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Mission to Mars!

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Suppose that we have on the moon a space station which is the only source of fuel and on it there are unlimited number of identical spaceships, We want to send one of them to Mars, each spaceship has a fuel capacity to allow it to fly exactly 1/4 way to Mars (without fuel the spaceship will miss its direction and may explode!).
Each spaceship has the ability to refuel by a second spaceship through a specific connection between them without loss of speed . What is the fewest number of spaceships necessary to accomplish this mission without losing any one of them?

Note:1- Each spaceship must have enough fuel to return safe to the base space station .

2- The time and fuel consumption of refueling can be ignored. (so we can also assume that one spaceship can refuel more than one spaceship at the same time).

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The thing about the space travel is that it doesn't require fuel to cover the distance. The fuel is required to break free of the gravity and to establish the direction and speed. Once you're far enough from any large celestial bodies you can cover arbitrarily large distances without any fuel consumption. So the fuel consumption is not proportionate to the distance, but is proportionate to the masses of the celestial bodies you're launching from or you're in gravitational field of.

Anyway, assuming that the fuel is consumed at a constant rate for the distance travelled I have a couple of questions:

1) Does the spaceship that reaches Mars also have to return to the station?

2) Can the spaceships that return to the base be used again, and if so, then do we count the number of spaceships used or do we count the number of spaceship launches?

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The thing about the space travel is that it doesn't require fuel to cover the distance. The fuel is required to break free of the gravity and to establish the direction and speed. Once you're far enough from any large celestial bodies you can cover arbitrarily large distances without any fuel consumption. So the fuel consumption is not proportionate to the distance, but is proportionate to the masses of the celestial bodies you're launching from or you're in gravitational field of.

Anyway, assuming that the fuel is consumed at a constant rate for the distance travelled I have a couple of questions:

1) Does the spaceship that reaches Mars also have to return to the station?

2) Can the spaceships that return to the base be used again, and if so, then do we count the number of spaceships used or do we count the number of spaceship launches?

k-man makes a good point. For the sake of the puzzle, let's assume that the fuel spent is proportional to distance travelled. Let's also assume that the ship that gets to Mars will need to return home.

It's not clear whether the OP wants the minimum unique number of ships required or the minimum unique number of ship launches.

Let's assume that we want the minimum unique number of ships (a ship can launch more than once). I'm getting an answer of 9 ships required if we allow a ship to turn off its engine and wait for fuel; see below for an outline of the solution. This will take a *long* time, though.

Let's assume that we want the minimum number of ship launches (also the one that takes the least time to get to Mars). Here are where things get interesting.

Let the distance to Mars be 12 unit. On a full tank a ship can travel 3 units. We define a function s(d) as the number of unique launches required to put 1/3 gas tank at distance d (and have all the ships return home to base). It is easy to see that

s(1) = 1

s(2) = 3

s(3) = 9

And in general,

s(N) = 2 * ( s( 1) + s( 2 ) + … s( N - 1 ) ) + 1 = 3(N - 1 )

The above is derived by assuming for distance N, there is a special ship designated to reach distance N. At every integer distance less than N on the trip out and on the trip back, it gets (1/3) of a tank from a supporting network on other ships.

So, if we want to put (1/3) gas tank on Mars, then we would need 311 launches. Since we don't want to put any gas on Mars, we would only need 2*310.

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The thing about the space travel is that it doesn't require fuel to cover the distance. The fuel is required to break free of the gravity and to establish the direction and speed. Once you're far enough from any large celestial bodies you can cover arbitrarily large distances without any fuel consumption. So the fuel consumption is not proportionate to the distance, but is proportionate to the masses of the celestial bodies you're launching from or you're in gravitational field of.

Anyway, assuming that the fuel is consumed at a constant rate for the distance travelled I have a couple of questions:

1) Does the spaceship that reaches Mars also have to return to the station?

