Why is the work done by a rocket engine greater at higher speeds?

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From this comment by orlp:




If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?











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  • Net work done on an object such as a rocket always equals its change in kinetic energy.
    – David White
    22 hours ago










  • Hint: You're ignoring part of the system. There isn't just a rocket flying in space when you press the button to burn.
    – The_Sympathizer
    15 hours ago










  • Notice that "but in the former case the rocket travels a much greater distance during this period" is also true if you don't fire the rocket during this time. (That is, the rocket travels a much greater distance when it is flying at any nonzero speed than when it is stationary.) Does the phrase I'm quoting precisely capture what you intend?
    – Eric Towers
    11 hours ago







  • 1




    Neither the question nor any of the answers so far points out that this fact for rockets is called the Oberth Effect. Doing a search on that term will lead to many tutorials explaining how this works. Players of Kerbal Space Program or other space simulators know very well that you always burn your rockets at the lowest, fastest point of your orbit if you want to save fuel!
    – Eric Lippert
    3 hours ago











  • Possible duplicate of Where does the extra kinetic energy of the rocket come from?
    – knzhou
    24 mins ago














up vote
15
down vote

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From this comment by orlp:




If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?











share|cite|improve this question









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Pilchard123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Net work done on an object such as a rocket always equals its change in kinetic energy.
    – David White
    22 hours ago










  • Hint: You're ignoring part of the system. There isn't just a rocket flying in space when you press the button to burn.
    – The_Sympathizer
    15 hours ago










  • Notice that "but in the former case the rocket travels a much greater distance during this period" is also true if you don't fire the rocket during this time. (That is, the rocket travels a much greater distance when it is flying at any nonzero speed than when it is stationary.) Does the phrase I'm quoting precisely capture what you intend?
    – Eric Towers
    11 hours ago







  • 1




    Neither the question nor any of the answers so far points out that this fact for rockets is called the Oberth Effect. Doing a search on that term will lead to many tutorials explaining how this works. Players of Kerbal Space Program or other space simulators know very well that you always burn your rockets at the lowest, fastest point of your orbit if you want to save fuel!
    – Eric Lippert
    3 hours ago











  • Possible duplicate of Where does the extra kinetic energy of the rocket come from?
    – knzhou
    24 mins ago












up vote
15
down vote

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2









up vote
15
down vote

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2






2





From this comment by orlp:




If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?











share|cite|improve this question









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From this comment by orlp:




If I strap a rocket booster to a rocket in space and fire it for one second, then the work provided is much higher when the rocket is flying fast compared to when the rocket was stationary. In both cases the rocket fires for the same duration but in the former case the rocket travels a much greater distance during this period. What gives?








newtonian-mechanics forces work rocket-science distance






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edited 24 mins ago









knzhou

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  • Net work done on an object such as a rocket always equals its change in kinetic energy.
    – David White
    22 hours ago










  • Hint: You're ignoring part of the system. There isn't just a rocket flying in space when you press the button to burn.
    – The_Sympathizer
    15 hours ago










  • Notice that "but in the former case the rocket travels a much greater distance during this period" is also true if you don't fire the rocket during this time. (That is, the rocket travels a much greater distance when it is flying at any nonzero speed than when it is stationary.) Does the phrase I'm quoting precisely capture what you intend?
    – Eric Towers
    11 hours ago







  • 1




    Neither the question nor any of the answers so far points out that this fact for rockets is called the Oberth Effect. Doing a search on that term will lead to many tutorials explaining how this works. Players of Kerbal Space Program or other space simulators know very well that you always burn your rockets at the lowest, fastest point of your orbit if you want to save fuel!
    – Eric Lippert
    3 hours ago











  • Possible duplicate of Where does the extra kinetic energy of the rocket come from?
    – knzhou
    24 mins ago
















  • Net work done on an object such as a rocket always equals its change in kinetic energy.
    – David White
    22 hours ago










