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B Energy to achieve constant acceleration - F=ma and 1/2mv^2

  1. Oct 22, 2018 #1
    I was in a discussion regarding the time it took to get to the speed of light with a constant acceleration of 1g, and the time was something close to a year. (ignoring all the other major factors ). Just from a Newtonian perspective, if you have a 100kg rocket putting out 900N of thrust, you will get 1g of acceleration. now, say you get up to 10,000mps and you shut down... you are in spacing going 10,000mps.......(ignoring any pull of gravity)
    now, you fire up the engine... you instantly produce 900N of thrust , but do you still have 1g of acceleration? if so, how does the rocket know it is going 10,000mps or if it is at a stand still? and if it doesn't... then how does it figure that if you are going 10,000mps, it going to take a LOT more energy to produce that thrust at speed than if it was at standstill, relative to its starting point. the equations say this ... we know this to be on the surface of the earth with vehicles with limits to power (acceleration and force go down with velocity if power is constant) So, do rockets have the same issues in following laws . and if so, can i get a good explanation?
    It just seems like if im at 10,000mps and i push a bowling balll rear ward with 100N of force, i should cause 1m/s/s of acceleration of my 100kg rocket added to however fast we are traveling. ... i know this is not correct, so im asking you all why.
    thanks!
     
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  3. Oct 22, 2018 #2

    phyzguy

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    From a Newtonian perspective (ignoring relativity), work, or rate of energy input is F.v. So it takes a lot more energy input to achieve 900N of thrust when you are going 10,000 m/s than when you are standing still. Another way to look at it is that if a certain rocket exhaust velocity gives you 900N of thrust when you are standing still, the rocket exhaust velocity needs to be increased by 10,000 m/s to give you 900N of thrust when you are moving at 10,000 m/s. This takes a lot more energy.


    Edit: Please ignore this post. What I said about needing to increase the exhaust velocity in order to get the same thrust is incorrect. Dale's statements below are correct.
     
    Last edited: Oct 23, 2018
  4. Oct 22, 2018 #3

    Dale

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    A rocket engine produces a constant thrust. The same thrust results in a larger change in KE if the rocket is already traveling fast than if the rocket is at rest or traveling slowly.

    To me this is easiest to see with the formula ##P=F\cdot v##
     
  5. Oct 22, 2018 #4
    But the part im trying to grasp, is if you are traveling through space at 10,000mps, and you pushed something (100kg) out the back of the rocket with a force of 100N what would be the acceleration imparted on the spacecraft. wouldnt it be the same acceleration generated at 0 velocity? also, if it is the same thrust, how can then the energy required to make the same thrust be higher?
     
  6. Oct 22, 2018 #5

    A.T.

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    Because that's how energy is defined. Note that energy is frame dependent.
     
  7. Oct 23, 2018 #6

    PeroK

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    As @A.T. said energy is frame dependent. For example, if a car accelerates along a road, then in the reference frame of the surface of the Earth, that represents a certain increase in KE of the car. But, if you view the same acceleration from an external reference frame, where the Earth is spinning, then that is a very different change in KE of the car. Note that, in fact, as the Earth is spinning East and if the car accelerates West, then that will be a reduction in KE of the car! And, of course, from the reference frame of the Solar system the Earth is orbiting the Sun and you'd get again a different change in KE of the car.

    In general, all reference frames will agree on the total change in energy, but not the change in energy/KE for each part of the system. In the above examples, if you take the change in energy of the Earth into account, then you have a closed system where energy is conserved.

    In your example, if you take the change in KE of the rocket expellant into account, then you'll find that the total change in KE is equal to the energy generated by the engines and that this change in KE will be the same for all reference frames, including the rest frame of the rocket and the original rest frame of the rocket.
     
  8. Oct 23, 2018 #7

    sophiecentaur

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    A noddy way of approaching this is to think of how much damage the rocket can do when it collides with something that's stationary in the reference frame. People on the rocket will detect 1g with their weighing machine all the time and that's all they can know. They would know how fast they were going, relative to a spec of dust they fly into because of the energy / damage of the collision and they would know how fast they are going relative to the launch location , using SUVAT equations.
    You need to distinguish between Momentum change and Kinetic Energy change.
     
  9. Oct 23, 2018 #8

    Dale

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    The energy required is not higher. What is happening is that you are looking only at the energy of the rocket and you are ignoring the energy of the exhaust. At low speeds most of the energy goes into the exhaust. As you increase the speed more of the energy goes into the rocket and less into the exhaust.
     
