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

zanick
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!

## Answers and Replies

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.

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Mentor
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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##

• bob012345 and sophiecentaur
zanick
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##
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?

also, if it is the same thrust, how can then the energy required to make the same thrust be higher?
Because that's how energy is defined. Note that energy is frame dependent.

• bob012345
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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?

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.

• bob012345
Gold Member
how does the rocket know it is going 10,000mps or if it is at a stand still?
Because that's how energy is defined. Note that energy is frame dependent.
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.

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also, if it is the same thrust, how can then the energy required to make the same thrust be higher?
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.

• sophiecentaur and russ_watters
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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.
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.

• sophiecentaur and Dale
Gold Member
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

• Dale
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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.

Gold Member
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.

zanick
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.
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?

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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?

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.

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• bob012345
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Then, can you help me do an "energy inventory"? Because of P=Fs. of power is constant, force will go down proportional with velocity.
@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.

• bob012345
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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?

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.

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• PeroK
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And, if he pitches in the opposite direction, the fan sees him pitch a stationary ball!

• bob012345
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And, if he pitches in the opposite direction, the fan sees him pitch a stationary ball!
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.

zanick
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.
I had my son , who now asks lots of physics questions now.... ask me something relating to this. (and what you initially said above).. say you are on a 1000kg rocket traveling at 20,000mph and you fired off a gun with a projectile and force of 100N for 1 second, with 1kg mass. would this accelerate the rocket? most would say, "yes", by the acceleration from the force of 100N on a 1kg mass, or 100m/s/s . (F/m=a) then, say you put up some solar sails and got to 30,000mph and did the same thing... would the acceleration be the same because the force of the gun firing the same projectile, be the same?... (minus or accounting for the mass that was ejected) if so, im trying to explain the squared factor of KE, based on energy that would be required to accelerate the same from a higher velocity vs the lower velocity

Mentor
say you are on a 1000kg rocket traveling at 20,000mph and you fired off a gun with a projectile and force of 100N for 1 second, with 1kg mass. would this accelerate the rocket? most would say, "yes", by the acceleration from the force of 100N on a 1kg mass, or 100m/s/s . [emphasis added]
I think you tripped-over your example: the rocket is 1,000 kg, not 1kg. Since you've declared the force to be 100N for 1sec, you don't need to care about the mass of the projectile.
(F/m=a) then, say you put up some solar sails and got to 30,000mph and did the same thing... would the acceleration be the same because the force of the gun firing the same projectile, be the same?... (minus or accounting for the mass that was ejected) if so, im trying to explain the squared factor of KE, based on energy that would be required to accelerate the same from a higher velocity vs the lower velocity
f=ma, f=ma, f=ma, f=ma, f=ma. Nothing on your post says anything about kinetic energy; only f=ma.

Mentor
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im trying to explain the squared factor of KE, based on energy that would be required to accelerate the same from a higher velocity vs the lower velocity
That seems like a bad way to explain the squared factor in KE. I would just show him the work energy theorem to explain where the squared factor comes in.

• russ_watters
zanick
I had my son , who now asks lots of physics questions now.... ask me something relating to this. (and what you initially said above).. say you are on a 1000kg rocket traveling at 20,000mph and you fired off a gun with a projectile and force of 100N for 1 second, with 1kg mass. would this accelerate the rocket? most would say, "yes", by the acceleration from the force of 100N on a 1kg mass, or 100m/s/s . (F/m=a) then, say you put up some solar sails and got to 30,000mph and did the same thing... would the acceleration be the same because the force of the gun firing the same projectile, be the same?... (minus or accounting for the mass that was ejected) if so, im trying to explain the squared factor of KE, based on energy that would be required to accelerate the same from a higher velocity vs the lower velocity
Yes Dale... was rushing out and didn't finish my description. so yes, the projectile doesn't matter other than reducing the mass of the rocket which is now just under 1000kg....., So, the rocket has pushed out the 1k mass a with 100N of force for a second and there for it has accelerated itself ….100/1000= .1m/s/s or .1m/s faster. is this correct? then, later when the rocket achieves 30,000mph via the "alternate propellent force", the same gun is fired rearward again. will the same acceleration be achieved?

