## Observatory

In a solar system far, far away, 2 planets are exactly in line with their sun (as above).

Icerix, the colder planet farther away, takes 350 days to obrit the sun.

Cetacearth, the watery planet closer to the sun, only takes 100.

Assuming the orbits are in the same plane, in how many days will the 2 planets next be in line with their sun again?

*Puzzle concept submitted by Ian Pedder*

No complicated angular/rotational geometry is needed to solve!

[Can you figure this one out? Enter your answer in the comment section and we'll reveal all on Friday!]

jasc| Profile March 19th, 2008 - 11:11 pm700d

zxo| Profile March 19th, 2008 - 11:20 pmAlso assuming the planets orbit in the same direction

If we flash forward 1 Cetacearth year, Cetacearth is back where it started and Icerix has traveled 100 days, or 2/7 or its orbit. For an alignment to occur, the planets must each have traveled the same fraction of their orbit. Keeping in mind Icerix’s 100 day head start, we can write the following equation:

(100+X)/350 = X/100

Rearranging a bit, and reducing fractions:

100 + X = 7X/2

100 = 5X/2

X = 40

Checking: (100+40)/350 = 2/5 –> 40/100 = 2/5

Thus, 140 will elapse before the next alignment.

LadyInsomnia| Profile March 19th, 2008 - 11:36 pmAssuming the orbits are in the same plane makes it simple: just find the least common multiple of 350 and 100. That’s 700 days until they’re in line again.

gizbot| Profile March 20th, 2008 - 2:48 amCute but not deep.

The first estimate, as a check, says it’s going to happen between a hundred and two hundred days. The fast planet circles the sun once and then catches up with the slow planet. After 100 days, the slow planet will have gone about a third the distance around the sun, so an answer should be about 135.

Now, dig out 7th grade algebra:

Let x = the number of days, after the first 100, until the planets line up.

Let x/100 = the portion around the orbit the inner planet until they line up.

Let (100+x)/350 = the portion around the orbit the outer (slow) planet has gone when they line up.

So:

x/100 = (100+x)/350; planets line up with fast and slow in same place.

3.5x = 100+x; multiply both sides by 350.

2.5x = 100; subtract x from both sides

x = 40

So, they meet up at 100+x days, or 140 days.

hex| PUZZLE MASTER | Profile March 20th, 2008 - 4:51 amAligned means on same side of sun or on opposite sides.

Equations are:

Alpha1=t/100 x 2 x Pi where t is in days

Alpha2=t/350 x 2 x Pi

We have 2 cases for alignment:

a- Alpha1 = Alpha2 + 2 x K x Pi where k is an integer

b- Alpha1 = Alpha2 + Pi + 2 x K x Pi where k is an integer

Solving for t:

t=0, 70, 140, 280, 350, 560, 630, …

da Moose| Profile March 20th, 2008 - 7:56 am103.5 days

ag3121| Profile March 20th, 2008 - 8:18 am35 days

Shawn| PUZZLE GRANDMASTER | Profile March 20th, 2008 - 9:12 amAfter one revolution of 100 days by Cetacearth, the outer planet Icerix has moved along its orbit by a percentage of 100/350.

As Cetacearth proceeds on its second revolution, it will catch Icerix after some percentage (x/100) of the Cetacearth orbit has been fulfilled, where x = the number of days IN THE SECOND REVOLUTION.

Icerix will have traced a path on its orbit’s circumference of 100/350, plus some percentage (x/350) of its orbit.

(x/100) = (100/350) + (x/350)

x=40

Cetacearth and Icerix will be re-aligned on day 140.

lograh| Profile March 20th, 2008 - 9:57 amwell, for each “day” that passes (are these days on the inner planet or the outer? to get the visual angle needed for the above picture the two would be drastically different in size, which would imply very different day lengths — but I’ll presume that falls in the category of “over-thinking the problem” the inner gets 1% of the way around, and the outer gets .285714% the way around. Clearly the inner planet will take off “ahead” of the outer planet from this point, and will come around to lap it at some point in the future. The only question is when.