  10. Oct 23, 2018 #9

    jbriggs444

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    And if you go fast enough (faster than 1/2 exhaust velocity) then the books still balance because you are removing energy from the exhaust rather than adding it.
     
  11. Oct 23, 2018 #10

    bob012345

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    Yes, the rocket engine appears to do different amounts of work depending on the reference frame you observe it from since different observers see the same force acting through different distances. So one can say the engine does more work as observed in certain reference frames but at the same time, the energy release at the engine itself is constant. As has been said, both are true and all observers who do a complete energy inventory will agree. But the simple fact that the same amount of fuel burned in a rocket is far more effective from the perspective of the mission planners in the Earth reference frame if the rocket is already moving fast is highly useful in space flight. The effect is called the Oberth effect after Hermann Oberth. It can be enhanced further by using the gravity well of a planet to increase the speed before a burn is done which allows for a much greater speed increase for the same fuel use than one could get by itself even though the effect works from any starting speed.


    https://en.m.wikipedia.org/wiki/Oberth_effect

    https://en.m.wikipedia.org/wiki/Hermann_Oberth
     
  12. Oct 23, 2018 #11

    bob012345

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    One caveat is that at highly relativistic speeds, the constant acceleration of the rocket will be seen to be reduced by 1/gamma^3 thus the force observed will be reduced by observers at a high relative velocity.
     
  13. Oct 24, 2018 #12

    bob012345

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    I forgot to mention the effect is used in virtually all rocket launches from Cape Canaveral. Since orbital speed is with respect to the center of the earth, not the surface, and the earth is spinning, launching with the rotation of the earth uses less fuel to achieve orbit than launching against the rotation of the earth since the rocket is already moving at about 1000mph around the enter of the earth before it even starts. If the earth weren't spinning all orbits woud require the same energy to launch to from the perspective of providing the fuel to burn.
     
  14. Oct 31, 2018 #13
    Then, can you help me do an "energy inventory"? Because of P=Fs. of power is constant, force will go down proportional with velocity. and if power is constant, that means the energy released or used in its ability to "do work" is constant. If acceleration was constant that means KE would be going up with the square of speed and force would have to be kept constant.. energy required do to the ever increasing rate of doing work, would go up with the square of speed.

    earlier someone alluded to the fact that the faster velocity would require more energy to create the same force, but that's the part I'm having a hard time understanding.

    quite simply, can a rocket in space at 10,000mps vs 100,000mps, accelerate at the same rate , given the same force. and if so, the energy required would be MUCH higher, but why?
     
  15. Oct 31, 2018 #14

    PeroK

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    Let's take two different scenarios.

    1) You are cycling along a road. Let's assume the resisting forces of air etc. are small. You get up to, perhaps, 30km/h and then it gets almost impossible to go any faster. Why?

    The force of propulsion in this case is the friction between the tyres and the road. As you move faster (relative to the road against which you are generating the force), the key equations are:

    ##W = Fd##

    Where ##W## is the work done by that force. And:

    ##P = W/t = Fd/t = Fv##

    Hence:

    ##F = P/v##

    Where ##P## is the power you need to generate the force and ##v## is the speed, relative to the road.

    Given that your muscles are the source of the energy and have a maximum power output, you can see that the accelerating force ##F## reduces in proportion to your speed. This is the main reason that it becomes harder to accelerate the faster you are going.

    If we then think about a small resisting force ##f##, then as the speed increases, the accelerating force ##F## reduces to ##f## and at that point the forces are balanced and the bicycle is no longer accelerating.

    Note that this also explains why you have the slightly paradoxical situation on a bike that you are going at your maximum speed (no matter how hard you pedal, you don't go faster), yet when you stop pedaling you slow down only gradually (i.e there is only a small resisting force).

    The same argument generally explains any motion through a medium where your force of propulsion is generated using the medium itself: road, air, water.

    Let's now look at the second scenario:

    2) A spaceship is moving through space (relative to the Earth, say). Space is not a physical medium that you can push on. And, in fact, the spaceship has no way to measure a speed "relative to space". Travelling at 100m/s or 100,000m/s relative to the Earth is fundamentally no different (as far as space is concerned). Motion through space is all relative to the observer.

    The question in this case is what is your means of propulsion if you want to accelerate the spaceship? The simplest method is to fire things out the back and use conservation of momentum as your means to accelerate the ship.

