How Does Power Affect Force in Space Travel?

[jack action]

Apparently you're not understanding it either, because as the rocket goes faster, the exhaust velocity with respect to the rocket does NOT go down. And that's the whole point of this discussion and the whole reason for all the confusion.

It's not something that makes sense unless you put some time to think about it, and/ or start a thread to get some help figuring it out.

I shouldn't have brought back the rocket. I don't know how to tell this simpler than it is:

P = Fv
F = ma
a = dv/dt

3 possible cases:

1- Object at constant velocity:

dv = 0; a = 0; F = 0 ---> P = 0

2- Object at constant acceleration:

F = constant; P/v = constant ----> P is proportional to v

3- Object at constant power (which I think is the subject under discussion):

Fv = constant ----> F is inversely proportional to v. Which leads to a is inversely proportional to v

If you have a rocket under constant power - same amount of energy spent on the rocket's motion in a given time - the thrust will be smaller as the speed of the rocket goes up, whether you decrease the velocity of the exhaust or its mass flow rate: something has to give.

Why is that? Because for the rocket that goes faster, the displacement will be greater in the same time dt. And since we assume that the energy spent will be the same and that energy = force times displacement, therefore the force must go down for the faster rocket to respect the initial assumption that power is constant.

And that is true for any type of object going through any type of motion.

Right now, you all have a problem similar to this https://www.physicsforums.com/showthread.php?t=408584", except that the person couldn't accept the first case (a constant v means zero power). And just like I said in that thread, I will give up at this point, as you are not even trying.

 

[Lsos]

Yeah, something does have to give. What happens is that, to an observer on the earth, the rocket's power IS increasing.

However, the paradox, or source of confusion, is that despite the power increasing the amount of propeller burned per second is NOT increasing. If you ask the pilot of the rocket, he'll tell you that the throttle is at the same position, the pumps are pumping the same amount of fuel, and the rocket is experiencing the same thrust...and yet, to the guy on the earth, the rocket is going faster and faster.

To the guy on the earth, the power is increasing. To the guy on the rocket, the power is the same.

The reason for this is, of course, because we're talking about a rocket. By taking it's propellant with it, it's essentially taking it's frame of reference with it. As such, we don't care that it's going 100000mph away from the earth. The propellant doesn't know this. It doesn't know how fast it's moving in relation to the earth, the sun, the galaxy, light, or whatever. It just knows that it's sitting still on a rocket, and when it is ignited it will go out at a given velocity, regardless of how fast other objects in the universe are moving. It will be expelled out the back at the same relative speed to the rocket, and will continue to provide the same thrust.

It seems pretty obvious that this paradox is why the OP started this thread.

 

[rcgldr]

If you have a rocket under constant power - same amount of energy spent on the rocket's motion in a given time.

For constant power the energy spent and forces involved on the rocket and the spent fuel remain constant. The initial condition could be rocket in space with zero speed, and all of the energy goes into the spent fuel exhaust.

Yeah, something does have to give. What happens is that, to an observer on the earth, the rocket's power is increasing.

Not if that observer also takes into account the energy put into the spent fuel exhaust. The total energy gain of rocket and spent fuel per unit time will be constant for a constant power (throttle) setting. Fuel will be consumed and chemical potential energy converted into mechanical energy and heat at the same rate, regardless of the rockets speed relative to any inertial (non-accelerating) frame of reference.

Imagine a rocket and it's unspent fuel in outer space away from any gravitational effects. It's a closed system with no external forces, the center of mass of rocket and fuel will not move, and the momentum of the rocket and it's fuel will remain contant (zero if the frame of reference is the rockets initial velocity).

 

[Lsos]

Not if that observer also takes into account the energy put into the spent fuel exhaust.

Ah, true. So to an Earth observer the total power remains the same, but as the rocket goes faster, more of this power goes into the rocket rather than into the exhaust. Is this the right way to think of it?

 

[K^2]

No, the power that pushes the rocket traveling at considerable velocity may greatly exceed the power of the engine.

The point is that in the beginning, the engine works to accelerate both the rocket and the propellant it is carrying. Only the energy used to accelerate the rocket is useful at this point. The rest is stored in the propellant. Once the rocket builds up to considerable speed, that energy stored in the propellant helps you push the rocket.

 

[rcgldr]

No, the power that pushes the rocket traveling at considerable velocity may greatly exceed the power of the engine.

The point is that in the beginning, the engine works to accelerate both the rocket and the propellant it is carrying. Only the energy used to accelerate the rocket is useful at this point. The rest is stored in the propellant. Once the rocket builds up to considerable speed, that energy stored in the propellant helps you push the rocket.

True, but the total energy includes the acceleration of the fuel from within the rocket to out of the exhaust nozzle of the rocket engine. If you take the energy change of the spent fuel into account, then the total rate of increase in energy of rocket and fuel (both spent and expelled) are constant with a constant amount of power.

As mentioned before the power is related to Force x (V_rocket - V_exhaust), and (V_rocket - V_exhaust) will be a constant regardless of the rockets speed, for any given throttle (and power output) setting.

 

[K^2]

Yes. That's right. So it's a matter of coordinate system choice. If you choose rocket-bound system, then the exhaust always has constant kinetic energy, supplied by constant power of the engine. This works the same way regardless of whether it's the chemical rocket or ion driven one. In this frame, the rocket is always still, and so requires no power to accelerate. 100% of the power goes to accelerate propellant.

In any inertial frame, however, the rocket is going to have velocity over any finite time interval. So the power requirement to apply constant force to the rocket is always changing. With the engine still supplying constant power, it's not something you can explain without considering energy stored in propellant.