  • Hint: You're ignoring part of the system. There isn't just a rocket flying in space when you press the button to burn.
    – The_Sympathizer
    15 hours ago










  • Notice that "but in the former case the rocket travels a much greater distance during this period" is also true if you don't fire the rocket during this time. (That is, the rocket travels a much greater distance when it is flying at any nonzero speed than when it is stationary.) Does the phrase I'm quoting precisely capture what you intend?
    – Eric Towers
    11 hours ago







  • 1




    Neither the question nor any of the answers so far points out that this fact for rockets is called the Oberth Effect. Doing a search on that term will lead to many tutorials explaining how this works. Players of Kerbal Space Program or other space simulators know very well that you always burn your rockets at the lowest, fastest point of your orbit if you want to save fuel!
    – Eric Lippert
    3 hours ago











  • Possible duplicate of Where does the extra kinetic energy of the rocket come from?
    – knzhou
    24 mins ago















Net work done on an object such as a rocket always equals its change in kinetic energy.
– David White
22 hours ago




Net work done on an object such as a rocket always equals its change in kinetic energy.
– David White
22 hours ago












Hint: You're ignoring part of the system. There isn't just a rocket flying in space when you press the button to burn.
– The_Sympathizer
15 hours ago




Hint: You're ignoring part of the system. There isn't just a rocket flying in space when you press the button to burn.
– The_Sympathizer
15 hours ago












Notice that "but in the former case the rocket travels a much greater distance during this period" is also true if you don't fire the rocket during this time. (That is, the rocket travels a much greater distance when it is flying at any nonzero speed than when it is stationary.) Does the phrase I'm quoting precisely capture what you intend?
– Eric Towers
11 hours ago





Notice that "but in the former case the rocket travels a much greater distance during this period" is also true if you don't fire the rocket during this time. (That is, the rocket travels a much greater distance when it is flying at any nonzero speed than when it is stationary.) Does the phrase I'm quoting precisely capture what you intend?
– Eric Towers
11 hours ago





1




1




Neither the question nor any of the answers so far points out that this fact for rockets is called the Oberth Effect. Doing a search on that term will lead to many tutorials explaining how this works. Players of Kerbal Space Program or other space simulators know very well that you always burn your rockets at the lowest, fastest point of your orbit if you want to save fuel!
– Eric Lippert
3 hours ago





Neither the question nor any of the answers so far points out that this fact for rockets is called the Oberth Effect. Doing a search on that term will lead to many tutorials explaining how this works. Players of Kerbal Space Program or other space simulators know very well that you always burn your rockets at the lowest, fastest point of your orbit if you want to save fuel!
– Eric Lippert
3 hours ago













Possible duplicate of Where does the extra kinetic energy of the rocket come from?
– knzhou
24 mins ago




Possible duplicate of Where does the extra kinetic energy of the rocket come from?
– knzhou
24 mins ago










3 Answers
3






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up vote
20
down vote













The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?



To understand this it is useful to consider a “toy model” rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).



Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning” the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.



So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?



Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.



At low speeds most of the fuel’s energy is “wasted” in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.






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  • 2




    At really high speeds the exhaust will in fact be losing energy as it changes from a positive velocity to a slightly less positive velocity. The energy it loses is of course transferred to the rocket.
    – Arcanist Lupus
    18 hours ago










  • Yes, exactly. Since $Delta KE_rocket+Delta KE_exhaust+Delta PE_fuel=0$, when $Delta KE_exhaust<0$ the amount of KE gained by the rocket is greater than the amount of chemical PE lost by the fuel!
    – Dale
    6 hours ago

















up vote
5
down vote













There are multiple ways to view this.



