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Thread: Rocket Thought Experiment

  1. #1 Rocket Thought Experiment 
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    OK, I have presented this "thought experiment" maybe a year ago, had a couple of responses that gave me some more contemplation (which I will include here) but I still am stumped:

    Imagine we have two identical rockets that have a thrust that allows them each to accelerate at 11 G's. We put them on the surface of the earth (or better yet, on a planet with the same mass as the earth but with no atmosphere), and we launch them. The first rocket's engines go for one second before shutting off, and the second rocket goes for two seconds before shutting off. As I understand it, they both would accelerate at approximately 320 feet/sec/sec. The paradox for me here is that while the second rocket would reach a height of four times the first rocket before falling back to the planet, it seems that it would only use twice the amount of fuel. I reason this by Newton's famous equation F=ma. If the mass of both rockets is the same, and the force from both engines is the same, the acceleration should be constant, especially if there is no friction from the atmosphere. There is nothing in the equation F=ma that qualifies the rockets' acceleration based on relative speed. In my previous post, someone clued me into the equation W=Fs, or Work equals Force times distance. Using this equation, I think the second rocket may only go up twice as high; however, I am not able to reconcile the two equations together.

    Any help or further insights into this quandary of mine would be greatly appreciated.


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  3. #2  
    Universal Mind John Galt's Avatar
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    You seem to have ignored, among other things, the changing mass of the rockets as they consume fuel. i.e. m is a variable in the equation, so to maintain the same acceleration the force applied must constantly reduce. Considering the rocket equation may be of some help to you.


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  4. #3  
    Genius Duck Moderator Dywyddyr's Avatar
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    The distance covered by an accelerating rocket is predicated on the "initial" conditions at each stage of acceleration: s=ut+1/2a.t2.
    Thus each increment of time that the rocket accelerates "starts" from a different base: accelerate twice as long and the distance covered isn't just twice as far simply because the second second of acceleration doesn't start from zero speed.
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    Thanks, GD, for replying. I am not able to follow what you said, though. Would you be able to bear with me as I try to understand this in simple layman's terms? I really do want to understand this.

    It seemed like you were saying that the acceleration of the rocket is relative to its current speed. Like starting at speed zero, it might have a different rate of acceleration than say, at 320 ft/s, after one second of rocket burn. Is this accurate? And if so, it seems to me that F=ma is inaccurate, because it seems clear to me that this equation does not consider speed to be a factor.

    Thanks for your indulgence.
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  6. #5  
    Genius Duck Moderator Dywyddyr's Avatar
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    Quote Originally Posted by Wundergeist View Post
    It seemed like you were saying that the acceleration of the rocket is relative to its current speed. Like starting at speed zero, it might have a different rate of acceleration than say, at 320 ft/s, after one second of rocket burn. Is this accurate?
    That's not quite correct.
    The rate of acceleration is the same, but the "initial" speed also adds to the distance covered.
    In the case of the 2 rockets the longer-accelerating one is already moving at 352 ft/ sec (v=u+at) by the time the second second starts. Additional acceleration will make it increase speed even more, thus covering more distance per unit time compared to the first rocket.
    Either way both accelerate at 352 ft/ sec/ sec.

    And if so, it seems to me that F=ma is inaccurate, because it seems clear to me that this equation does not consider speed to be a factor.
    F=ma is accurate because the resulting acceleration is regardless of the current speed (i.e. it's not a factor).
    Apply a (given) force to a (given) mass and you get a specified acceleration. If something is moving at 10 m/ sec then it will now be at (say) 20 m/ sec. The same force applied to something moving at 50 m/ sec will now move at 60 m/ sec. (Assuming it's applied in the direction of motion, of course).
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  7. #6  
    Universal Mind John Galt's Avatar
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    Duck, you are also ignoring the change in mass as fuel is consumed.
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    Genius Duck Moderator Dywyddyr's Avatar
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    The change in mass doesn't come into when using the basic equations of motion (which is where I'm deliberately keeping things for the time being).
    If that change were taken into account then the distance covered would be more than the noted 4 times in the OP (due to, as you pointed out, reduced mass of the rocket itself and therefore increasing acceleration).
    For this simple example we can take it as having a motor that throttles back to produce a constant acceleration (thus accounting for, and "cancelling out", the mass reduction).