Maybe changing the scale of the values might make it more visual. say we want to experience 1g of acceleration. (everyone can relate to this) this 1000kg rocket would need 10,000N of force from the "gun". 10,000/1000=10m/s/s ( 1g) . the rocket fires the gun which applies the force for a second, from a velocity of 20,000mph. now its traveling at 20,021mph...……… later, at 30,000mph, the gun is fired again.... is the acceleration still 1g for that 1 second? or , because of KE, would the acceleration be less for the same energy applied? ( with regards to the projectile, the force x velocity =power and the watt-seconds, or joules would be the same as when fired from the lower velocity of travel) however, when considering the higher velocity of the rocket, does that same energy result in a lower acceleration? (i.e. less than the 1g experience at 20Kmph vs the new velocity of 30kmph) is this so because the same force applied over the same time, but greater distance ? that seems to imply more energy. I know that KE will go up with the ^2 of velocity, but im getting stuck wondering how the same "gun firing causing 10,000N" doesn't produce the same acceleration of 1g regardless of the velocity in space.

Once you answer this for me, I can explain it to my curious son... its making my head hurt because I have forgotten so much! thanks!

Homework Helper
then, later when the rocket achieves 30,000mph via the "alternate propellent force", the same gun is fired rearward again. will the same acceleration be achieved?
Yes. The same mass ejected with the same relative velocity produces the same force and results in the same acceleration.

One could dispense with the "alternate propellant" and simply adopt a frame of reference within which the rocket was already moving at 30,000 mph. The result of ejecting a 1 kg mass is the same.

As has been pointed out, once you account properly for energy, including the change in kinetic energy of the ejected mass, the books balance and no contradiction is obtained.

• sophiecentaur
zanick
Yes. The same mass ejected with the same relative velocity produces the same force and results in the same acceleration.

One could dispense with the "alternate propellant" and simply adopt a frame of reference within which the rocket was already moving at 30,000 mph. The result of ejecting a 1 kg mass is the same.

As has been pointed out, once you account properly for energy, including the change in kinetic energy of the ejected mass, the books balance and no contradiction is obtained.
ok, this is where Im having the problem understanding...……… say we have this same rocket, and it now fires this projectile every second resulting in a force of 10,000N for 1 second.... doesn't the rocket now accelerate at 1g, for as long as it has projectiles? I'm having the problem understanding seeing how the energy matches the KE gained as the rocket accelerates. every firing of the rocket , is the same projectile, same force, same amount of fuel. it seems like the energy stays constant and the KE goes up with velocity, but I know this is not the case.
However, in re-reading your post.. you say, "same relative velocity". so, since the projectile would be exiting the rocket at lesser and lesser relative velocities, the force then might not be the same with each firing...…. then, if that is the case, the firing of the projectile, would produce less force and its seems like its net reaction would be independent of the rocket's velocity. I keep on going back to a comparison to pushing bowling balls out the back of the ship..... every time I apply the force to push the bowling ball out the rear of the ship, isn't the force the same, regardless of how fast the ship is going? (if in space) if so, how is the energy required going up for each "push" if the energy I'm using is the same each time?

Mentor
every firing of the rocket , is the same projectile, same force, same amount of fuel. it seems like the energy stays constant and the KE goes up with velocity, but I know this is not the case.
However, in re-reading your post.. you say, "same relative velocity". so, since the projectile would be exiting the rocket at lesser and lesser relative velocities, the force then might not be the same with each firing...…. then, if that is the case, the firing of the projectile, would produce less force and its seems like its net reaction would be independent of the rocket's velocity. I keep on going back to a comparison to pushing bowling balls out the back of the ship..... every time I apply the force to push the bowling ball out the rear of the ship, isn't the force the same, regardless of how fast the ship is going? (if in space) if so, how is the energy required going up for each "push" if the energy I'm using is the same each time?
I think you might be looking at it backwards: the speed of the reaction mass relative to the rocket is the same every time.

But the speed of the reaction mass relative to the starting point (Earth?) is different every time because the rocket's speed is changing. To an observer on Earth, the first reaction projectile moves to the left and the rocket moves to the right. The second reaction projectile moves to the left a little slower. After a while, the reaction projectiles are actually moving to the right even while accelerating the rocket to the right.

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• sophiecentaur and PeroK
zanick
I think you might be looking at it backwards: the speed of the reaction mass relative to the rocket is the same every time.

But the speed of the reaction mass relative to the starting point (Earth?) is different every time because the rocket's speed is changing. To an observer on Earth, the first reaction projectile moves to the left and the rocket moves to the right. The second reaction projectile moves to the left a little slower. After a while, the reaction projectiles are actually moving to the right even while accelerating the rocket to the right.
That makes sense, but can you address the rocket accelerating to the right and what that acceleration would be? isn't it still 1g with every firing? and I so, why would it require 4x the energy if the velocity of the rocket was doubled, to create this same force, and thus same acceleration, by firing the projectile.

