$. It is this part which makes use of assumptions. :) $\endgroup$ Oct 15, 2015 at 14:30
  • $\begingroup$ It is left to the reader :) $\endgroup$
    – A.F.23
    Oct 15, 2015 at 15:38
  • $\begingroup$ @A.F.23: Thank you for you concise response! It would appear to me, however, that your solution suggests that it is only the initial direction of the velocity that is of importance, not the magnitude. I have not seen you make use of the magnitude at any point. Am I missing something? $\endgroup$
    – 2good4this
    Oct 19, 2015 at 16:55
  • $\begingroup$ Yes you are, read last two comments. I also wrote a "why?" In my proof for you, that is it :) $\endgroup$
    – A.F.23
    Oct 19, 2015 at 18:35

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