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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?

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    Provide a figure, please. – 2017-02-18
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    This is the exact question. You have to sketch a picture of the planets. – 2017-02-18
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    It is not because you haven't a picture in your book that you cannot draw one for us... – 2017-02-18
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    This is the full question extracted from an exam paper. I do not know if this question comes from a book or not. – 2017-02-18
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    @LittleRookie As far as i know the trajectory of moon $B$ is a cycloid. – 2017-02-18
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    @MyGlasses No, it can as well be a convex curve. – 2017-02-18
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    @LittleRookie: Finally I solved your problem. – 2017-02-18

1 Answers 1

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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

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

and

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

The sum of the two terms above is

$$\cos(\phi).$$

So,

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

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 $R+r$.

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    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 $\vec m$. – 2017-02-18
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    @CiaPan: Thank you for warning me. I edited my answer now. – 2017-02-18