I know here on earth with cars, and with their most optimistic capability to accelerate at any velocity, would operate at constant max HP. But, they are acting against the ground to accelerate and constant power, means force goes down with velocity (inversely proportional), so, 0-50mph would use a lot less energy than 50mph to 100mph, due to the fact that the HP-seconds would be much greater, because it would be spending more time from 50-100mph getting to that top speed, due to less force available for acceleration. the rate of change of KE along the way would be constant, but in the end there would be 4x the energy stored at 100mph vs 50mph for these reasons. IN space and with a rocket with near constant thrust, power goes up with velocity... and so would energy required and stored as KE.

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Homework Helper
the energy matches the KE gained as the rocket accelerates
The rocket is not the only place you need to look for a change in kinetic energy. What about the ejected mass? That changes kinetic energy also. Perhaps surprisingly, that change can be a reduction.

cars, and with their most optimistic capability to accelerate at any velocity, would operate at constant max HP. But, they are acting against the ground to accelerate and constant power, means force goes down with velocity
In the case of cars, the "exhaust velocity" is the velocity of the earth relative to the car. Rockets behave the same way. If you double the exhaust velocity, it takes twice as much power (and half as much mass flow rate) to obtain a given thrust.

In the case at hand we are contemplating a fixed exhaust velocity relative to the rocket. If exhaust velocity and mass flow rate are both kept constant, consumed power(1) , delivered power(2) and thrust are also constant regardless of the velocity of the rocket.

(1) Consumed power is, for example, the rate of consumption of the chemical potential energy that powers the rocket.

(2) Delivered power is the rate of increase in kinetic energy of the entire system, including rocket and exhaust.

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Mentor
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doesn't the rocket now accelerate at 1g, for as long as it has projectiles?
Yes, except as the mass of the rocket diminishes the acceleration actually increases.

I'm having the problem understanding seeing how the energy matches the KE gained as the rocket accelerates. every firing of the rocket
Every problem that anyone has with energy conservation in a rocket is due to failure to consider the kinetic energy of the exhaust. 100%.

I encourage you to actually do the math. Work it out, including both the change in energy of the rocket and the exhaust.

s the same projectile, same force, same amount of fuel. it seems like the energy stays constant and the KE goes up with velocity, but I know this is not the case.
In fact it is the case, and yet energy is conserved.

isn't the force the same, regardless of how fast the ship is going? (if in space)
Yes.

how is the energy required going up for each "push" if the energy I'm using is the same each time?
You have been told multiple* times: You are neglecting the change in the KE of the exhaust. That is where the extra energy comes from.

All that remains is for you to work it out mathematically for yourself so that you can understand and believe the answers you have already received.

* I count 9 separate posts where someone has told you that you need to consider the change in KE of the exhaust and 0 posts where you did so. This is very frustrating to respondents and discourages people from helping you in the future. Put yourself in our shoes: When talking with your son, how frustrating do you find it when you have to repeat yourself literally 9 times? And presumably you are more attached to your son than we are to you.

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• sophiecentaur
Gold Member
with 100N of force for a second
every time I apply the force to push the bowling ball out the rear of the ship, isn't the force the same, regardless of how fast the ship is going?
Try this:
You seem to have been asking the same basic questions throughout this thread and I think that the answers you have been getting may 'too hard'. Perhaps a step back to more basic basics could help again.
The Momentum increase is the same if the Impulse (force times time) is the same. I think there is still some confusion between Momentum and KE. Your last post implies equal amounts of Impulse are applied (Force for one second). You have been mixing up the two quantities (Momentum and KE) in your thinking and they are not the same.
how is the energy required going up for each "push" if the energy I'm using is the same each time?
You are using your intuition here and it's not just an intuitive thing.
Kinetic Energy is not an Absolute Thing - the formula involves v which is a relative quantity (say relative to Earth). Your bursts of rocket fuel / bowling ball energy may be the same from the point of view of the ship but they affect the Kinetic Energy of the ship, relative to the Earth, differently, depending on the ship's velocity.

• jbriggs444
Homework Helper
You are using your intuition here and it's not just an intuitive thing.
It would help if the intuitive argument being made were stated explicitly. It is difficult to debunk vague hand-waving. I think the argument goes something like this:

"If we apply the same force and produce the same relative exhaust velocity we are clearly using the same amount of potential energy. That might be energy in our muscles, chemical energy in our rocket fuel or electrical energy in a battery that we are using to pump out an ion stream.