The easiest, I think, is that kinetic energy scales with the square of velocity
$$K=frac 12 m v^2$$



If we assume that the rocket booster supplies a constant acceleration, then comparing initial and final velocities we find that
$$Delta v=v_final-v_init=at$$



So for the same amount of time, the change in the velocity is the same, regardless of what the starting velocity actually is. Since we have a squared dependence on the velocity in $K$, this means that the kinetic energy increases more if we started out with a larger velocitiy. i.e.
$$Delta K=frac 12 m v_final^2-frac 12 m v_init^2=frac 12 m (v_final-v_init)(v_final+v_init)=frac 12 m Delta v (Delta v+2v_init)$$



So as we can see, the sum of the velocities in the expression for $Delta K$ is what contributes to a larger change in kinetic energy. Since the work done is equal to the change in kinetic energy, it must be that the rocket does more work when we start off at a larger velocity.



The second way to view this, which you could argue is the same as the first, is to look at the definition of work
$$W=intvec F cdot dvec x$$



Or in one dimension with a constant force
$$W=FDelta x$$



Now, once again assuming a constant acceleration, we know that
$$Delta x = frac 12 a t^2 + v_initt$$



So that the work done is
$$W=Fleft(frac 12 a t^2 + v_inittright)$$



Once again, we see that the initial velocity determines the work. A qualitative explanation from this is that when the velocity is larger, then the object covers more distance in the same amount of time. So if we look at the time the force is being applied, the faster it is moving the larger distance the force is applied over. Therefore, we get more work done if the object is initially moving faster.



The supposed issue behind all of this is that is seems like we are getting more energy from applying the same force for the same amount of time. But if you work through it you find that this is no issue at all. This is even true for objects falling near the Earth's surface. Even though the force is constant, gravity does more and more work on the object while it falls. Or in other words, the rate of energy conversion from potential to kinetic energy increases as the object falls.






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    In both cases (stationary and flying fast), the change in total kinetic energy of the rocket+propellant system is the same, equal to the chemical energy released during a one-second burn. If the rocket was stationary, the propellant goes from rest to moving backward, increasing its kinetic energy. If the rocket was flying (very) fast, the propellant goes from moving forward fast (with the rocket) to moving forward slower (trailing behind), decreasing its kinetic energy. This is enough to understand qualitatively why the kinetic energy of the rocket increases more in the second case.



    The two cases are related by a Galilean transformation (choice of uniformly moving reference frame). Consistency is assured by the fact that, for any isolated system (like the rocket+propellant), the change in total kinetic energy between one time and another is Galilean invariant (the same in any uniformly moving reference frame).






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      3 Answers
      3






      active

      oldest

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      3 Answers
      3






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes








      up vote
      20
      down vote













      The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?



      To understand this it is useful to consider a “toy model” rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).



      Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning” the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.



      So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?



      Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.



      At low speeds most of the fuel’s energy is “wasted” in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.






      share|cite|improve this answer
















      • 2




        At really high speeds the exhaust will in fact be losing energy as it changes from a positive velocity to a slightly less positive velocity. The energy it loses is of course transferred to the rocket.
        – Arcanist Lupus
        18 hours ago










      • Yes, exactly. Since $Delta KE_rocket+Delta KE_exhaust+Delta PE_fuel=0$, when $Delta KE_exhaust<0$ the amount of KE gained by the rocket is greater than the amount of chemical PE lost by the fuel!
        – Dale
        6 hours ago














      up vote
      20
      down vote













      The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?



      To understand this it is useful to consider a “toy model” rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).



      Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning” the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.



      So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?



      Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.



      At low speeds most of the fuel’s energy is “wasted” in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.






      share|cite|improve this answer
















      • 2




        At really high speeds the exhaust will in fact be losing energy as it changes from a positive velocity to a slightly less positive velocity. The energy it loses is of course transferred to the rocket.
        – Arcanist Lupus
        18 hours ago










      • Yes, exactly. Since $Delta KE_rocket+Delta KE_exhaust+Delta PE_fuel=0$, when $Delta KE_exhaust<0$ the amount of KE gained by the rocket is greater than the amount of chemical PE lost by the fuel!
        – Dale
        6 hours ago












      up vote
      20
      down vote










      up vote
      20
      down vote









      The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?