After 100 days, the inner planet has finished 100% a revolution while the outer has only done 28.57% of a revolution. A bit more guess-and-check of the numbers shows that at about day 140, the outer planet has done roughly 40% of it’s revolution, while the inner would have done 140%, which would put it also at the 40% mark. So, my back-of-the-napkin guess would be at sometime on day 140 they line up again.

mroth01| Profile March 20th, 2008 - 11:16 amThere’s probably a shorter way to do this, but here’s what I got:

1) New position of Icerix (in degrees) = Velocity of Icerix (deg/day) * Time (days)

2) New position of Cetacearth (in degrees) = Velocity of Cetacearth * Time – 360 degrees [b/c Cetaearth would have to make a complete revolution]

Set equation1 equal to equation 2, solve for T

(360/350)T = (360/100)T – 360

T = 140 days

MarkS| Profile March 20th, 2008 - 11:40 amClearly the two will coincide during Cetacearth’s second

trip around the sun, ie. in the second 100 days. To figure when both have traveled the same fraction of their orbit’s circumference during Cetacearth;s second, I think the formula becomes:

(d-100)/100 = d/350 and gives the number of days as:

140

Kira| Profile March 20th, 2008 - 12:57 pmI just assumed you’d use the lowest common multiple to 100 and 350, which would be 700, so in 700 days, Icerix would’ve gone around the sun twice and Cetacearth 7 times, and that’s the shortest time for this alignment to be achieved

is that correct?

sharhan| Profile March 20th, 2008 - 1:25 pmWould say in the neighborhood of 138.8 days.

theWizof6| Profile March 20th, 2008 - 1:51 pm700 days

The answer is the least common multiple of the orbit durations.

The LCM of 350 and 100 is 700, thus two orbits for Icerix and seven for Cetacearth.

Ari| Profile March 20th, 2008 - 2:50 pmThe rotational speeds of the bodies are: V_Cetacearth=1/100 revolutions/day,

and V_Icerix=1/350 revolutions/day.

Cetacearth is gaining Icerix with the speed of (V_Cetacearth-V_Icerix)

So the time it takes for the planets to align again is

1 / ((1/100)-(1/350)) = 700/5 = 140 days.

pieman| Profile March 20th, 2008 - 2:53 pm140?

vj2181| Profile March 20th, 2008 - 3:07 pmi think the answer is 140 days after the first alignment.

michaelc| Profile March 20th, 2008 - 4:39 pmCetacearth makes a little over 1 revolution while Icerix makes less than one. We have to determine how long it takes Cetacearth to “catch up” to Icerix after one complete revolution. The easiest way to do it is just set up a formula to solve.

350 days/revoluiton = x days/R revolutions

100 days/revolution = x days/(1+R) revoluitons

2 equations and 2 unknowns. Solving reveals

x= 140 days and

R= 0.4 revolutions

The rotational angle geometry might be more fun to do however.

YukinoMiko| Profile March 20th, 2008 - 5:53 pm700 days

Tommy| Profile March 20th, 2008 - 7:30 pmGuess: 140

nitwit88| Profile March 21st, 2008 - 12:26 am140 days

runninfool| Profile March 21st, 2008 - 3:48 am140 days later.

wally| Profile March 21st, 2008 - 8:43 am140 days. the closer planet would have already orbited once and gone 40% through it’s next orbit. the further planet will be 40% through it’s first orbit at that point.

RK| Profile March 21st, 2008 - 10:45 amFor those who said 140, you are very close to the right answer, but you’ll need to go back and read the question very carefully to catch the 2nd twist to the problem :twisted:

Zxo, Michaelc, Ari, MarkS, Shawn, Gizbot, Lograh, Mroth1 provide very nice explanations of how to go about solving the 1st (and major) part of the question.

wally| Profile March 21st, 2008 - 11:02 amI guess technically the planets could be inline on opposite sides of the sun instead of the same side, so 70 days could be the correct answer.

daniel.leonard| Profile March 21st, 2008 - 2:49 pmThe answer is 70 days. After 70 days, the planet that orbits in 350 days will be 70/350 or 2/10 of the way around the sun, and the planet that orbits in 100 days will be 70/100 or 7/10 of the way around it. They are separated by 5/10 or 1/2 of an orbit, so they are diametrically across from one another, with the star in the middle, forming a line.

