# Alignment of planets that orbit at different speeds [closed]

If a planet completes its circle around the sun in 120 years and another planet in 70 years then when will they come in a straight line as if they are now?

In the simplest, most likely intended interpretation of this problem, the planets have perfectly circular, coplanar orbits, like runners around a track.

In one year, the inner planet completes $\frac{1}{70}$ of a circle, and the outer planet completes $\frac{1}{120}$ of a circle. Therefore, the planets' angular distance widens by $\frac{1}{70} - \frac{1}{120} = \frac{1}{168}$ of a circle.

This means that the planets will realign again they way they do now in 168 years.

If they're rotating in opposite directions, however, the angular distance widens by the sum of their orbital speeds, rather than their difference, which is $\frac{19}{840}$ of a circle per year. This makes the planets realign in $\frac{840}{19} \doteq 44.21$ years.

The next time they will be in any alignment is in

168 years.
Effectively, every rotation of the fast planet it gains 50 "years" on the slower one. So, after 2 rotations, it's 20 "years" behind the slower. Since the rate of gain is 50/rotation, it will require .4 rotations to catch up. .4 rotations is 28 years for the fast planet. 2.4 rotation is 168 years.
To check our math, 168 years of the slow planet is 1.4 rotations, so they are aligned.

• Not any alignment. If they orbit for 84 years, they will be across from each other relative to the sun. – Joe Z. Mar 17 '15 at 22:28

The question never stated how the 2 planets are rotating. Therefore by assuming they are rotating on the same plane in opposite direction, it'll take less than 60 years for them to become a straight line again, and many more until in 840 years they'll be aligned again in the exact location as they are now

There are a couple key pieces of information missing from this question.

1) Are the two planets' orbits co-planar?

2) Are the two planets orbiting in the same direction?

# Not co-planar orbits

If the answer to (1) is no, then the answer to (2) doesn't matter. The two planets can only align if they have each undergone an integer number of half-orbits. The slower completes a half-orbit in 60 years--the faster in 35, so: $60x = 35y$ with $x$ and $y$ both integers. The smallest solution to this is $x = 7$, or 420 years.

# Co-planar orbits

The slower planet orbits at the speed of $3$ degrees per year, the faster at $36/7$ degrees per year.

## Orbiting the same direction

How long before the faster planet has gone $180$ degrees further than the slower?

$180 + 3t = (36/7)t$

$t = 84$

## Orbiting in opposite direction

How long before the two planets together sweep out an arc of 180 degrees?

$180 = 3t + (36/7)t$

$t = 22.1$ years approx.

• You forgot the mention whether the orbits are circular or even whether the planets always travel the same path. – Joe Z. Mar 17 '15 at 22:35
• @JoeZ. I took "completes its circle around the sun" to mean circular orbits. – dmitch Mar 17 '15 at 22:51

In

840 years

Because

840 is the least common multiple of 120 and 70.

Although

These planets will be aligned in a straight line much more often, it will take them 840 years to be in a straight line as if they are now

• "As they are now" just means that they're in a line of sun, inner planet, outer planet. The bearing of the line doesn't matter. – Joe Z. Mar 17 '15 at 22:29

Never.

Let's call the planets $A$ the faster and $B$ the slower. In 120 years $B$ completed a loop while $A$ completed $1 + \frac{50}{70} = \frac{12}{7}$ of loop. For this purpose we can remove full loops since they don't add information on the relative position of the planets, so we can say that $A$ completed $\frac{5}{7}$ of loop vs. $1$ loop of $B$.

Now in the time $A$ completes the remaining $\frac{2}{7}$ of loop to catch $B$ (in ${70}\cdot{\frac{2}{7}} = 20$ years), $B$ moved away another $\frac{20}{120} = \frac{1}{6}$ of loop, so $A$ has to move more to catch it.

Now in the time $A$ completes the remaining $\frac{1}{6}$ of loop to catch $B$ (in ${70}\cdot{\frac{1}{6}} = 11,67$ years), $B$ moved away another $\frac{\frac{70}{6}}{120} = \frac{7}{72}$ of loop, so $A$ has to move more to catch it.

Now in the time $A$ completes the remaining $\frac{7}{72}$ of loop to catch $B$ (in ${70}\cdot{\frac{7}{72}} = 6,805$ years), $B$ moved away another $\frac{\frac{7}{72}}{120} = \frac{7}{8640}$ of loop, so $A$ has to move more to catch it.

Now in the time $A$ completes the remaining $\frac{7}{8640}$ of loop...

• There is a reason this is called a paradox... – dmg Mar 17 '15 at 15:02
• @dmg Yes, because it is a paradox (??). What's your point? Obviously this question is solely a math problem, not interesting and easy to solve (as downvotes denote), I only wanted to add some Puzzling SE spice in it. Can't fully understand the downvote though, this is a valid response... – Narmer Mar 17 '15 at 15:08
• @Narmer His point is that the logic seems to support your answer of "never", but your answer is plainly wrong. – reo katoa Mar 17 '15 at 16:34