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Constant acceleration requires continous power increase!

  1. Apr 12, 2009 #1
    Hi everyone,

    According to the equation Power = Force x Velocity or P = mav or a= P/mv, in order to keep say a rocket at constant accelration would require a CONSTANT INCREASE IN ENERGY INPUT PER UNIT TIME. This is somehow problematic to visualize for me. It would mean that accelerating a rocket with g from 10000-20000 km/h would require THREE TIMES more energy than accelrating it with g from 0-10000 km/h both relative to the earth(due to KE= .5mv^2) . This means as the rocket's accelration is to be kept constant the astronaut would observe that he/she needs to burn more and more fuel per unit time while yet expeiencing the same accelration. To demonstarate this problem further. SAY now the earth is also accelrated towards the rocket until it cathces up with it and is then again slowed down to a conatant speed with the rocket. One would now require more OF THE SAME EARTH'S FUEL/time on this accelrated earth(OR DOES PE OF EARTH'S FUEL INCREASE LINEARLY WITH IT'S BEING ACCELRATED? i dont think so and if tht were the case then the rocket's fuel PE should also have increased) to speed from 0-10000 km/h with the rocket using the same accelration g as one did before from the old earth. But does that make sense and has it been verified experimentally?
    Last edited: Apr 12, 2009
  2. jcsd
  3. Apr 12, 2009 #2
    To give a simpler example lets say a rocket starts off from an asteroid weighing around 100T and 50% of it being fuel for the rocket. The rocket accelerates constantly with g away from the asteroid until it has reached a speed of say 10000 km/h. Then it stops accelerating. Now the asteroid is also accelerated until it is once again in contact with the rocket. Then it too is stopped from accelrating. Now the rocket should according to the above formulas require three times more energy to once again accelerate away from the asteroid with constant g until it reaches the same 10000 km/h relative to the asteroid. BUT this would also mean that any jumping by the astronaut, also with the same accelration as he/shed done before, from the now accelerated asteroid would cost him/her three times as much energy from the body as before! How can that be?
  4. Apr 12, 2009 #3


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    This is a good little puzzle problem. If you do it carefully, you'll find it really does take about the same amount of fuel for each increment in velocity, even though, as you say, the energy is going up as the square of velocity. The key to the problem is that the exhaust gets almost all the energy; not the rocket! Put the exhaust into the equation as well. Assume that the exhaust velocity is very high. Typical values are around 2 to 4 km/s. Remember, this is the velocity of exhaust relative to the rocket!

    10000 km/h is close to the exhaust velocity. It will be easier to start with if you think of smaller increments in velocity, significantly less than exhaust velocity.

    Try this exercise. A rocket is in deep space, next to a small drifting asteroid. The rocket has exhaust velocity of 3000 m/s. As the rocket starts to leave, it burns a small amount of fuel in a short fast burn, and ends up moving away at 10 m/s. Then it burns the same amount of fuel again, and ends up moving at about 20 m/s (actually, a fraction more than this, because it is lighter now having burnt some fuel. You can ignore this difference and still get close to the right answer.)

    You can assume the rocket and fuel together weighed 10000 kg if it helps.

    Total kinetic energy is initially zero. Calculate the total kinetic energy of the rocket AND the exhaust, after the first burn, and after the second burn, all from the perspective of the floating asteroid. Use conservation of momentum to see how much exhaust is expelled.

    Cheers -- Sylas
  5. Apr 12, 2009 #4


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    Consider an accelerating car: it takes longer to accelerate from 30-60 than 0-30 because of this principle.
  6. Apr 12, 2009 #5


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    Very true. It's an interesting converse to the rocket. A car really does need much more fuel to get from 30-60 than from 0-30. A rocket doesn't. So... why is the behaviour so different between the car and the rocket?

    Both the car and the rocket accelerate by pushing on something, and in both cases there's an equal and opposite reaction of momentum. The answer lies in looking at what gets pushed in each case.

    Cheers -- Sylas

    PS. And here's a puzzle to make the car seem counter intuitive. Suppose a car burns a certain amount of fuel to accelerate from 30-60. Observed from an inertial observer on the ground, the car gains a certain amount of energy. But observed from another inertial observer who is moving at a steady 30 km/hr, the car has gained only 1/3 as much. Yet both observers see the same amount of fuel being consumed.

    Where did the extra energy seen from ground end up, when observed by a moving observer? It must have gone somewhere. And it did. But where?
    Last edited: Apr 12, 2009
  7. Apr 12, 2009 #6


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    Hi jonnyk,

    I would like to point you to D_H's excellent rocket tutorial page: https://www.physicsforums.com/showthread.php?t=199087

    D_H is one of our resident rocket scientists. But I will try to answer the basic question.