      To understand this it is useful to consider a “toy model” rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).



      Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning” the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.



      So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?



      Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.



      At low speeds most of the fuel’s energy is “wasted” in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.






      share|cite|improve this answer












      The key point of this question is that it intuitively seems like conservation of energy is not working right. A rocket is powered by a chemical reaction that releases chemical energy at a constant rate. So how can a constant rate of energy release lead to a greater increase in KE when going fast?



      To understand this it is useful to consider a “toy model” rocket that operates on the same principles, but is easier to analyze. Specifically, let’s consider a 10 kg ball (rocket) and a 1 kg ball (exhaust) which is attached to a massless spring (fuel).



      Suppose this spring has enough energy stored that when the rocket is initially at rest it can propel it to 1 m/s, and by conservation of momentum the exhaust is propelled to -10 m/s. Conversely, if the rocket starts at 5 m/s then after “burning” the fuel the rocket is propelled to 6 m/s and the exhaust moves at -5 m/s.



      So now let’s check energy. In the first case the KE of the rocket increased from 0 J to 5 J, while in the second case it increased from 125 J to 180 J. The spring stores the same amount of energy in both cases, so why does the KE increase by 5 J at the low speed and by 55 J at the high speed?



      Notice that we forgot to calculate the energy that went into the exhaust. This is the pivotal mistake of most such analyses. In the first case the KE of the exhaust increased from 0 J to 50 J, while in the second case the KE was 12.5 J before and after. So in both cases the total change in KE (both the rocket and the exhaust) was 55 J.



      At low speeds most of the fuel’s energy is “wasted” in the KE of the exhaust. At higher speeds more goes into the rocket and less into the exhaust. For a real rocket, the same thing happens on a continuous basis. Both energy and momentum are conserved, and in fact more power is delivered to the vehicle as the speed increases under constant thrust.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 21 hours ago









      Dale

      1,232213




      1,232213







      • 2




        At really high speeds the exhaust will in fact be losing energy as it changes from a positive velocity to a slightly less positive velocity. The energy it loses is of course transferred to the rocket.
        – Arcanist Lupus
        18 hours ago










      • Yes, exactly. Since $Delta KE_rocket+Delta KE_exhaust+Delta PE_fuel=0$, when $Delta KE_exhaust<0$ the amount of KE gained by the rocket is greater than the amount of chemical PE lost by the fuel!
        – Dale
        6 hours ago












      • 2




        At really high speeds the exhaust will in fact be losing energy as it changes from a positive velocity to a slightly less positive velocity. The energy it loses is of course transferred to the rocket.
        – Arcanist Lupus
        18 hours ago










      • Yes, exactly. Since $Delta KE_rocket+Delta KE_exhaust+Delta PE_fuel=0$, when $Delta KE_exhaust<0$ the amount of KE gained by the rocket is greater than the amount of chemical PE lost by the fuel!
        – Dale
        6 hours ago







      2




      2




      At really high speeds the exhaust will in fact be losing energy as it changes from a positive velocity to a slightly less positive velocity. The energy it loses is of course transferred to the rocket.
      – Arcanist Lupus
      18 hours ago




      At really high speeds the exhaust will in fact be losing energy as it changes from a positive velocity to a slightly less positive velocity. The energy it loses is of course transferred to the rocket.
      – Arcanist Lupus
      18 hours ago












      Yes, exactly. Since $Delta KE_rocket+Delta KE_exhaust+Delta PE_fuel=0$, when $Delta KE_exhaust<0$ the amount of KE gained by the rocket is greater than the amount of chemical PE lost by the fuel!
      – Dale
      6 hours ago




      Yes, exactly. Since $Delta KE_rocket+Delta KE_exhaust+Delta PE_fuel=0$, when $Delta KE_exhaust<0$ the amount of KE gained by the rocket is greater than the amount of chemical PE lost by the fuel!
      – Dale
      6 hours ago










      up vote
      5
      down vote













      There are multiple ways to view this.