    Or have I missed something more fundamental?
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    OK, that is my understanding, too. So if what you say is correct, then the second rocket will indeed cover more distance than the first one by a factor of four. This is disregarding a lot of small things, like John Galt mentioned: Yes, the mass of the second rocket will be less after the first second and accelerate faster. And then the gravity will slightly lessen as it goes further away from the planet, etc. These really don't have relevance to the problem.

    So, GD, if we are seeing the same page, then the first rocket will go for one second before the engine stops burning, moving at a speed of 320 f/s. It will take 10 seconds for gravity (32 ft/s/s) to decelerate the first rocket, reaching a height of 1760 feet before it starts to fall back to the surface of the planet. The second rocket will stop burning after two seconds, attaining a speed of 640 f/s. It will take 20 seconds for gravity to overcome the second rocket's acceleration, reaching a height of 7040 feet before it starts to come back down, which is four time higher than the first rocket, amounting to four times the potential energy than the first rocket, while only using twice the amount of fuel.

    What am I missing?!?
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  10. #9  
    Genius Duck Moderator Dywyddyr's Avatar
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    What you're missing is, like I said, the "start speed" at the second second.
    The additional 352 ft/ sec when the second second of acceleration starts means that the second rocket is already covering distance at a high rate.
    This goes back to the F=ma.
    The acceleration (in the second case) is being applied to something already moving.

    Time v s
    0.1 11 0.55
    0.2 22 2.2
    0.3 33 4.95
    0.4 44 8.8
    0.5 55 13.75
    0.6 66 19.8
    0.7 77 26.95
    0.8 88 35.2
    0.9 99 44.55
    1 110 55
    1.1 121 66.55
    1.2 132 79.2
    1.3 143 92.95
    1.4 154 107.8
    1.5 165 123.75
    1.6 176 140.8
    1.7 187 158.95
    1.8 198 178.2
    1.9 209 198.55
    2 220 220
    Units are metric (metres and m/ sec) with an 11 g (rounded to 110 m/ sec2) acceleration.
    Although the speed rises linearly (11 m/ sec per 0.1 sec) the distance covered MUST account for the speed the rocket already has PLUS any speed added during that acceleration.
    Last edited by Dywyddyr; October 8th, 2014 at 10:28 AM.
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    I think what you are missing is the kinetic energy of the exhaust. Starting from rest (in the planet's reference frame) the exhaust gas will have a high velocity and will get most of the kinetic energy from the rocket engine - much more than what is imparted to the rocket ship. Later on, the exhaust gas is going much slower relative to earth and gets less of the kinetic energy, while the ship receives more KE. Of course, all this will change if you calculate the KE from a different reference frame.

    Take this example.
    Case 1 - You have a gun at rest with a mass of 100 and shoots a projectile with a mass of 1 and a muzzle velocity of 100. After firing the projectile, the gun has velocity 1, and kinetic energy .5*100*1^2 = 50. The bullet has kinetic energy .5*1*100^2=5000. The total system kinetic energy is 5050.
    Case 2 - If the initial velocity is 1, the gun plus bullet weigh 101 and have a kinetic energy of .5*101*1= 50.5. After firing the gun has a speed of 2 and a KE of .5*100*2^2=200 so it has gained 150 units of kinetic energy. The bullet is going 99, so its KE is 0.5*1*99^2=4900.5 so the total KE of bullet +gun = 4900.5+200=5100.5. The delta-KE for the system is 5100.5-50.5=5050, same as Case 1. But the gun has gained more energy than it did after the first shot.
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    GDuck, it appears we are saying the same thing: At one second on your chart, 55 meters have been traveled, and at two seconds, 220 meters have been traveled, exactly four times the distance.
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    Harold,
    I find it interesting what you say about the gun example, and am honestly a little confused: I thought that for every action (force) there is an equal and opposite reaction. So wouldn't the gun in case 1 have the exact same amount of kinetic energy as the bullet?

    I think I might have figured out the flaw in my thinking with the rocket conundrum: I have assumed that something moving at twice the speed has four times the energy. I have also assumed that if the second rocket is four times higher than the first one, then it must have four times the potential energy. I am now beginning to think that that is incorrect. If something with the same mass as another object is moving twice as fast, I think it only takes twice the amount of energy to make it stop, as compared to the first object, not four times as much energy.