"By conservation of momentum, the same relative exhaust velocity and the same mass fraction being expelled should produce a consistent acceleration for the rocket.

"But the energy gained by the rocket is different depending on how fast it is moving. Energy is also a conserved quantity. If the same energy input can produce a varying energy output, conservation of Energy fails. This seems to be a contradiction"

One way out [an incorrect way] is to speculate that momentum is not conserved and that the same energy input produces a reduced momentum output depending on velocity.

The correct way out is to do a proper accounting for energy, taking into account the kinetic energy of the mass that is expelled out the back. As has been pointed out ad nauseum.

• Dale
zanick
Thanks Dale, I'm not ignoring conservation of energy, I'm just trying to understand it and apply it to this thought experiment. I seem to be ignoring the KE of the exhaust gases because im not fully understanding the relation. Yes, I have seen it mentioned and I know the math works out...but im still coming back to a constant thrust being applied and therefor a constant acceleration of the rocket.
I see your point though..... I'm creating thrust, but the same amount of KE goes into the exhaust. but to fully understand it, are we saying you cant have constant thrust unless you have enough energy to feed both the exhaust and the rocket forces? Then there is the fact that the mass of the rocket is going down, increasing the acceleration.... im getting there , but still not quite. Im fighting the intuition that the energy is going up 2x to velocity, but not squared. help me get there so I can conceptually get it.... thanks!

another way I could look at it, is if we had a fixed energy release rate feeding the rocket and exhaust, would acceleration go down proportional to velocity?

In your frustration of helping me here, you reference my attachment to my son as well as your possible lack to me, I will offer what was told to me about explaining complex ideas to others.... the more you understand, the easier it is to explain in the most simple terms. When my son doesn't understand something in science, I blame myself for not knowing it well enough to simply it even further if he doesn't get it after my help.... so I try again. It forces me to learn the topic better. I appreciate the effort to help me with this example...… its been a long time since I studied it , that I'm sure ill never know as much as you and the group have forgotten! So, thanks!

Yes, except as the mass of the rocket diminishes the acceleration actually increases.

Every problem that anyone has with energy conservation in a rocket is due to failure to consider the kinetic energy of the exhaust. 100%.

I encourage you to actually do the math. Work it out, including both the change in energy of the rocket and the exhaust.

In fact it is the case, and yet energy is conserved.

Yes.

You have been told multiple* times: You are neglecting the change in the KE of the exhaust. That is where the extra energy comes from.

All that remains is for you to work it out mathematically for yourself so that you can understand and believe the answers you have already received.

* I count 9 separate posts where someone has told you that you need to consider the change in KE of the exhaust and 0 posts where you did so. This is very frustrating to respondents and discourages people from helping you in the future. Put yourself in our shoes: When talking with your son, how frustrating do you find it when you have to repeat yourself literally 9 times? And presumably you are more attached to your son than we are to you.

Homework Helper
I'm creating thrust, but the same amount of KE goes into the exhaust.
No, it does not.

For a fixed fuel burn rate, you have a fixed budget of input power. The more that goes into adding to the kinetic energy of the rocket, the less is available to add to the kinetic energy of the expelled exhaust.

If the rocket is moving faster than 1/2 of its exhaust velocity, you might want to think about whether speaking of adding KE to the exhaust is sensible.

zanick
No, it does not.

For a fixed fuel burn rate, you have a fixed budget of input power. The more that goes into adding to the kinetic energy of the rocket, the less is available to add to the kinetic energy of the expelled exhaust.

If the rocket is moving faster than 1/2 of its exhaust velocity, you might want to think about whether speaking of adding KE to the exhaust is sensible.
can you explain further?
so, like with cars on earth, if power is kept constant, acceleration goes down proportional to velocity and this means in space the same is true? ….. net thrust (force) would have to go down with velocity as well, correct?

Homework Helper
can you explain further?
so, like with cars on earth, if power is kept constant, acceleration goes down proportional to velocity and this means in space the same is true? ….. net thrust (force) would have to go down with velocity as well, correct?
In space you carry your reaction mass with you. On Earth you use the road. It is an entirely different proposition. Using the one to reason about the other is not leading you toward correct conclusions.

No, thrust does not go down with increasing velocity if you carry your reaction mass with you.

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