      The easiest, I think, is that kinetic energy scales with the square of velocity
      $$K=frac 12 m v^2$$



      If we assume that the rocket booster supplies a constant acceleration, then comparing initial and final velocities we find that
      $$Delta v=v_final-v_init=at$$



      So for the same amount of time, the change in the velocity is the same, regardless of what the starting velocity actually is. Since we have a squared dependence on the velocity in $K$, this means that the kinetic energy increases more if we started out with a larger velocitiy. i.e.
      $$Delta K=frac 12 m v_final^2-frac 12 m v_init^2=frac 12 m (v_final-v_init)(v_final+v_init)=frac 12 m Delta v (Delta v+2v_init)$$



      So as we can see, the sum of the velocities in the expression for $Delta K$ is what contributes to a larger change in kinetic energy. Since the work done is equal to the change in kinetic energy, it must be that the rocket does more work when we start off at a larger velocity.



      The second way to view this, which you could argue is the same as the first, is to look at the definition of work
      $$W=intvec F cdot dvec x$$



      Or in one dimension with a constant force
      $$W=FDelta x$$



      Now, once again assuming a constant acceleration, we know that
      $$Delta x = frac 12 a t^2 + v_initt$$



      So that the work done is
      $$W=Fleft(frac 12 a t^2 + v_inittright)$$



      Once again, we see that the initial velocity determines the work. A qualitative explanation from this is that when the velocity is larger, then the object covers more distance in the same amount of time. So if we look at the time the force is being applied, the faster it is moving the larger distance the force is applied over. Therefore, we get more work done if the object is initially moving faster.



      The supposed issue behind all of this is that is seems like we are getting more energy from applying the same force for the same amount of time. But if you work through it you find that this is no issue at all. This is even true for objects falling near the Earth's surface. Even though the force is constant, gravity does more and more work on the object while it falls. Or in other words, the rate of energy conversion from potential to kinetic energy increases as the object falls.






      share|cite|improve this answer
























        up vote
        5
        down vote













        There are multiple ways to view this.



        The easiest, I think, is that kinetic energy scales with the square of velocity
        $$K=frac 12 m v^2$$



        If we assume that the rocket booster supplies a constant acceleration, then comparing initial and final velocities we find that
        $$Delta v=v_final-v_init=at$$



        So for the same amount of time, the change in the velocity is the same, regardless of what the starting velocity actually is. Since we have a squared dependence on the velocity in $K$, this means that the kinetic energy increases more if we started out with a larger velocitiy. i.e.
        $$Delta K=frac 12 m v_final^2-frac 12 m v_init^2=frac 12 m (v_final-v_init)(v_final+v_init)=frac 12 m Delta v (Delta v+2v_init)$$



        So as we can see, the sum of the velocities in the expression for $Delta K$ is what contributes to a larger change in kinetic energy. Since the work done is equal to the change in kinetic energy, it must be that the rocket does more work when we start off at a larger velocity.



        The second way to view this, which you could argue is the same as the first, is to look at the definition of work
        $$W=intvec F cdot dvec x$$



        Or in one dimension with a constant force
        $$W=FDelta x$$



        Now, once again assuming a constant acceleration, we know that
        $$Delta x = frac 12 a t^2 + v_initt$$



        So that the work done is
        $$W=Fleft(frac 12 a t^2 + v_inittright)$$



        Once again, we see that the initial velocity determines the work. A qualitative explanation from this is that when the velocity is larger, then the object covers more distance in the same amount of time. So if we look at the time the force is being applied, the faster it is moving the larger distance the force is applied over. Therefore, we get more work done if the object is initially moving faster.