    Would anybody out there be able to confirm that?
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  14. #13  
    Genius Duck Moderator Dywyddyr's Avatar
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    Quote Originally Posted by Wundergeist View Post
    I have assumed that something moving at twice the speed has four times the energy.
    KE = 1/2mv2.
    So yes, double the speed four times the energy.
    Same table but KE added (nominal mass of 2).
    Time v s KE
    0.1 11 0.55 121
    0.2 22 2.2 484
    0.3 33 4.95 1089
    0.4 44 8.8 1936
    0.5 55 13.75 3025
    0.6 66 19.8 4356
    0.7 77 26.95 5929
    0.8 88 35.2 7744
    0.9 99 44.55 9801
    1 110 55 12100
    1.1 121 66.55 14641
    1.2 132 79.2 17424
    1.3 143 92.95 20449
    1.4 154 107.8 23716
    1.5 165 123.75 27225
    1.6 176 140.8 30976
    1.7 187 158.95 34969
    1.8 198 178.2 39204
    1.9 209 198.55 43681
    2 220 220 48400

    I think it only takes twice the amount of energy to make it stop, as compared to the first object, not four times as much energy.
    No, you were right first time - 4 times.

    I think that what you're missing is that the continued acceleration adds, at an increasing rate, to existing energy at each stage.
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    Quote Originally Posted by Wundergeist View Post
    Harold,
    I find it interesting what you say about the gun example, and am honestly a little confused: I thought that for every action (force) there is an equal and opposite reaction. So wouldn't the gun in case 1 have the exact same amount of kinetic energy as the bullet?
    It's equal and opposite force, not equal and opposite energy. The kinetic energy is the same in case 1 and 2 if you always measure it from the reference frame of the shooter. It's different measured in different reference frames. From the shooter's reference frame the gun and bullet are initially at rest so KE=0. This is not true from any other reference frame.
    I think I might have figured out the flaw in my thinking with the rocket conundrum: I have assumed that something moving at twice the speed has four times the energy. I have also assumed that if the second rocket is four times higher than the first one, then it must have four times the potential energy. I am now beginning to think that that is incorrect. If something with the same mass as another object is moving twice as fast, I think it only takes twice the amount of energy to make it stop, as compared to the first object, not four times as much energy.

    Would anybody out there be able to confirm that?
    No, you were right the first time. You can see this by the formula for kinetic energy, KE=one half the mass multiplied by the velocity squared.

    Here's another example that doesn't involve rockets. Let's say you are accelerating a vehicle at a constant rate with an internal combustion engine. The formula for power is torque multiplied by rotational speed. So if you double the vehicle's speed while maintaining torque constant, you double the power. Does this mean you get free extra power from your engine by speeding up? No. The engine has to turn twice as fast, or else you change to a higher gear, but then the torque on your engine doubles to get the same thrust at the wheels. Either way, you're burning more gasoline.
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    Harold,
    I have been thinking about your case 1 example of the bullet (thanks for the great analogy), and have a question about it: So are you saying that the bullet absorbed 100 times the amount of energy from the blast (the gunpowder explosion) than the gun did? Let us say that it was a rifle with a pretty good kickback. Would that mean that if someone were to (very stupidly) hold up a thick steel plate that someone else shot the bullet at, that the impact of the bullet on the steel plate would be 100 times more powerful than the kickback from the rifle?
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    Quote Originally Posted by Wundergeist View Post
    Harold,
    I have been thinking about your case 1 example of the bullet (thanks for the great analogy), and have a question about it: So are you saying that the bullet absorbed 100 times the amount of energy from the blast (the gunpowder explosion) than the gun did? Let us say that it was a rifle with a pretty good kickback. Would that mean that if someone were to (very stupidly) hold up a thick steel plate that someone else shot the bullet at, that the impact of the bullet on the steel plate would be 100 times more powerful than the kickback from the rifle?
    In this collision, momentum would be conserved. Some of the recoil momentum is due to burnt powder gases, but the bullet would have close to the same momentum as the recoil of the gun. But a lot of the bullet's energy will be dissipated by the bullet deforming on impact, and the bullet fragments going off with some remaining kinetic energy. So the guy holding the plate wouldn't feel much more than the guy shooting the gun. Maybe less if the plate is heavy enough. If the guy didn't have a steel plate in front of him, and the bullet expended its energy in his body, then there would be a very noticeable difference.