    When you are doing conservation problems you need to be careful to account for all of the energy. As sylas mentioned above, in the case of a rocket, most of the chemical energy in the fuel does not go into changing the KE of the rocket, but rather into changing the KE of the exhaust. As the rocket accelerates the KE of the exhaust is reduced and a larger fraction of the chemical energy in the fuel goes into KE of the rocket.

    Please realize, rockets work by conservation of momentum. Energy is also conserved, of course, but it is generally easier to analyze a rocket from the perspective of momentum.
    Last edited: Apr 12, 2009
  8. Apr 12, 2009 #7
    Hi everyone,
    First of all thankyou very much for all the informative answers. I tried to workout this problem with another example regarding a robot floating in front of a large rock which initally pushed it and are now moving with constant velocity inside of and relative to a large spacestation. Let the mass of the rock be 1000 kg, mass of the robot be 10 kg and say their velocity is 1 m/s. Now robot and rock are floating at 1 m/s relative to the spacestions walls. The KE of the robot should be .5(10)(1)^2J= 5J in it's own frame. NOW the robot starts to jump using its legs which are in contact with the rock until it reaches a velocity of 1 m/s relative to the rock and thus 2 m/s relative to the spacestation. This would make his KE wrt ss = .5(10)(2)^2=20J. BUT the robot only used 5J from his own body to do the jump because it wouldve accelrated from 0-1 m/s in it's own frame. However wrt the ss it has a KE of 20J. If now something from the spaceship's walls were to catch the robots KE by slowing it back to 1 m/s it should derive 15J NOT 5J. This means part of the KE mustve come from the rock and indeed the rock would keep on slowing down with every addtional jump of the robot.
  9. Apr 12, 2009 #8

    D H

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    Because cars don't work like rockets.

    I'll present three apparently conflicting yet completely consistent views on how the fuel requirements for a rocket.

    It takes the same amount of fuel to make a rocket accelerate from 0 to some velocity v as it does to make it accelerate from v to 2v.

    Suppose we have two identical rockets with equal amounts of the same kind of fuel. One rocket, rocket A, is stationary with respect to some observer while the other rocket, rocket B, is moving at some velocity v. The amount of fuel rocket A needs to accelerate from 0 to v is exactly equal to the amount of fuel rocket B needs to accelerate from v to 2v. All that matters is the Δv -- if all other things are equal, that is. All other things are never equal. The next two viewpoints address all other things not being equal.

    It takes a lot more fuel to make a rocket accelerate from 0 to some velocity v than it does to make it accelerate from v to 2v.

    Suppose a rocket accelerates such that it increases its velocity by 2v. From the perspective of an observer initially at rest with respect to the rocket, the rocket accelerates from 0 to 2v. The amount of fuel needed to get the rocket from rest velocity v compared to the amount needed to get the rocket from v to 2v.

    Some numbers: Suppose the final velocity, 2v, is four times the rocket's exhaust velocity. (This means 98.2% of the initial vehicle mass must be fuel.) Almost all of the fuel is needed to achieve the first half of the total gain in velocity. In this example, about 88.1% of the fuel is used in the first half (Δv-wise) of the trip, with only 11.9% of the fuel needed to achieve the latter half of the acceleration. This is just one example. Fuel consumptions is highly non-linear thanks to the rocket equation.

    It takes a lot less fuel to make a rocket accelerate from 0 to some velocity v than it does to make it accelerate from v to 2v.

    Now suppose two identical rockets are filled with different amounts of fuel. Rocket A is filled with just enough fuel to enable a change in velocity of Δv while rocket B is filled with just enough fuel to enable a twice that change in velocity. The amount of fuel in rocket A is less than the half the amount of fuel in rocket B. Using the same velocities as in the previous example, one has to put 8.4 times much fuel on rocket B as in rocket A.

    So what is going on here?

    The first case is just a matter of reference frames. The latter two cases are flip sides of the same coin, the ideal rocket equation. Think of the Saturn V. That huge machine spent an incredible amount of fuel just getting the vehicle off the ground. At first stage separation, the vehicle had already consumed 4.5 million pounds of propellant but had only increased the vehicle's velocity by about 6,000 miles per hour. The rocket equation is a very mean SOB.
  10. Apr 12, 2009 #9


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    Sure; but they still meet conservation of energy and momentum.