2) Can the spaceships that return to the base be used again, and if so, then do we count the number of spaceships used or do we count the number of spaceship launches?

Thanks...I know that the space travel doesn`t need fuel,but I wanted to be a kind of energy used by the machine to keep it working.

for your 1st question: The spaceship or ships reaching mars should have an amount of energy keeping them doing their mission.

2- Yes, the spaceship that returns can be sent again and again, so I need number of ships used.

Edited by wolfgang
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The thing about the space travel is that it doesn't require fuel to cover the distance. The fuel is required to break free of the gravity and to establish the direction and speed. Once you're far enough from any large celestial bodies you can cover arbitrarily large distances without any fuel consumption. So the fuel consumption is not proportionate to the distance, but is proportionate to the masses of the celestial bodies you're launching from or you're in gravitational field of.

Anyway, assuming that the fuel is consumed at a constant rate for the distance travelled I have a couple of questions:

1) Does the spaceship that reaches Mars also have to return to the station?

2) Can the spaceships that return to the base be used again, and if so, then do we count the number of spaceships used or do we count the number of spaceship launches?

k-man makes a good point. For the sake of the puzzle, let's assume that the fuel spent is proportional to distance travelled. Let's also assume that the ship that gets to Mars will need to return home.

It's not clear whether the OP wants the minimum unique number of ships required or the minimum unique number of ship launches.

Let's assume that we want the minimum unique number of ships (a ship can launch more than once). I'm getting an answer of 9 ships required if we allow a ship to turn off its engine and wait for fuel; see below for an outline of the solution. This will take a *long* time, though.

Let's assume that we want the minimum number of ship launches (also the one that takes the least time to get to Mars). Here are where things get interesting.

Let the distance to Mars be 12 unit. On a full tank a ship can travel 3 units. We define a function s(d) as the number of unique launches required to put 1/3 gas tank at distance d (and have all the ships return home to base). It is easy to see that

s(1) = 1

s(2) = 3

s(3) = 9

And in general,

s(N) = 2 * ( s( 1) + s( 2 ) + … s( N - 1 ) ) + 1 = 3(N - 1 )

The above is derived by assuming for distance N, there is a special ship designated to reach distance N. At every integer distance less than N on the trip out and on the trip back, it gets (1/3) of a tank from a supporting network on other ships.

So, if we want to put (1/3) gas tank on Mars, then we would need 311 launches. Since we don't want to put any gas on Mars, we would only need 2*310.

I want the fewest number of ships required , and at least one of them should launch with enough energy keeping it working.

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I want the fewest number of ships required , and at least one of them should launch with enough energy keeping it working.

If you want the number of unique ships required, then

You would need 9 different ships. Let's label the ships 1-9. Let the distance to Mars be 1 unit. Let the base space station be located at (x=0) and Mars be at (x=1). The idea is as follows

1) Let ships 2-9 go to the location (x = 1/9) and stay there. Ship 1 would then shuttle back and forth between the starting point (x = 0) and (x = 1/9) to deliver gas to the ships 2-9 until all 8 ships are full of gas. Each trip ship 1 can deliver 1/36 of a tank.

2) Let ships 3-9 go to the point (x=2/9). Ship 2 now shuttles back and forth between (x=1/9) and (x=2/9) to deliver gas to ships 3-9 until all are full of gas. Each trip ship 2 will have to wait at (x=1/9) for ship 1 to deliver gas to it, and then it can deliver (1/36) of a tank at (x=2/9).

3) Ship 4-9 will then proceed to point 3/9, and then ship 3 shuttles back and forth as above. Eventually ship 9 will be able to reach Mars and go back.

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Well...9 ships are too much...try to reduce the number


I want the fewest number of ships required , and at least one of them should launch with enough energy keeping it working.