    One of the ways to measure recoil is called free recoil energy. You put the rifle in a ballistic pendulum, and see how high it swings after firing it. A typical free recoil energy for something like a high powered hunting rifle (308 win) would be 14 to 15 foot-pounds, whereas the bullet kinetic energy would be on the order of 2800 foot pounds.
    Free recoil - Wikipedia, the free encyclopedia
    308 Winchester Ballistics Chart | Ballistics 101
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    Quote Originally Posted by Wundergeist View Post
    OK, I have presented this "thought experiment" maybe a year ago, had a couple of responses that gave me some more contemplation (which I will include here) but I still am stumped:

    Imagine we have two identical rockets that have a thrust that allows them each to accelerate at 11 G's. We put them on the surface of the earth (or better yet, on a planet with the same mass as the earth but with no atmosphere), and we launch them. The first rocket's engines go for one second before shutting off, and the second rocket goes for two seconds before shutting off. As I understand it, they both would accelerate at approximately 320 feet/sec/sec. The paradox for me here is that while the second rocket would reach a height of four times the first rocket before falling back to the planet, it seems that it would only use twice the amount of fuel. I reason this by Newton's famous equation F=ma. If the mass of both rockets is the same, and the force from both engines is the same, the acceleration should be constant, especially if there is no friction from the atmosphere. There is nothing in the equation F=ma that qualifies the rockets' acceleration based on relative speed. In my previous post, someone clued me into the equation W=Fs, or Work equals Force times distance. Using this equation, I think the second rocket may only go up twice as high; however, I am not able to reconcile the two equations together.

    Any help or further insights into this quandary of mine would be greatly appreciated.
    I think you are trying to blow a gasket over this. Reduce the problem. Instead of a rocket use a cannon and assume you fire the cannon ball from different heights. Consider escape velocity to simplify the issues. If you look at the escape velocity it reduces as r gets larger ( r being the radius from central mass). Look at the following…


    https://en.wikipedia.org/wiki/Escape_velocity


    Since a cannon ball can not accelerate after leaving the cannon, its escape velocity is only dependent on the initial charge of the cannon and the height of the cannon.


    A higher cannon needs a lesser escape velocity (velocity to be gained) thus a smaller charge (lesser energy).


    As a rocket rises from the surface of a mass distance r increases thus decreasing the needed velocity to be gained (its needed energy consumption).


    So goes the dynamics of the polaris missile and its velocity to be gained in three axises (all the computer needs to evaluate cut off to target).
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    OK, here is the gist of what doesn't make sense to me:
    If it is true that the kinetic energy of an object is related to the square of its speed, or displacement, then when the two rockets are accelerating, the acceleration must be in relation to its speed if the thrust of both rockets is constant. So that would mean that the acceleration would decrease over the one second period for the first rocket and over the two seconds for the second rocket. Otherwise, the second rocket would gain four times the energy of the first rocket while only using twice the fuel, which of course, doesn't make sense. However, if the acceleration does decrease in relation to their speed, then F=ma is inaccurate.

    We can also put both rockets' starting point in deep space, so that the variable that Harold mentioned with the gas molecules escaping the rockets' engine are constant. So, does anyone see what my paradox is, or am I just somewhere out in the left field of deep space myself?
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    Quote Originally Posted by Wundergeist View Post
    OK, here is the gist of what doesn't make sense to me:
    If it is true that the kinetic energy of an object is related to the square of its speed, or displacement, then when the two rockets are accelerating, the acceleration must be in relation to its speed if the thrust of both rockets is constant. So that would mean that the acceleration would decrease over the one second period for the first rocket and over the two seconds for the second rocket. Otherwise, the second rocket would gain four times the energy of the first rocket while only using twice the fuel, which of course, doesn't make sense.
    Why doesn't it make sense? I already showed you how energy was conserved when you consider the exhaust gas kinetic energy.
    However, if the acceleration does decrease in relation to their speed, then F=ma is inaccurate.
    The acceleration does not decrease.
    We can also put both rockets' starting point in deep space, so that the variable that Harold mentioned with the gas molecules escaping the rockets' engine are constant. So, does anyone see what my paradox is, or am I just somewhere out in the left field of deep space myself?
    Even if you put the rocket in deep space, you are still measuring velocity in some inertial reference frame, and you still get the results that I calculated.

    Do you understand that the kinetic energy is relative to the reference frame in which you calculate it? What you seem to want to do is measure the acceleration at each interval based on the last position and velocity of the space ship. If you want to use conservation of energy, you have to pick a reference frame and stick with it.
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