[Guest]

Question 8: A satellite moving at constant speed can orbit the moon one time in eight hours. After each complete cycle, the satellite instantly reverses direction a quarter of the way back around the moon before immediately continuing forward again for another complete cycle. Beginning in the forward direction above a particular spot of the moon, how many hours does it take the satellite to orbit that spot three times?

36 Hours

32 Hours

24 Hours

56 Hours

44 Hours

How is it not 32 hours ?

One complete cycle = 8 hours. But since it goes back 1/4 a way ( 2 hours) it'll take 4 hours just to start again.

So before starting the 2nd cycle, 12 hours has passed.

2nd cycle, takes 8 hours to complete But then again, it goes back 1/4 way , so another 4 hours to start the 3rd cycle. Total hours

first cycle : 12

second cycle 12 = 24 hours.

Third cycle, only needs to return to that point. So 8 hours. Yet 32 hours is wrong. The answer is 36, why ?

Thanks

[Guest]

I'm thinking (i didn't do the math but)...maybe the moon's rotation?

[Guest]

I'm thinking (i didn't do the math but)...maybe the moon's rotation?

I was thinking that but it seems too complicated to be solved in your head which youre suppose to do.

[Guest]

First turn: 8 hours (first pass over the point)

back: + 2 hours = 10 hours

another turn: + 8 = 18 hours (second pass over the point)

back: + 2 = 20 hours

another turn: + 8 = 28 hours (third pass over the point)

back: + 2 = 30 hours

another turn: + 6 (thir turn completes 1/4 of turn beyond the start, so I just took 6 hours) = 36 hours.

Try doing the path with your finger in the air, you will get it easily

[Guest]

I'm thinking (i didn't do the math but)...maybe the moon's rotation?

You refer to the difference between an apparent orbit as seen by an observer on the lunar surface and what might be considered to be a sidereal orbit. The question setter is clearly thinking in terms of an apparent orbit.

Edited August 7, 2009 by jerbil

[Guest]

Skolnick i dont think that answer is correct because you have made more than 3 turns past the point.

Does a full cycle include going back the 1/4 turn then gouing back to the starting point before actually beginning the next cycle? If so that would mean that one full cycle actually is equivalent to 12 hrs rather than 8 hrs like stated in the question which would then mean that in order to officially complete the third cycle you would still have to go back a quarter turn then forward a quarter turn. So in essence the math stated in question would just be incomplete, No?

[Guest]

It only makes the first pass over the start point at 12 hrs, yes... then the second pass at 24 hrs, yes... then the third pass at 36 hrs. IF after reversing direction the quarter turn the next full 8 hr cycle is at that point initiated.

[Guest]

It only makes the first pass over the start point at 12 hrs, yes... then the second pass at 24 hrs, yes... then the third pass at 36 hrs. IF after reversing direction the quarter turn the next full 8 hr cycle is at that point initiated.

Its like this..

1st rotation -> For one complete rotation, it takes 8 hrs. after every cycle, it takes a quarter cycle back. the time taken for it is 2 hrs. (8+2 = 10hrs)

2nd rotation -> It has to move these 2 hours forward to complete another cycle. Also it moves a quarter step back. so it moves back 4 hrs. (8+2+2 = 12 hrs)

3rd rotation -> Again, it has to move 4 hrs in additional to complete another rotation apart from the 8hrs for a rotation from where it started and 4hrs to cope up with the reverse action and move forward. (8+2+4 = 14hrs).

totally its 36 hrs.

The catch here is the rotation does not complete when it has reached the spot where it was because of the reversal. it has to move to the position where it started to complete one full cycle.

[Guest]

If the moon is rotating at half the speed as the sattelite, then it may take 36 hrs. to cross the spot on the moon 3x if the forward direction of the sattelite is in the opposite direction of the moon's rotating direction...

so the sattelite takes off going the opposite way the spot on the moon is 'going'...at 8 hrs. they meet, and the sattelite reverses it course, and is now going the same direction as the spot...so it will take twice as long (4 hrs.) to gain the 1/4 distance from the spot plus the normal 2 hrs. back to the spot to complete the cycle...so it would actually take 14 hours per cycle for the first 2 cycles...14+14=28 then the last cycle is only 8 hrs. so you get 36 hrs...hope I did my math and calculations right..

[Guest]

I think it's the way the question is worded. It took me drawing this out three times before I reread the question and got it! This, of course, is ignoring the obvious fact that the moon DOES move, but since the question did not address it, neither did I.

For the moment, let's imagine space is 2D. The moon is a circle, okay? Let's say the satellite begins at the very top, and it travels clockwise for a full rotation (eight hours). Then, accordiing to the question, it will IMMEDIATELY REVERSE and travel for a quarter of a turn (two hours), at which point it willl continue back clockwise. This will happen twice more (a total of thirty hours) at which point the satellite will be a quarter of a turn away from its starting point IF IT CONTINUES GOING COUNTER-CLOCKWISE. But, it has to go back clockwise, making it six more hours. If you don't believe me, draw it out.

