Where is the energy coming from?

[LePenguino]

For the following, let's assume that a rocket can apply a constant force on itself, and ignoring the fact that it loses mass from its exhaust, it must therefore have a constant acceleration.

If the force from the engines is constant, this means that a constant amount of fuel in the rocket is being consumed, and therefore, a constant amount of chemical energy from the propellant(s) is being released.

However, the equation for kinetic energy is E = 1/2m(v*v). This means that if velocity changes at a constant rate, the amount of kinetic energy grows exponentially.

This forms a conflict; if the chemical energy being released by the fuel is constant, but the kinetic energy that the rocket gains grows exponentially, where is this extra energy coming from as the rocket speeds up?

[spudman2]

You're ignoring the fact that it's mass decreases.

[Brenok]

The kinetic energy isn't increasing exponentially, it's increasing quadratically.

But what you're describing is the Oberth Effect, so mentioned in KSP. Basically, you gain more energy per unit of fuel because the exhaust that exits the engine gains less engine. When you're at rest, the exhaust exits the rocket at high speed. If the rocket is moving fast, close to the exhaust speed, the exhaust smoke would appear to exit the engine with no speed. (I don't know if that makes sense, maybe I'll need to rewrite it later to clarify.)

[PrivateFlip]

I believe most of the time your question is closely related the answer given to a popular question on these forums "what is the oberth effect?" Wikipedia link.

The same amount of energy you 'get' is 'taken' from the exhausted propellant which is traveling with a lower velocity in the backward direction because you where already moving forward.

[Ralathon]

The fuel in the rocket goes faster as well, so it has kinetic energy. That's your missing energy.

Say you have a 4kg rocket with 2kg of fuel and 2kg payload. The fuel contains enough energy to push it back with 10m/s relative to the payload.

If you burn while stationary you start out with 0 kinetic energy and end with both the payload and the fuel having 25 Joules of kinetic energy. This means our fuel contains 50 Joules of energy.

If you burn while the rocket is moving at 5m/s both the fuel and the payload start with 25 Joules. After the burn the fuel is stationary (0 Joules) and the payload is moving at 10m/s and has 100 Joules of energy. The payload started out with 25, the rocketfuel had 25 and the fuel contained 50. So it all adds up with no energy lost.

[Soda Popinski]

This forms a conflict; if the chemical energy being released by the fuel is constant, but the kinetic energy that the rocket gains grows exponentially, where is this extra energy coming from as the rocket speeds up?

That's incorrect. The energy the fuel reaction releases is at a constant rate, and so is the energy the rocket is GAINING*. The speed, on the other hand is increasing, but at the same rate that energy is being dumped into it (constant Force). Acceleration increases, because F=ma, and Force is constant, mass decreases, so acceleration must increase.

* Say the fuel burning is releasing 1000 kJoules per second, the rocket gains 500 kJoules per second and the propellant gains the other 500 kJoules per second. The rocket already has a certain amount of speed that keeps getting added to (kinetic energy).

[K^2]

That's incorrect. The energy the fuel reaction releases is at a constant rate, and so is the energy the rocket is GAINING*. The speed, on the other hand is increasing, but at the same rate that energy is being dumped into it (constant Force). Acceleration increases, because F=ma, and Force is constant, mass decreases, so acceleration must increase.

* Say the fuel burning is releasing 1000 kJoules per second, the rocket gains 500 kJoules per second and the propellant gains the other 500 kJoules per second. The rocket already has a certain amount of speed that keeps getting added to (kinetic energy).

No, the energy of the rocket does increase at a faster rate than chemical energy is depleted. The energy required to give a rocket a certain delta-V increases as velocity does. But chemical energy released from expended fuel is exactly the same.

Furthermore, the reason why it works this way has been addressed in this thread several times. It's worth reading the whole thing before adding in another reply, especially, an incorrect one.

[LePenguino]

The fuel in the rocket goes faster as well, so it has kinetic energy. That's your missing energy.

