i need help regarding the following question.

Planet $A$ circles counterclockwise around Star $O$ at a constant angular velocity $\theta$. The radius of the circle is $R$. A moon $B$ circles counterclockwise around Planet $A$ at the same angular velocity $\theta$. The radius of the circle is $r$.

Assume that $A,B,O$ are in the same plane. Show that the trajectory of moon $B$ is a circle centered at Star $O$.

Can we solve this question just by knowledge on conic section and trigonometry?

  • $\begingroup$ Provide a figure, please. $\endgroup$
    – Jean Marie
    Feb 18, 2017 at 12:28
  • $\begingroup$ This is the exact question. You have to sketch a picture of the planets. $\endgroup$ Feb 18, 2017 at 14:20
  • $\begingroup$ It is not because you haven't a picture in your book that you cannot draw one for us... $\endgroup$
    – Jean Marie
    Feb 18, 2017 at 14:31
  • $\begingroup$ This is the full question extracted from an exam paper. I do not know if this question comes from a book or not. $\endgroup$ Feb 18, 2017 at 14:33
  • $\begingroup$ @LittleRookie As far as i know the trajectory of moon $B$ is a cycloid. $\endgroup$
    – Nosrati
    Feb 18, 2017 at 14:41

1 Answer 1


A vector pointing from the sun to the planet is

$$\vec{p(t)}=R(\cos(t\theta)\vec i+\sin(t\theta)\vec j).$$

The vector pointing from the planet to the moon is

$$\vec {m(t)}=r(\cos(t\theta+\phi)\vec i+\sin(t\theta+\phi)\vec j).$$

That is, the moon orbits the planet with the same angular speed and with a phase shift $\phi.$

The sum of these vectors points from the sun to the moon:

$$\vec{p(t)}+\vec{m(t)}=(R\cos(\theta t)+r\cos(\theta t+\phi))\vec i+(R\sin(\theta t)+r\sin(\theta t+\phi))\vec j.$$

In order to check if the moon orbits about the sun along a circle, let's take the squares of the $x$ and the $y$ coordinates.

For the $x$ coordinate:

$$(R\cos(\theta t)+r\cos(\theta t+\phi))^2=R^2\cos^2(\theta t)+2Rr\cos(\theta t)\cos(\theta t+\phi)+r^2\cos^2(\theta t+\phi).$$

For the $y$ coordinate

$$(R\sin(\theta t)+r\sin(\theta t+\phi))^2=R^2\sin^2(\theta t)+2Rr\sin(\theta t)\sin(\theta t+\phi)+r^2\sin^2(\theta t+\phi).$$

And the sum of the squares of the coordinates

$$x^2+y^2=R^2+r^2+2Rr(\cos(\theta t)\cos(\theta t+\phi)+\sin(\theta t)\sin(\theta t+\phi)).$$

I claim that the multiplier of $2Rr$ depends only on $\phi$. Indeed

$$\cos(\theta t)\cos(\theta t+\phi)=\cos(\theta t)(\cos(\theta t)\cos(\phi)-\sin(\theta t)\sin(\phi))=$$ $$=\cos^2(\theta t)\cos(\phi)-\cos(\theta t)\sin(\theta t)\sin(\phi)$$


$$\sin(\theta t)\sin(\theta t+\phi)=\sin(\theta t)(\sin(\theta t)\cos(\phi)+\cos(\theta t)\cos(\phi))=$$$$=\sin^2(\theta t)\cos(\phi)+\sin(\theta t)\cos(\theta t)\sin(\phi).$$

The sum of the two terms above is




Which is the equation of a circle with radius $\sqrt{R^2+r^2+2Rr\cos(\phi).}$ If the phase is $0$ then the radius is exactly $R+r$.

  • $\begingroup$ It's wrong. The problem formulations defines both angular velocity of a planet with respect to the star and the moon with respect to the planet equal.– but it does not say the phases of both circular movements are equal. You may use, say, $(t\theta)$ for $\vec p$, assuming you can arbitrarily settle the zero of $t$, but then necessarily $(t\theta + \phi)$ with some constant $\phi$ for $\vec m$. $\endgroup$
    – CiaPan
    Feb 18, 2017 at 14:18
  • $\begingroup$ @CiaPan: Thank you for warning me. I edited my answer now. $\endgroup$
    – zoli
    Feb 18, 2017 at 21:49

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.