runninfool| Profile March 21st, 2008 - 9:23 pmIf the “twist” you are referring to is that they need to be aligned with their sun as well, how would they not be in line with it at 140 days if it is the center of both orbits? The only thing you need to find is where the orbits line up because each planet is always in line with the sun.

suineg| PUZZLE MASTER | Profile March 22nd, 2008 - 12:37 pmRk is from earth so i think his question was in earth days so i think the answer should be in earth days:

140—>350

X—–>365

so i think that would be 146 days, but hey it was just a guess, another curiosity in the picture they were aligned in this order: Icerix, Cetaearth and th sun. In 140 days in would be Icerix, sun, Cetaearth. If you would like to know wer th planets will be alignd with the sun the same way as the picture again will beeee 700 days, i mean 730 earth days. Coooool

mlee| Profile March 22nd, 2008 - 4:27 pm280 days?

tumeke| Profile March 23rd, 2008 - 10:54 pmI dont know how to work this out..

but i’d say at about 66 days they should be opposite each other in line with the sun.

kris fenton| Profile March 24th, 2008 - 6:04 pmBased on orbit alone, the planets would take 140 days to be aligned again, but this is assuming that axis rotation of both planets is the same.

Therefore we suggested that because of the different rotation of the planets that one day on the Icerix one day is equal to on 3.5 days Cetacearth.

Or simply put, the planets are always in line with one another.

RK| Profile March 24th, 2008 - 6:37 pm70 is the answer we were looking for. As Hex originally noted above, in line means on the same side of sun or on opposite sides.

lograh| Profile March 25th, 2008 - 6:02 pmd’oh! I thought it just meant in line as pictured! I should have caught that it was just asking for co-linear in any order. didn’t read close enough, I guess.

michaelc| Profile March 26th, 2008 - 3:23 pmSneaky with mis-leading photagraphy!!

Ok you got me.

Shawn| PUZZLE GRANDMASTER | Profile March 27th, 2008 - 11:54 amI disagree. “…exactly in line with their sun (as above).” clearly means that the two planets are on the same side of the star when the timeline begins. There can be no misinterpretation of the picture, for if the star as pictured is between the planets, it is a very small star indeed.

The only way 70 days can be an acceptable answer is if the parenthetical “as above” is eliminated.

RK| Profile March 27th, 2008 - 10:27 pmHi Shawn- The planets start off as shown in the picture.

You can compare this to a clock where the planets are attached to the tip of the hour and minute hand. A similar question can be asked: Starting from 12 o’clock, when is the next time the hands of the clock in line again?

The hands of the clock will be together again at 1 hour 5 minutes 27 seconds. This is when they are on the ‘same side of the track’, like the answer of 140.

However, at 32 minutes 43 seconds past 12 o’clock, the hands will also be in line with each other (opposite side of the track). This would correspond with the answer of 70.

Is the question/answer reasonable in this light?

RK| Profile March 27th, 2008 - 10:35 pmsorry; for above, it should read- ‘when is the next time the hands of the clock are in line again’

suineg| PUZZLE MASTER | Profile March 29th, 2008 - 11:43 amNice answer, I made too much assumptions, that the has to be in line in the exact same position, and the scale was unnecesary i guess because the frame of reference was earth days. 70–> is a cool answer

suineg| PUZZLE MASTER | Profile March 29th, 2008 - 11:46 amBy the way Hex got the answer way before the 2nd twist comment was made so congratulation Hex

Shawn| PUZZLE GRANDMASTER | Profile March 31st, 2008 - 2:41 pmYes, I see it now.

Thank you, RK.

Tsopi| Profile October 18th, 2010 - 3:42 pmI for Icerix and C for Caracearth d for days

I makes a full round every 100 d

C makes a full round every 350 d Meaning that when I will have done a round, C will have covered the 1/3,5 of its root

==> 2 I rounds (2/3,5) C rounds

==> 3 I rounds (3/3,5) C rounds

==> (At this point C needs 50 d to make a full round witch is the half of I rounds)

==> 3,5 I rounds (3,5/3,5)= 1 C rounds

Notice:C makes a round for every 3,5 I rounds

Conlcusion:When We will have 7 I rounds we will have 2 C rounds=> 7* 100 or 2*350 = 700 d