        The supposed issue behind all of this is that is seems like we are getting more energy from applying the same force for the same amount of time. But if you work through it you find that this is no issue at all. This is even true for objects falling near the Earth's surface. Even though the force is constant, gravity does more and more work on the object while it falls. Or in other words, the rate of energy conversion from potential to kinetic energy increases as the object falls.






        share|cite|improve this answer






















          up vote
          5
          down vote










          up vote
          5
          down vote









          There are multiple ways to view this.



          The easiest, I think, is that kinetic energy scales with the square of velocity
          $$K=frac 12 m v^2$$



          If we assume that the rocket booster supplies a constant acceleration, then comparing initial and final velocities we find that
          $$Delta v=v_final-v_init=at$$



          So for the same amount of time, the change in the velocity is the same, regardless of what the starting velocity actually is. Since we have a squared dependence on the velocity in $K$, this means that the kinetic energy increases more if we started out with a larger velocitiy. i.e.
          $$Delta K=frac 12 m v_final^2-frac 12 m v_init^2=frac 12 m (v_final-v_init)(v_final+v_init)=frac 12 m Delta v (Delta v+2v_init)$$



          So as we can see, the sum of the velocities in the expression for $Delta K$ is what contributes to a larger change in kinetic energy. Since the work done is equal to the change in kinetic energy, it must be that the rocket does more work when we start off at a larger velocity.



          The second way to view this, which you could argue is the same as the first, is to look at the definition of work
          $$W=intvec F cdot dvec x$$



          Or in one dimension with a constant force
          $$W=FDelta x$$



          Now, once again assuming a constant acceleration, we know that
          $$Delta x = frac 12 a t^2 + v_initt$$



          So that the work done is
          $$W=Fleft(frac 12 a t^2 + v_inittright)$$



          Once again, we see that the initial velocity determines the work. A qualitative explanation from this is that when the velocity is larger, then the object covers more distance in the same amount of time. So if we look at the time the force is being applied, the faster it is moving the larger distance the force is applied over. Therefore, we get more work done if the object is initially moving faster.



          The supposed issue behind all of this is that is seems like we are getting more energy from applying the same force for the same amount of time. But if you work through it you find that this is no issue at all. This is even true for objects falling near the Earth's surface. Even though the force is constant, gravity does more and more work on the object while it falls. Or in other words, the rate of energy conversion from potential to kinetic energy increases as the object falls.






          share|cite|improve this answer












          There are multiple ways to view this.



          The easiest, I think, is that kinetic energy scales with the square of velocity
          $$K=frac 12 m v^2$$



          If we assume that the rocket booster supplies a constant acceleration, then comparing initial and final velocities we find that
          $$Delta v=v_final-v_init=at$$



          So for the same amount of time, the change in the velocity is the same, regardless of what the starting velocity actually is. Since we have a squared dependence on the velocity in $K$, this means that the kinetic energy increases more if we started out with a larger velocitiy. i.e.
          $$Delta K=frac 12 m v_final^2-frac 12 m v_init^2=frac 12 m (v_final-v_init)(v_final+v_init)=frac 12 m Delta v (Delta v+2v_init)$$



          So as we can see, the sum of the velocities in the expression for $Delta K$ is what contributes to a larger change in kinetic energy. Since the work done is equal to the change in kinetic energy, it must be that the rocket does more work when we start off at a larger velocity.



          The second way to view this, which you could argue is the same as the first, is to look at the definition of work
          $$W=intvec F cdot dvec x$$



          Or in one dimension with a constant force
          $$W=FDelta x$$



          Now, once again assuming a constant acceleration, we know that
          $$Delta x = frac 12 a t^2 + v_initt$$



          So that the work done is
          $$W=Fleft(frac 12 a t^2 + v_inittright)$$



          Once again, we see that the initial velocity determines the work. A qualitative explanation from this is that when the velocity is larger, then the object covers more distance in the same amount of time. So if we look at the time the force is being applied, the faster it is moving the larger distance the force is applied over. Therefore, we get more work done if the object is initially moving faster.