[Guest]

orbit i'll guess the other guys are right *shrug* pass over ... it could be as little as 10 hours 1 second ... If the moon is 2d like a clock face ... the point could be a microscopic fraction past the 9 ... hence it would go round from 12 back to 12 making it 1 pass ... then reverse to 9 making it 2 passes ... then just a tiny little bit to pass back over the point on the moon making it 3 passes ... So ... in theory ... all your options are wrong ... :-) breaking questions is more entertaining

[Guest]

Kyiela--

But if you keep going, start counting the time the first moment it hits the spot. I think that's what the question was asking.

[Guest]

one complete cycle involves 8 hours clockwise (a complete rotation) AND a 2 hour quarter travel anti clockwise. Each cycle therefore lasts 10 hours, and ends up a quarter revolution back from where it started. After two cycles (20 hours) you'll be half way round, and after 3 cycles (30 hours) you'll be 3 quarters away from start.

However you'll have already pass over the start point 3 times. If you take your first cycle, if you count landing on it and reversing as a pass over, then the second cycle passes over it the second time, and during the third cycle you pass over it for the third time - half a rev from the start point, ie 4 hours - so that's 2 complete cycles plus 4 hours - 24 hours.

On the other hand if you have to 'pass completely over the start point' - then you don't do this on the first cycle - you hit it, and reverse, without passing completely over it, so in this case the answer is 3 cycles plus 3/4 of a rev - 36 hours.

[Guest]

Now that I think about it, the question is ambiguously worded (as JonR pointed out above). Of course, that doesn't take into consideration the OP (and my) misinterpretation of the words "complete cycle"....complete cycle means "full orbit" and not "pass over the same point".

t0 start

orbit 1 complete at time t1 (8 hours)

4 hours later, orbit 2 starts. (2 hours for reversal, 2 hours to reach the point)

6 hours later, a reversal takes place (the orbit after the first reversal ends)

6 hours later, (2 hours to reverse, 2 hours to rereach the first reversal, and 2 hours to rereach the op) orbit 2 ends, orbit 3 begins

4 hours later, a reversal (halfway around the moon) takes place

4 hours later, you are back to halfway around.

4 hours later, you are back to your original point.

Orbit 1 takes 8 hours

Orbit 2 takes 16 hours

Orbit 3 takes 12 hours

Total: 36 hours.

Edited August 7, 2009 by tpaxatb

[Guest]

Orbit 1 takes 8 hours

Orbit 2 starts 4 hours later

Orbit 2 takes 12 hours

Orbit 3 starts immediately

Orbit 3 takes 12 hours

For a total of 36 hours

[Guest]

Skolnick i dont think that answer is correct because you have made more than 3 turns past the point.

Does a full cycle include going back the 1/4 turn then gouing back to the starting point before actually beginning the next cycle? If so that would mean that one full cycle actually is equivalent to 12 hrs rather than 8 hrs like stated in the question which would then mean that in order to officially complete the third cycle you would still have to go back a quarter turn then forward a quarter turn. So in essence the math stated in question would just be incomplete, No?

Hmmm...with my solution, the satellite passes four times over the point, which is exactly three turns over the moon. When it gives one round, it passes two times over the starting point, doesn't it?

[Guest]

Assuming it starts at the spot it see it 3 times in 12hrs. Think about it: it sees it at time equals zero then 8hrs later it sees it again then it goes backwards and for two hours and continues forward for two more that's 4 more than the original 8 which is only 12. Assuming the Moon is static (not rotating) the satelite can spot the same spot three times in as little as 12hrs. Time zero is always when the satelite first sees the spot so the answer is 12hrs for three sightings. Try it have a friend stand still and walk around them... LOL

[Guest]

This solution meets the criteria of passing over (or hitting) the spot only three times and doesn't require reinterpretation of the OP's question. Yielding 36 hours.

I'm thinking that the answer has to do with the rotation of the moon. The moon rotates once in about 27 days. So the spot on the moon that the satellite started over has moved a little more than 1/27th of a rotation in 32 hours (Point being that it has moved further along on each cycle). Assuming that the satellite is rotating in the same direction that the moon rotates, by the time it completes one cycle, that spot has moved further along than one cycle. So on each cycle, the satellite has yet to reach the spot. Therefore, it has to go back a quarter and then forward again. It will probably take slightly more than 36 hours because it has to travel an additional 1/27ths (plus a little extra) of a rotation over its entire trip, but this is negligent (a few minutes).

Edited August 7, 2009 by Kriil

[Guest]

the 12hrs mentioned still works but the satelite doesn't see the spot at time zero. The words "start again" it must go back to the origin to begin the timing cycle. Therefore each cycle is 12 hours and 12 x 3 is 36?? Assuming the object is always lit of course...