    We can do a simple analysis in the same way for both cases, using the energy, and momentum, of the reaction mass. For the rocket, it's the exhaust. For the car, it's the Earth. If anyone is still interested in the puzzle I proposed of the accelerating car, it is solved with the same underlying physics as the rocket. It doesn't matter that the engine doesn't work the same way; you still use the same physics.

    The puzzle: a car accelerates from 30 to 60km/hr, seen by an inertial observer at rest. Another inertial observer in steady motion initially matched to the car sees the car accelerate only from 0 to 30. This observer sees the car gain only 1/3 as much kinetic energy, while both see it consuming the same fuel. Where's the other 2/3 gone?

    Compare: a rocket accelerates from 30 to 60 km/hr, seen by a floating astronaut. Another floating astronaut sees it accelerate from 0 to 30 km/hr, while consuming the same fuel. Where's the other 2/3 of the energy gone?

    We've already answered the second question. You have to look at the energy terms for the reaction mass, which is the exhaust for the rocket. The same answer applies for the car, It's a bit more subtle in that the car is moving wrt its reaction mass, but there's still the one set of equations that works in both cases.


    Solution, simplified for a "pulse" or short burn of the rocket of a small amount of fuel, so we can give all the exhaust the same velocity.

    Let M be a reaction mass, and m be a vehicle mass. Let v be the relative velocity of the reaction mass and vehicle. Let the vehicle change velocity by δ, by pushing on the reaction mass.

    For a rocket, M << m. For a car, m << M. For a rocket, v=0, because it is carrying its own exhaust. For a car moving at 30 km/hr, v = 30 km/hr.

    Let u be the initial velocity of the vehicle as seen by some inertial observer. For a car, we usually think of an observer at rest wrt to the Earth, so that u=v. But the solution works in general.

    The reaction mass must have a change in velocity Δ, so that MΔ + mδ = 0. This is conservation of momentum, and it is used in the equations below to rewrite energy terms. Note that Δ and δ have opposite signs.

    The observer initially sees the vehicle moving at velocity u, and the reaction mass at velocity (u-v). After the acceleration, the vehicle is moving at (u+δ), and the reaction mass at (u-v+Δ).

    The change in kinetic energy is:
    0.5M(u-v+\Delta)^2 - 0.5M(u-v)^2 + 0.5m(u+\delta)^2 - 0.5mu^2
    & = &
    M(u-v)\Delta + 0.5M\Delta^2 + mu\delta + 0.5m\delta^2 \\
    & = & (M\Delta + m\delta)u - Mv\Delta + 0.5M\Delta^2 + 0.5m\delta^2 \\
    & = & 0.5M\Delta^2 + 0.5m\delta^2 - Mv\Delta \\
    & = & 0.5M\Delta^2 + 0.5m\delta^2 + mv\delta \\
    & = & m\delta(0.5(\Delta+\delta) + v)
    Τhe change in energy is independent of an observer's velocity. Using the final form, we can ignore the reaction mass, and just look at its velocity.

    Case 1. Rocket

    v = 0, because reaction mass initially moves with the rocket. M << m, since it is a short burn, in which all the exhaust gets the same velocity wrt an inertial observer. Δ >> δ, by conservation of momentum. The exhaust velocity is high. For a small course corrections, you can say that Δ+δ ~ Δ.

    The total energy gain is about 0.5mδΔ. Note that term includes the exhaust velocity Δ. You can't ignore the exhaust velocity when looking at a rocket.

    Case 2. Car

    In this case M >> m, because the Earth is much heavier than the car. Also Δ+δ ~ δ, because the velocity given to the Earth is tiny. The velocity of the Earth with respect to the car is non-zero, for a moving car, so the v becomes relevant.

    The total kinetic energy gain is thus about 0.5mδ2 + mvδ

    This time, the term includes the relative velocity of car and Earth. You can't ignore the initial velocity when looking at a car.

    That energy gain is the same as 0.5m(v+δ)2 - 0.5mv2, as you should expect.

    That extra term of mvδ is the solution to the puzzle as I phrased it. For a moving observer, initially moving at velocity v with the car, they only see the car obtaining a kinetic energy gain of 0.5mδ2. The extra energy mvδ is the same as -MvΔ, and it corresponds to a gain in the kinetic energy of the Earth itself, which is seen to be moving at -v, so that even a tiny increment in velocity makes a significant energy difference.