    In this case, it makes no difference what speed the spaceship is travelling relative to the Earth: the process (as far as the spaceship is concerned) is always exactly the same. If we assume non-relativistic speeds, and the spaceship has mass ##M## and the propellant has mass ##m## and is fired out at speed ##u## in the ship's initial reference frame, then the ship increases its speed by ##v## according to:

    ##Mv = mu##

    Hence:

    ##v = \frac{mu}{M}##

    You get the same increase in speed of the ship independent of its speed relative to whoever is looking at it.

    Let's look at the energy involved. To do this, we assume the ship is initially moving as some speed ##w## relative to the Earth.

    The initial Kinetic Energy (KE) is given by:

    ##KE_i = \frac12 (M+m)w^2##

    As above, we have the final KE:

    ##KE_f = \frac12 M(w+v)^2 + \frac12 m(w-u)^2##

    The faster that the ship is moving relative to the observer, the greater the increase in KE of the ship, but the greater the decrease in energy of the propellant. And, if we take the change in KE, we find:

    ##\Delta KE = KE_f - KE_i = \frac12 M(w^2 + v^2 + 2vw) + \frac12 m(w^2 + u^2 - 2uw) - \frac12 Mw^2 - \frac12 mw^2 = \frac12 Mv^2 + \frac12 mu^2 + Mvw - muw = \frac12 Mv^2 + \frac12 mu^2##

    The cross terms in ##w## disappear because we have ##Mv = mu##.

    This is a computational illustration of what we have been saying about energy being frame dependent. The initial and final KE are different depending on your frame of reference, but the change in KE is the same in all frames. And this change in KE, of course, represents the energy released by whatever engine (or muscles) fired the propellant from the rocket.
     
    Last edited by a moderator: Oct 31, 2018
  16. Oct 31, 2018 #15

    Dale

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    @PeroK did an excellent energy inventory in his scenario 2 above. So I won’t repeat that here. But I will address your power comment.

    Again you are neglecting the exhaust in your comments about power. A rocket engine provides a constant thrust, so F is constant. If the velocity of the rocket is v and the velocity of the exhaust relative to the rocket is u then the velocity of the exhaust is v-u. So the power on the rocket is ##P_R=F\cdot v## and the power on the exhaust is ##P_E=-F\cdot(v-u)##. So the total power is ##P_R+P_E= Fu## which is independent of v. So the power on the rocket does increase with v but the power provided by the engine is independent of v as long as you account for the power that goes into the exhaust also.
     
  17. Oct 31, 2018 #16

    bob012345

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    It's been explained but let me add my two cents worth.

    The answer is yes. The rocket in space will accelerate at the same rate given the same force and given the same release of energy from the same amount of fuel burned. The observation that one case (100,000m/s) requires a much higher energy that the 10,000m/s case is simply because kinetic energy is defined by the observer with whatever relative velocity that observer has. Consider the Chicago Cubs pitcher Yu Darvish practicing his 100mph pitches. On the field Darvish, the catcher, the fans all see the same speed and compute the same energy for the ball. Now consider the team is on a 100mph train to St. Louis to play the Cardinals. Darvish and the catcher decide to practice pitches along the isle of the train car he's in. Pitching in the direction the train is moving, everyone on the train agrees on the energy of the ball which is the same as on the field. Now say a fan is watching the team train pass by and sees the pitch. The fan see the ball moving at 200mph and computes four times the energy for the ball. We can agree Darvish didn't pitch any harder so he exerted and imparted the same energy to the ball. The 4X energy of the ball as seen by the observer not on the train is real, not an illusion. If the ball were allowed to escape the train and not allowing for air resistance, the ball would proceed forward of the train 100mph faster than the train. Where did all the extra energy come from? If you go through all the math you will find it borrowed it from the kinetic energy of the train itself. The train slows down a bit not very noticeable but real enough. The rocket exhaust serves the same purpose as the train. Some of the kinetic energy of the exhaust mass is borrowed by the rocket which gives it the extra energy it needs at the higher speed relative to some observer. In this way every observer reckons the rocket can increase its speed by the same amount for the same fuel. Of course, if the ball is caught on the train by the catcher, the energy is returned to the train.
     
    Last edited: Oct 31, 2018
  18. Oct 31, 2018 #17

    PeroK

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    And, if he pitches in the opposite direction, the fan sees him pitch a stationary ball!
     
  19. Nov 1, 2018 #18

    bob012345

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    Stationary in the horizontal direction but it still falls under gravity. Of course if the pitch is very fast it doesn't fall much before its caught so it might look like its hovering in space.
     
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