If you want the number of unique ships required, then


You would need 9 different ships. Let's label the ships 1-9. Let the distance to Mars be 1 unit. Let the base space station be located at (x=0) and Mars be at (x=1). The idea is as follows

1) Let ships 2-9 go to the location (x = 1/9) and stay there. Ship 1 would then shuttle back and forth between the starting point (x = 0) and (x = 1/9) to deliver gas to the ships 2-9 until all 8 ships are full of gas. Each trip ship 1 can deliver 1/36 of a tank.

2) Let ships 3-9 go to the point (x=2/9). Ship 2 now shuttles back and forth between (x=1/9) and (x=2/9) to deliver gas to ships 3-9 until all are full of gas. Each trip ship 2 will have to wait at (x=1/9) for ship 1 to deliver gas to it, and then it can deliver (1/36) of a tank at (x=2/9).

3) Ship 4-9 will then proceed to point 3/9, and then ship 3 shuttles back and forth as above. Eventually ship 9 will be able to reach Mars and go back.

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Can one ship transfer fuel to multiple ships simultaneously? Can one ship receive fuel from multiple ships simultaneously?

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Well...9 ships are too much...try to reduce the number

I don't think the number of ships can be reduced below 9. Here is why

Let's say that the moon is at location (x = 0), and Mars is at (x=1). Each ship can travel a distance of (1/4) on a full tank.

The moon base has unlimited fuel, so if we only have 1 space-ship, the max range we can reach (and return safely) is (1/4)*(1/2) = 1/8.

We can imagine that each spaceship we use can extend the fuel capacity of the base by a distance up to (but not including) 1/8. For instance, at a lesser distance of (1/9), a single spaceship can take as many trips as required to shuttle back and forth to deposit fuel at location (x=1/9). Each trip the ship can deliver 1/36 of a tank, but we can take as many trip as necessary. Note that the single ship *cannot* extend the fuel-supply to the precise distance of (1/8), as it would be not able to deposit any fuel at (x=1/8) and return home safely.

So, if a single space-ship can extend the fuel-supply line by 0 < d < 1/8, then we would need minimum of 8 ships to extend the fuel-supply line past (x=7/8). We then need 1 more ship to go to Mars and return. That makes 8+1 = 9 ships.

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Dear Bushindo....you should notice that I have concidered the moon as our base station(!) and not the earth.So you should think a little bit (laterally),and try to reduce the number to 5 ships only!...note : all of them will reach mars

Well...9 ships are too much...try to reduce the number

I don't think the number of ships can be reduced below 9. Here is why


Let's say that the moon is at location (x = 0), and Mars is at (x=1). Each ship can travel a distance of (1/4) on a full tank.

The moon base has unlimited fuel, so if we only have 1 space-ship, the max range we can reach (and return safely) is (1/4)*(1/2) = 1/8.

We can imagine that each spaceship we use can extend the fuel capacity of the base by a distance up to (but not including) 1/8. For instance, at a lesser distance of (1/9), a single spaceship can take as many trips as required to shuttle back and forth to deposit fuel at location (x=1/9). Each trip the ship can deliver 1/36 of a tank, but we can take as many trip as necessary. Note that the single ship *cannot* extend the fuel-supply to the precise distance of (1/8), as it would be not able to deposit any fuel at (x=1/8) and return home safely.

So, if a single space-ship can extend the fuel-supply line by 0 < d < 1/8, then we would need minimum of 8 ships to extend the fuel-supply line past (x=7/8). We then need 1 more ship to go to Mars and return. That makes 8+1 = 9 ships.


Edited by wolfgang
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If the fuel is used for guidance and anti-explosibility measures only, and not for propulsion; then all five spaceships could be flung from the moon towards Mars starting at the same time. All the ships would be tethered in such a way that adjusting the course of one would alter the course of the fleet as a whole. For the first quarter of the trip, Spaceship #1 does all of the course correction. For the second quarter, Spaceship #2; and so on until landing on Mars. Once on Mars, Spaceship # 5 redistributes it's fuel equally between all of the spaceships to power their lasers and defeat the Martians.