    Kinetic energy of the earth for a moving inertial observer:
    0.5M(-v+\Delta)^2 - 0.5Mv^2 & = & -Mv\Delta + 0.5M\Delta^2 \\
    & \approx & -Mv\Delta \\
    & = & mv\delta

    Cheers -- Sylas
  11. Apr 13, 2009 #10

    D H

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    In short, cars don't work like rockets. Cars push on the Earth while rockets push on themselves.
  12. Apr 13, 2009 #11


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    Consider the rocket and it's fuel as a closed system without any external forces. The only force is internal, thrust at the rocket engine that accelerates spent fuel in one direction and the rocket and its remaining fuel in the other direction. Using the orignal center of mass of this closed system of fuel and rocket as a frame of reference, the center of mass doesn't move, and the total momentum of fuel + rocket will remain zero. Chemical potential energy is converted into kinetic energy of the spent fuel and the rocket.

    Each component of spent fuel's terminal velocity is relative to the rockets current velocity at the time the spent fuel is released. Using the original center of mass as the frame of reference, as rocket velocity increases, spent fuel is released at ever decreasing speeds, and if you take the increasing mass of spent fuel and decreasing mass of rocket into account, then with a constant amount of thrust, the kinetic energy of the system increases at a linear rate; the rate of change of kinetic energy is constant, and the power involved is constant.

    Power equals force times speed, but the speed is with respect to the point of application of that force. In the case of a rocket, "speed" is the difference in speed between the rocket and the terminal velocity of the spent fuel (at that moment), and this speed is independent of an external (to rocket and spent fuel) frame of reference. For a constant mass flow rate and thrust, this "speed" would be a constant.
    Last edited: Apr 13, 2009
  13. Apr 23, 2009 #12
    @Jeff Reid
    You say that the KE increases linearly because mass of fuel decreases. However KE is .5 m x v^2 so for every increase in velocity there is much greater aount of mass that needs to decrease in order for the KE to increase only linearly. NOW 1 kg of matter collided with anti matter of the same mass could produce enormous amount of energy. Enough to accelerate a 1000kg rocket to more than 1000000 m/s(i.e. KE is always around (500+negligible fuel mass) X v^2)). BUT HOW can such a rocket have a constant acceleration all the way from 0-1000000 m/s without continuously, as velocity increases, also increasing the fuel output per unit time? Thnx
    Last edited: Apr 23, 2009
  14. Apr 23, 2009 #13


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    You're still ignoring the exhaust, where the majority of the kinetic energy goes as a rocket accelerates. Include the exhaust in your calculations, and everything works just fine.
  15. Apr 23, 2009 #14


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    I posted a link to wiki below, but will try to explain what is going on. Assume that the spent fuel exits the rocket at 10,000 m/s. Initially the rocket isn't moving so the spent fuel gets all of the energy. Take a short period where only 1 kg of fuel is spent from a 100kg rocket.

    From the rocket equation from the wiki link below
    After 1kg of fuel is spent:

    Rocket velocity = 10000 m/s x ln(100/(100-1)) = 100 m/s
    Average speed2 of the spent fuel = 1/(10000-9900) Integral from 9900 to 10000 of V2dv
    Average speed2 of the spent fuel = 99,003,333 (m/s)2 (correlates to average speed of 9950.04 m/s)

    KE1 of system = 1/2 (99kg (100 m/s)2 + 1 kg (9950.04 m/s)2)
    KE1 of system = 1/2 (990,000 + 99,003,333) = 49,996,666 J

    After 2kg of fuel spent:

    Rocket velocity = 10000 m/s x ln(100/(100-2)) = 202 m/s
    Average speed2 of the spent fuel = 97,003,668 (m/s)2 (correlates to average speed of 9849.04 m/s)

    KE2 of system = 1/2(98kg (202 m/s)2 + 1 kg (9950.04 m/s)2 + 1 kg (9849.04 m/s)2)
    KE2 of system = 1/2(3,998,792 + 99,003,333 + 97,003,668) = 100,002,896 J

    KE2 / KE1 = 2.000

    So double the fuel consumption resulted in double the KE of the system. Note how the KE of the 2nd kg of spent fuel was less than the KE of the 1kg of spent fuel. Also, note that when the rocket speed is 5000m/s the fuel accelerates from +5000 m/s to -5000m/s, no change in KE of the fuel. At faster speeds, the KE of the spent fuel is less than when it was in the rocket.