[Guest]

Sorry guys, I go for the 24h-answer:

- we have to be on the vertical of the initial point 3x;

- a "full orbit" is a complete circling of the Moon (forget the Moon's movement);

Satellite starts at point "P";

Makes a full orbit (8h), and is over P again;

Goes back 1/4 (another 2h, total 10h), it's over point P-2h now, and starts another orbit;

After 1/4 (more 2h, total 12h), makes second pass over P;

6h later (total 18h), finishes second orbit over point P-2h;

Goes back 1/4 (2h, total spent 20h), to point P-4h, on the opposite side of the Moon, to start 3rd orbit;

Half an orbit later (4h, total spent 24h), is over P again (3rd time);

[Guest]

Sorry guys, I go for the 24h-answer:

- we have to be on the vertical of the initial point 3x;

- a "full orbit" is a complete circling of the Moon (forget the Moon's movement);

Satellite starts at point "P";

Makes a full orbit (8h), and is over P again;

Goes back 1/4 (another 2h, total 10h), it's over point P-2h now, and starts another orbit;

After 1/4 (more 2h, total 12h), makes second pass over P;

6h later (total 18h), finishes second orbit over point P-2h;

Goes back 1/4 (2h, total spent 20h), to point P-4h, on the opposite side of the Moon, to start 3rd orbit;

Half an orbit later (4h, total spent 24h), is over P again (3rd time);

We don't have to be over P 3 times. We have to start and stop over P three times. The second time we reach P only begins the second orbit.

The third time it passes over the point P (the original point), it has only orbited that point TWICE.

The first time it is over P again, it has completed its first orbit.

The second time it is over P (after it has reversed and restarted) only STARTS the second orbit over P.

The third time it is over P (after a reverse and restart to P-4H), it has COMPLETED its second orbit over P. Since it isn't reversing, this IMMEDIATELY starts its third orbit over P.

The fourth time it is over P, (reversing from P-4h to P-6h) it has now completed its 3rd orbit.

[Guest]

We don't have to be over P 3 times. We have to start and stop over P three times. The second time we reach P only begins the second orbit.

The third time it passes over the point P (the original point), it has only orbited that point TWICE.

The first time it is over P again, it has completed its first orbit.

The second time it is over P (after it has reversed and restarted) only STARTS the second orbit over P.

The third time it is over P (after a reverse and restart to P-4H), it has COMPLETED its second orbit over P. Since it isn't reversing, this IMMEDIATELY starts its third orbit over P.

The fourth time it is over P, (reversing from P-4h to P-6h) it has now completed its 3rd orbit.

You should refrase the problem, then: "reverses direction a quarter of the way back around the moon before immediately continuing forward again for another complete cycle" clearly states that after reverse a new start begins. So the start and "end before reverse" of each orbit is always different from the previous one.

[Guest]

I get the standard 36.

Visualize a clock. Starting point = "12".

1. Do one circuit, arriving back at "12". 8 hours. ONE ORBIT.

2. Back up to "9". 2 hours (total 10 hours). This is your new start point.

3. Do one more circuit, arriving back at "9". 8 hours (total 18 hours). (You do pass over "12" for the second time.)

4. Back up to "6". 2 hours (total 20 hours). This is your new start point.

5. Arrive at "12". 4 hours (total 24 hours). TWO ORBITS. (Third time over the "12".)

6. Assuming you don't like that one, continue on to "6". 4 hours (total 28 hours).

7. Back up to "3". 2 hours (total 30 hours). This is your new start point.

8. Go forward to "12". 6 hours (total 36 hours). THREE ORBITS (Fourth time over the "12").

Put another way: 3 orbits without back-ups: 24 hours. 3 back-ups,@ 4 hours each: 12 hours. Total: 36 hours.

[Guest]

I get the standard 36.

Visualize a clock. Starting point = "12".

1. Do one circuit, arriving back at "12". 8 hours. ONE ORBIT.

2. Back up to "9". 2 hours (total 10 hours). This is your new start point.

3. Do one more circuit, arriving back at "9". 8 hours (total 18 hours). (You do pass over "12" for the second time.)

4. Back up to "6". 2 hours (total 20 hours). This is your new start point.

5. Arrive at "12". 4 hours (total 24 hours). TWO ORBITS. (Third time over the "12".)

6. Assuming you don't like that one, continue on to "6". 4 hours (total 28 hours).

7. Back up to "3". 2 hours (total 30 hours). This is your new start point.

8. Go forward to "12". 6 hours (total 36 hours). THREE ORBITS (Fourth time over the "12").

Put another way: 3 orbits without back-ups: 24 hours. 3 back-ups,@ 4 hours each: 12 hours. Total: 36 hours.

I think that this a very sensible post, especially since you seem explicitly to acknowlege that reaching a point is not synonymous with "passing" it nor "orbiting" it.