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Dear Bushindo....you should notice that I have concidered the moon as our base station(!) and not the earth.So you should think a little bit (laterally),and try to reduce the number to 5 ships only!...note : all of them will reach mars

So I guess the lateral solution of this puzzle is contingent upon the properties of this 'fuel'. My previous solution assumes that "the fuel spent is proportional to distance travelled". Since that doesn't seem to be the case, I'd like some clarification about the conditions of the puzzle.

1) My impression is that the fuel is used to make constant minor course corrections during the trip (per OP: "without fuel the spaceship will miss its direction") and to maintain the life-support and ship computer (per OP: "… and may explode"). Is this correct? If that is not correct, what is the fuel being used for?

2) Consider the following two scenarios

* A single ship flies from base (x=0) to the one-fourth point (x=1/4)

* A ship flies from the halfway point (x=1/2) to the three-fourth point (x=3/4)

Does the ship use the same amount of fuel (1 tank) in both scenarios?

3) Are the ships supposed to land on the surface of Mars?

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Yes...thats right...Thanks

If the fuel is used for guidance and anti-explosibility measures only, and not for propulsion; then all five spaceships could be flung from the moon towards Mars starting at the same time. All the ships would be tethered in such a way that adjusting the course of one would alter the course of the fleet as a whole. For the first quarter of the trip, Spaceship #1 does all of the course correction. For the second quarter, Spaceship #2; and so on until landing on Mars. Once on Mars, Spaceship # 5 redistributes it's fuel equally between all of the spaceships to power their lasers and defeat the Martians.

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Yes...thats right...Thanks

If the fuel is used for guidance and anti-explosibility measures only, and not for propulsion; then all five spaceships could be flung from the moon towards Mars starting at the same time. All the ships would be tethered in such a way that adjusting the course of one would alter the course of the fleet as a whole. For the first quarter of the trip, Spaceship #1 does all of the course correction. For the second quarter, Spaceship #2; and so on until landing on Mars. Once on Mars, Spaceship # 5 redistributes it's fuel equally between all of the spaceships to power their lasers and defeat the Martians.

Dear wolfgang, the solution that you consider correct suffers from two problems: 1) It violates common understanding of thermodynamics and 2) even if we allow such violation, it does not satisfy the conditions in your original post. Here is why

From your original post, each ship needs to make constant trajectory corrections since "without fuel the spaceship will miss its direction". The solution that you accept states that 4 space-ships can "piggy-back" on the course correction ability of 1 ships. That is, the ships would be tethered in such a way that adjusting the course of 1 ship would alter the course of the whole.

Of course, a simple understanding of physics would indicate that we don't really save any fuel if we allow 1 ship to change the course for all 5. If a single ships needs to make a minor course correction, it would need to provide enough thrust (energy) to move its mass however much required. If a single ship is tethered to 4 other ships, Newton's equation (F=ma) implies that the navigating ship would need 5 times the energy to alter the current course. So, in reality, tethering 5 ships together would make the single navigating ship burn fuel 5 times as fast as it changes course.

That said, even if we allow such violation, how do you suppose to get the ships home. From your OP, "each spaceship has a fuel capacity to allow it to fly exactly 1/4 way to Mars" and "Each spaceship must have enough fuel to return safe to the base space station". Even allowing for the thermodynamics violation, all your 5 ships would end up on Mars with 1 tank left, which is only good enough for 1/4 of the way back to the moon base.

While I understand that lateral and out-of-box thinking are valued around here, in general such solutions have to be internally consistent (and avoid violating explicit conditions set forth in the original post). Also, in this forum, we obey the law of thermodynamics.

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Do we need to account for the fact that both the moon and Mars are in motion? I.e If it is a new moon (moon between earth and sun) the trip to mars is further than if it is a full moon (moon between Earth and Mars)

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I take it in such a way that 4 ships are connected to a 5th ship from 4 sides to give it the energy all the trip ,i.e. only the central ship will be functioning and the others adhering to it are as energy suppliers only.