Say you have a 4kg rocket with 2kg of fuel and 2kg payload. The fuel contains enough energy to push it back with 10m/s relative to the payload.

If you burn while stationary you start out with 0 kinetic energy and end with both the payload and the fuel having 25 Joules of kinetic energy. This means our fuel contains 50 Joules of energy.

If you burn while the rocket is moving at 5m/s both the fuel and the payload start with 25 Joules. After the burn the fuel is stationary (0 Joules) and the payload is moving at 10m/s and has 100 Joules of energy. The payload started out with 25, the rocketfuel had 25 and the fuel contained 50. So it all adds up with no energy lost.

I think you really answered this well. Thanks for clearing it up!

[Soda Popinski]

No, the energy of the rocket does increase at a faster rate than chemical energy is depleted. The energy required to give a rocket a certain delta-V increases as velocity does. But chemical energy released from expended fuel is exactly the same.

I was assuming he wasn't talking about the Oberth effect, since he didn't mention anything about the altitude. I've read the whole thread, and still think the energy released by 1kg of fuel will be the same amount of energy at the start or the end of a rocket burn. Please tell me how this isn't true.

I'll agree the speed of the rocket increases at a faster rate (acceleration increases), but energy doesn't, since mass is decreasing. With F=ma, the Force remains constant, but the mass decreases (as propellant is lost), so it accelerates at a faster rate. Hence increasing speed faster, but increasing kinetic energy at the same rate (E=0.5m*v^2 - the velocity for the given energy is increasing, but the mass less, giving a net same energy per unit time).

[Soda Popinski]

However, the equation for kinetic energy is E = 1/2m(v*v). This means that if velocity changes at a constant rate, the amount of kinetic energy grows exponentially.

This is assumption I think where the OP is wrong. Velocity change (acceleration) is not constant. As the mass of the rocket is depleted (due to propellant loss), the same amount of Force will yield a faster acceleration. So acceleration is not constant. That's why our TWR increases over time.

Edit: Rereading over this thread, I think my explanation isn't wrong, but incomplete, as I wasn't taking into account the KE of the propellant in the rocket. I'm still pretty sure acceleration isn't constant, and that's why speed is increasing at greater than a geometric rate (at least in part).

Edited February 9, 2014 by Soda Popinski

[K^2]

I was assuming he wasn't talking about the Oberth effect, since he didn't mention anything about the altitude. I've read the whole thread, and still think the energy released by 1kg of fuel will be the same amount of energy at the start or the end of a rocket burn. Please tell me how this isn't true.

I'll agree the speed of the rocket increases at a faster rate (acceleration increases), but energy doesn't, since mass is decreasing. With F=ma, the Force remains constant, but the mass decreases (as propellant is lost), so it accelerates at a faster rate. Hence increasing speed faster, but increasing kinetic energy at the same rate (E=0.5m*v^2 - the velocity for the given energy is increasing, but the mass less, giving a net same energy per unit time).

Forget about increasing acceleration for a moment. Take a rocket just starting to take off. It hasn't burned through enough fuel yet to make a big difference in mass. To go from 0m/s to 1m/s will take 0.5J of energy per kg of mass. To go from 1m/s to 2m/s takes 1.5J of energy per kg. And this keeps increasing.

There is a much simpler way to look at it. Power is equal to thrust times velocity. Rocket's thrust remains pretty much constant. It doesn't matter that the rocket gets lighter. But its velocity increases. So the rocket's power increases as its velocity does. Note that we've taken mass out of it completely now. So even with all of that taken into account, as the rocket goes faster, the rate at which its kinetic energy increases grows faster.

And again, it has been explained in this thread. And yes, it's Obereth effect, because it's not about altitude. It's about velocity. And the reason it works here is exactly the same as the reason it works in Obereth effect, and that has also been explained in this thread.

[Soda Popinski]

Forget about increasing acceleration for a moment. Take a rocket just starting to take off. It hasn't burned through enough fuel yet to make a big difference in mass. To go from 0m/s to 1m/s will take 0.5J of energy per kg of mass. To go from 1m/s to 2m/s takes 1.5J of energy per kg. And this keeps increasing.