          The supposed issue behind all of this is that is seems like we are getting more energy from applying the same force for the same amount of time. But if you work through it you find that this is no issue at all. This is even true for objects falling near the Earth's surface. Even though the force is constant, gravity does more and more work on the object while it falls. Or in other words, the rate of energy conversion from potential to kinetic energy increases as the object falls.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 21 hours ago









          Aaron Stevens

          2,934319




          2,934319




















              up vote
              4
              down vote













              In both cases (stationary and flying fast), the change in total kinetic energy of the rocket+propellant system is the same, equal to the chemical energy released during a one-second burn. If the rocket was stationary, the propellant goes from rest to moving backward, increasing its kinetic energy. If the rocket was flying (very) fast, the propellant goes from moving forward fast (with the rocket) to moving forward slower (trailing behind), decreasing its kinetic energy. This is enough to understand qualitatively why the kinetic energy of the rocket increases more in the second case.



              The two cases are related by a Galilean transformation (choice of uniformly moving reference frame). Consistency is assured by the fact that, for any isolated system (like the rocket+propellant), the change in total kinetic energy between one time and another is Galilean invariant (the same in any uniformly moving reference frame).






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              nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                up vote
                4
                down vote













                In both cases (stationary and flying fast), the change in total kinetic energy of the rocket+propellant system is the same, equal to the chemical energy released during a one-second burn. If the rocket was stationary, the propellant goes from rest to moving backward, increasing its kinetic energy. If the rocket was flying (very) fast, the propellant goes from moving forward fast (with the rocket) to moving forward slower (trailing behind), decreasing its kinetic energy. This is enough to understand qualitatively why the kinetic energy of the rocket increases more in the second case.



                The two cases are related by a Galilean transformation (choice of uniformly moving reference frame). Consistency is assured by the fact that, for any isolated system (like the rocket+propellant), the change in total kinetic energy between one time and another is Galilean invariant (the same in any uniformly moving reference frame).






                share|cite|improve this answer








                New contributor




                nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.



















                  up vote
                  4
                  down vote










                  up vote
                  4
                  down vote









                  In both cases (stationary and flying fast), the change in total kinetic energy of the rocket+propellant system is the same, equal to the chemical energy released during a one-second burn. If the rocket was stationary, the propellant goes from rest to moving backward, increasing its kinetic energy. If the rocket was flying (very) fast, the propellant goes from moving forward fast (with the rocket) to moving forward slower (trailing behind), decreasing its kinetic energy. This is enough to understand qualitatively why the kinetic energy of the rocket increases more in the second case.



                  The two cases are related by a Galilean transformation (choice of uniformly moving reference frame). Consistency is assured by the fact that, for any isolated system (like the rocket+propellant), the change in total kinetic energy between one time and another is Galilean invariant (the same in any uniformly moving reference frame).






                  share|cite|improve this answer








                  New contributor




                  nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  In both cases (stationary and flying fast), the change in total kinetic energy of the rocket+propellant system is the same, equal to the chemical energy released during a one-second burn. If the rocket was stationary, the propellant goes from rest to moving backward, increasing its kinetic energy. If the rocket was flying (very) fast, the propellant goes from moving forward fast (with the rocket) to moving forward slower (trailing behind), decreasing its kinetic energy. This is enough to understand qualitatively why the kinetic energy of the rocket increases more in the second case.



                  The two cases are related by a Galilean transformation (choice of uniformly moving reference frame). Consistency is assured by the fact that, for any isolated system (like the rocket+propellant), the change in total kinetic energy between one time and another is Galilean invariant (the same in any uniformly moving reference frame).







                  share|cite|improve this answer








                  New contributor




                  nanoman is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  share|cite|improve this answer






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                  answered 20 hours ago









                  nanoman

                  1411




                  1411




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