Yes...thats right...Thanks

If the fuel is used for guidance and anti-explosibility measures only, and not for propulsion; then all five spaceships could be flung from the moon towards Mars starting at the same time. All the ships would be tethered in such a way that adjusting the course of one would alter the course of the fleet as a whole. For the first quarter of the trip, Spaceship #1 does all of the course correction. For the second quarter, Spaceship #2; and so on until landing on Mars. Once on Mars, Spaceship # 5 redistributes it's fuel equally between all of the spaceships to power their lasers and defeat the Martians.

Dear wolfgang, the solution that you consider correct suffers from two problems: 1) It violates common understanding of thermodynamics and 2) even if we allow such violation, it does not satisfy the conditions in your original post. Here is why


From your original post, each ship needs to make constant trajectory corrections since "without fuel the spaceship will miss its direction". The solution that you accept states that 4 space-ships can "piggy-back" on the course correction ability of 1 ships. That is, the ships would be tethered in such a way that adjusting the course of 1 ship would alter the course of the whole.

Of course, a simple understanding of physics would indicate that we don't really save any fuel if we allow 1 ship to change the course for all 5. If a single ships needs to make a minor course correction, it would need to provide enough thrust (energy) to move its mass however much required. If a single ship is tethered to 4 other ships, Newton's equation (F=ma) implies that the navigating ship would need 5 times the energy to alter the current course. So, in reality, tethering 5 ships together would make the single navigating ship burn fuel 5 times as fast as it changes course.

That said, even if we allow such violation, how do you suppose to get the ships home. From your OP, "each spaceship has a fuel capacity to allow it to fly exactly 1/4 way to Mars" and "Each spaceship must have enough fuel to return safe to the base space station". Even allowing for the thermodynamics violation, all your 5 ships would end up on Mars with 1 tank left, which is only good enough for 1/4 of the way back to the moon base.



While I understand that lateral and out-of-box thinking are valued around here, in general such solutions have to be internally consistent (and avoid violating explicit conditions set forth in the original post). Also, in this forum, we obey the law of thermodynamics.

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I take it in such a way that 4 ships are connected to a 5th ship from 4 sides to give it the energy all the trip ,i.e. only the central ship will be functioning and the others adhering to it are as energy suppliers only.

Yes...thats right...Thanks

If the fuel is used for guidance and anti-explosibility measures only, and not for propulsion; then all five spaceships could be flung from the moon towards Mars starting at the same time. All the ships would be tethered in such a way that adjusting the course of one would alter the course of the fleet as a whole. For the first quarter of the trip, Spaceship #1 does all of the course correction. For the second quarter, Spaceship #2; and so on until landing on Mars. Once on Mars, Spaceship # 5 redistributes it's fuel equally between all of the spaceships to power their lasers and defeat the Martians.

Dear wolfgang, the solution that you consider correct suffers from two problems: 1) It violates common understanding of thermodynamics and 2) even if we allow such violation, it does not satisfy the conditions in your original post. Here is why

From your original post, each ship needs to make constant trajectory corrections since "without fuel the spaceship will miss its direction". The solution that you accept states that 4 space-ships can "piggy-back" on the course correction ability of 1 ships. That is, the ships would be tethered in such a way that adjusting the course of 1 ship would alter the course of the whole.

Of course, a simple understanding of physics would indicate that we don't really save any fuel if we allow 1 ship to change the course for all 5. If a single ships needs to make a minor course correction, it would need to provide enough thrust (energy) to move its mass however much required. If a single ship is tethered to 4 other ships, Newton's equation (F=ma) implies that the navigating ship would need 5 times the energy to alter the current course. So, in reality, tethering 5 ships together would make the single navigating ship burn fuel 5 times as fast as it changes course.