There is a much simpler way to look at it. Power is equal to thrust times velocity. Rocket's thrust remains pretty much constant. It doesn't matter that the rocket gets lighter. But its velocity increases. So the rocket's power increases as its velocity does. Note that we've taken mass out of it completely now. So even with all of that taken into account, as the rocket goes faster, the rate at which its kinetic energy increases grows faster.

And again, it has been explained in this thread. And yes, it's Obereth effect, because it's not about altitude. It's about velocity. And the reason it works here is exactly the same as the reason it works in Obereth effect, and that has also been explained in this thread.

Well at least tell me I was correct that acceleration isn't constant. That was my main point of contention with the OP.

Now back to the Oberth effect. Can you show me the math of how this works? I've been trying to wrap my head around it for some time now. I understand it conceptually (propellant already has kinetic energy, but need to see the math work out). You'll have to forgive me, as my most advanced physics was for biology majors.

Looking at your first paragraph, I can see 0 m/s to 1m/s taking 0.5J for 1kg of mass.

⌂KE = change in kinetic energy

⌂v = change in velocity = 1 m/s - 0 m/s = 1 m/s

m= 1kg

KE = ½m*v²

⌂KE = ½ (1kg)*(1m/s)²

⌂KE = 0.5 kg m²/s² = 0.5 J

Now for the second part

⌂KE = change in kinetic energy

⌂v = change in velocity = 2 m/s - 1 m/s = 1 m/s

m= 1kg

KE = ½m*v²

⌂KE = ½ (1kg)*(1m/s)²

⌂KE = 0.5 kg m²/s² = 0.5 J ≠ 1.5 J

[Ralathon]

Well at least tell me I was correct that acceleration isn't constant. That was my main point of contention with the OP.

You're correct that the rocket will accelerate faster as it burns more of its mass. It's just completely irrelevant for the problem in question

Looking at your first paragraph, I can see 0 m/s to 1m/s taking 0.5J for 1kg of mass.

⌂KE = change in kinetic energy

⌂v = change in velocity = 1 m/s - 0 m/s = 1 m/s

m= 1kg

KE = ½m*v²

⌂KE = ½ (1kg)*(1m/s)²

⌂KE = 0.5 kg m²/s² = 0.5 J

Yep, this is correct.

Now for the second part

⌂KE = change in kinetic energy

⌂v = change in velocity = 2 m/s - 1 m/s = 1 m/s

m= 1kg

KE = ½m*v²

⌂KE = ½ (1kg)*(1m/s)²

⌂KE = 0.5 kg m²/s² = 0.5 J ≠ 1.5 J

However, this isn't. The problem is that you're messing up reference frames in the bolded step. You indeed increase the velocity by 1m/s. But you can't plug it into the formula like that due to reference frames. If you look at it from the outside you see a rocket that goes 1m/s and thus has 0.5J of kinetic energy. After the burn you see a rocket that moves 2m/s and thus has 2J of energy. So the energy difference you see as an outside observer is 1.5J. This is why reference frames are so important, someone stationary will measure 1.5J while someone on a train moving alongside the rocket only measures 0.5J.

This is in fact, the whole reason of the oberth effect. From the point of view of the rocket it takes just 0.5J to accelerate by 1m/s while it takes 1.5J from the PoV of the earth.

[K^2]

In short, ÃŽâ€KE ≠ ½m ÃŽâ€v². That's because ÃŽâ€KE = KE_f - KE_i = ½m (v_f² - v_i²), and (v_f² - v_i²) ≠ (v_f - v_i)².

In terms of an infinitesimal increment in velocity, dv, one can write dKE = mv dv. So as v increases, dKE per dv increases as well.

Edit: Fixed a typo in the derivative.

Edited February 10, 2014 by K^2

[Soda Popinski]

Ok. Makes sense now. Thanks.