    If the rocket was 99kg of fuel and 1kg of payload, then it's terminal velocity is:
    Rocket velocity = 10000 m/s x ln(100/(100-99)) = 46051.7 m/s
    Based on the above the KE of the system = 99 x 49,996,666 J = 4,949,669,934 J
    The KE of the rocket = 1/2 x 1 kg x (46051.7 m/s)2 = 1,060,379,622 J
    The KE of the fuel = 3,889,290,312 J

    Last edited: Apr 23, 2009
  16. Apr 23, 2009 #15
    Thnx for the explanation.
  17. Apr 23, 2009 #16

    D H

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    In a reference frame moving faster than the rocket, the rocket loses kinetic energy as its velocity increases (comes closer to zero). The same conundrum occurs with an accelerating car. Consider two cars driving in the same direction but at different speeds. If the slower car starts accelerating, from the perspective of the faster car, that slower car is losing kinetic energy. A given expenditure of energy by the car or the rocket appears to yield a frame-dependent change in velocity. This doesn't make sense; a given expenditure of energy should result in the same change in velocity from the perspective of any inertial frame. So what gives?

    The answer to both conundrums is that you need to look at the whole picture. The car pushes against the Earth, so in the case of the car you need to look at the change in the Earth's kinetic energy as well as the car's. The rocket against the rocket's exhaust stream, so in the case of the rocket you need to look at the change in the exhaust's kinetic energy as well as the rocket's.

    In the case of a rocket, it is a bit easier if you look at things in terms of conservation of momentum rather than conservation of energy. I looked at things from both perspectives (momentum and energy) in this rocket tutorial thread, https://www.physicsforums.com/showthread.php?t=199087.
  18. Apr 23, 2009 #17
    @Jeff Reid and Sylas
    I forgot to ask. What if we propelled a rocket not by using exhaust but by say robots whod be using fuel to jerk towards the front continously inside the rocket? Will that not propel the rocket and will the KE increase liearly per unit time in that case too? Thnx.
  19. Apr 23, 2009 #18

    D H

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    No. Whatever momentum is gained by moving the vehicle forward is going to be lost as soon as the fuel hits the front of the rocket. Think about it for a bit.
  20. Apr 23, 2009 #19

    JK- Yes i thght of it for quite a bit but you probably know that we can sit on a a small toy car, lift our legs or without being in contact with anything but tht car can jerk forward and relax and the car will move in the direction of jerk. The process can then be repeated to make it move more and more.
  21. Apr 23, 2009 #20

    D H

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    That doesn't work on a frictionless surface.
  22. Apr 23, 2009 #21

    JK- Oh yes i think i get it now. On a frictionless surface the same act would cancel out and thered be no net movement. Thnx.
  23. Apr 23, 2009 #22


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    Note that I simplified the average speed^2 of the fuel. The intergral I used for average speed^2 assumed a linear decrease in spent fuel velocity, but the actual velocity decreases faster than this, as the rocket and unspent fuel's acceleration is faster than linear because of the decrease in mass with a constant amount of thrust. Still it was enough to explain the issue.

    The equations that DH linked to don't make this simplified assumption and are mathematically correct.

    Without doing all the math

    VE = relative exit velocity of spent fuel
    VS = spent fuel velocity
    VR = rocket velocity
    MS = mass of spent fuel
    M0 = initial mass of rocket and fuel
    M1 = final mass of rocket and fuel = M0 - MS
    VR = VE x ln(M0/M1)

    From the example I had before:

    VE = -10,000 m/s
    VR = VE x ln((100 - Ms)/100)
    MS0 = 0
    VS0 = -10,000 m/s
    VS1 = VE + VR = VE + VE x ln((100 - MS1)/100)

    average VS2 = 1/(VS1 - (-10,000)) x integral from -10000 to VS1 of (VE(1 + ln(100 - MS1)/100)))2

    at which point I stop with this approach, since I don't intend on figuring out that integral.

    From what I recall, and based on a couple of web sites to confirm it, you can consider the rocket as not moving (imagine it's held in place by a huge mass). Then all the energy goes in to the fuel, and you end up with:

    change of energy of spent fuel = 1/2 m v^2
    power = rate of change in energy = 1 /2 (mass_flow / unit_time) Ve^2.

    This power is independent of frame of reference, so now using a frame of reference where the rocket is accelerating, the energy of the system increases the same as it would if the rocket were not moving. The power is known, the mass flow rate of the spent fuel is known, and the kinetic energy of the rocket and the remaing onboard fuel can be calculated based on the rocket equations already mentioned. The energy in the spent fuel is then just the total energy minus the rocket's energy.
    Last edited: Apr 23, 2009
  24. Apr 24, 2009 #23


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    I created a spreadsheet, rocket.xls, contained in this zip:

    http://jeffareid.net/misc/rocket.zip [Broken]

    It calculates the energies via forumlas and simple numerical approximations for each step in time, with similar results.
    Last edited by a moderator: May 4, 2017
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