That said, even if we allow such violation, how do you suppose to get the ships home. From your OP, "each spaceship has a fuel capacity to allow it to fly exactly 1/4 way to Mars" and "Each spaceship must have enough fuel to return safe to the base space station". Even allowing for the thermodynamics violation, all your 5 ships would end up on Mars with 1 tank left, which is only good enough for 1/4 of the way back to the moon base.

While I understand that lateral and out-of-box thinking are valued around here, in general such solutions have to be internally consistent (and avoid violating explicit conditions set forth in the original post). Also, in this forum, we obey the law of thermodynamics.

I don't see how that helps. Since the controlling ship still needs 5x the amount of fuel to go the same distance, it would only get 1/8 of the way before having to turn back.

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I take it in such a way that 4 ships are connected to a 5th ship from 4 sides to give it the energy all the trip ,i.e. only the central ship will be functioning and the others adhering to it are as energy suppliers only.

Your proposal (tethering 4 ships to the 5th) is akin to increasing the mass of the 5th ship by five times while also increasing the fuel tank five times. Your solution assumes that fuel usage for the navigating ship is the same whether it is alone or whether it has 5 times the mass. In physics, as in life, there is no such thing as a free lunch. You can't carry 5 times the mass to Mars without expending extra energy.
Perhaps you are thinking that the navigating ship only has to set a course for the entire fleet, so the energy usage is independent of the mass. Recall that in space, if you need to change the course, you will need to change the momentum vector of the entire fleet. That means if you want to move 1 degree to the left, for example, the navigating ship will need to provide enough thrust (energy) to move its entire mass to the left that much.
Even if we allow the conservation-of-energy violation, the 5 ships still can not make it to Mars and back. From your OP, "each spaceship has a fuel capacity to allow it to fly exactly 1/4 way to Mars" and "Each spaceship must have enough fuel to return safe to the base space station". Since each tank will take you 1/4 of the way to Mars, 5 tanks are enough for 1.25 of the distance to Mars. How are your 5 ships supposed to get to Mars and get home on 5 tanks?
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I take it in such a way that 4 ships are connected to a 5th ship from 4 sides to give it the energy all the trip ,i.e. only the central ship will be functioning and the others adhering to it are as energy suppliers only.

Your proposal (tethering 4 ships to the 5th) is akin to increasing the mass of the 5th ship by five times while also increasing the fuel tank five times. Your solution assumes that fuel usage for the navigating ship is the same whether it is alone or whether it has 5 times the mass. In physics, as in life, there is no such thing as a free lunch. You can't carry 5 times the mass to Mars without expending extra energy.
Perhaps you are thinking that the navigating ship only has to set a course for the entire fleet, so the energy usage is independent of the mass. Recall that in space, if you need to change the course, you will need to change the momentum vector of the entire fleet. That means if you want to move 1 degree to the left, for example, the navigating ship will need to provide enough thrust (energy) to move its entire mass to the left that much.
Even if we allow the conservation-of-energy violation, the 5 ships still can not make it to Mars and back. From your OP, "each spaceship has a fuel capacity to allow it to fly exactly 1/4 way to Mars" and "Each spaceship must have enough fuel to return safe to the base space station". Since each tank will take you 1/4 of the way to Mars, 5 tanks are enough for 1.25 of the distance to Mars. How are your 5 ships supposed to get to Mars and get home on 5 tanks?

The fuel feeds a small generator on board producing enough electricity to send a signal to Mars' Sub-Ether Tractor Beam station. The signal relays precise position, velocity, and mass of the ships. The Tractor Beam on Mars supplies all required energy and does all the work in pulling and guiding the ships towards their destination.

Once arrived, the ships must bomb the Tractor Beam station, because such is their mission.

You might ask, why Marsians are playing along. Because they know the ship has enough fuel only for a fraction of the distance to send the signal. It's simple.

Edited by Prime
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