 

[rcgldr]

In any inertial frame, however, the rocket is going to have velocity over any finite time interval. So the power requirement to apply constant force to the rocket is always changing. With the engine still supplying constant power, it's not something you can explain without considering energy stored in propellent.

Just consider propellent to be part of the rocket until it's expelled. The point that seems to be missed here is that it is a constant power situation, regardless of the inertial frame, if you take into account that power is used to accelerate the expelled propellent, as well as the rocket. You can consider the another key frame, rocket moving at high speed, propellent expelled at zero speed, the power rate is the same as rocket not moving, and only prorpellent moving.

update -

force times distance versus speed

When the rocket is moving faster then (force_rocket x distance_rocket / unit time) is greater, but it's offset by the now slower moving expelled fuel (force_fuel x distance_fuel / unit time). The point here is that force x ( (distance_rocket + distance_fuel) / unit time) is constant (assuming constant power).

 

[K^2]

The rocket can be moving 10x faster than propellant exhaust speed. What do you say then?

It really isn't as simple as you seem to think.

 

[norrrbit]

Hello, thanks all for engagment.

Gerenuk wrote: "You cannot look up a random equation from the formula book and plug in some value for some letter."
That's why I'm asking the question: to understand the meaning of the letters.

"What about a helicopter in air? How much power does it need? " - Gerenuk this is the essence of my question, (In fact I bumped into the problem when wondering about personal, electrical VTOL). To avoid rocket propellant side effects let's reformulate the problem:

An electrical helicopter is hovering in Earth's gravitational field: it's mass doesn't change as it is powered by batteries and propellant is taken from the surrounding. The propulsion system has some constant efficiency (electric motor + rotor blades' efficency is perhaps 60%) - so power of the propulsion system is constant, ok?
But hovering counteracts constant force - gravitational field! And that is why I can't get rid of intuition that power is linearly related to force, Gerenuk!
If God or Mr/Ms Nature (whoever you prefer) would instantly turn off the gravity the helicopter would start to accelarate at constant g (as long as atmospheric pressure wouldn't change), though it is driven by constant power.


K^2 wrote:"Besides, the helicopter question can be as complicated as you want to make it, down to considering effects of vibrations of the blades on overall efficiency."

K^2elicopter case is not complicating things, it is simplifying as it preserves mass of the system. Anyway I can modify my spaceship (not rocket) question: what if it would be propelled by light pressure of external laser shining with constant light (=constant power)? Would it accelarate at constant or decreasing pace? (Of course I assume idealistically void space , so no efficency loss of laser light with distance.)

 

[K^2]

I gave you derivation for the helicopter. Read it. Understand it.

If you use light pressure to accelerate a craft, the force on the craft will be dropping. You can look at it from either of two perspective.

1) Power is constant, but force is applied to faster and faster traveling craft. So force must decrease.

2) The light source will get red-shifted as craft accelerates.

Either way, it's not a constant force.

 

[norrrbit]

K^2, sorry but in a hurry I didn't notice there are more pages of this thread. I think I am getting the point.

 

[rcgldr]

The rocket can be moving 10x faster than propellant exhaust speed. What do you say then?

If you do the math the result will be the same, if there's a constant force (note this is an expansive force, pushing rocket foward, pushing spent fuel backwards), then power is constant regardless of the rockets speed. I'll try to find the old thread that went through all the old math.

If the rocket is going 10x speed_exhaust, then the exhaust is going 9x speed_exhaust. The difference between rocket speed and exhaust speed will always remain 1x speed_exhaust.

I found DH's thread on rockets:

https://www.physicsforums.com/showthread.php?t=199087

 

[K^2]

Yeah. I actually derived these formulae earlier in this thread.

So why are you arguing with me? The engine power remains constant. Total power pushing the rocket depends on coordinate system. Any difference is made up by energy stored in the propellant. This is what I stated from the very beginning. Third post in the thread, actually.

 

[rcgldr]

So why are you arguing with me? The engine power remains constant. Total power pushing the rocket depends on coordinate system.

OK, the diference here is that I consider the total power to include both pushing the rocket forward and pushing the expelled propellant backwards, in which case the total power involving the rocket and the exhaust cloud is constant.

 

[norrrbit]

K^2, I went through your helicopter formulas - as an ignoramus in physics I am impressed, though not totally convinced.
It seems from your <br /> P = \frac{1}{2} \frac{(Mg)^{\frac{3}{2}}}{\sqrt{A\rho}}<br /> that by simply increasing disc area A one can easily decrease required power (increased disc mass is negligible - it raises with root and that is small portion of total mass). Do I get it right? At 1.204 air density the power for 100kg craft with disc diameter of 3 meters is just 0.03kW. That seems suspicious...

 

[K^2]

OK, the diference here is that I consider the total power to include both pushing the rocket forward and pushing the expelled propellant backwards, in which case the total power involving the rocket and the exhaust cloud is constant.

So you want to view the engine as doing negative work on the propellant? Technically true, sure, but... Alright.

I don't think it would help OP a whole lot with his confusion, though.

Do I get it right? At 1.204 air density the power for 100kg craft with disc diameter of 3 meters is just 0.03kW.

I'm getting over 5kW, which is a lot more reasonable. Check your numbers.

It goes to zero in the limit. That seems suspicious...

If it was resting on the ground, it'd take zero power. It doesn't take any energy to support something. It's only all the air you're moving. And with infinite area, you don't need blades. You can just make it a parachute, and it will be enough.

 

[norrrbit]

If you use light pressure to accelerate a craft, the force on the craft will be dropping. You can look at it from either of two perspective.

1) Power is constant, but force is applied to faster and faster traveling craft. So force must decrease.

2) The light source will get red-shifted as craft accelerates.

Either way, it's not a constant force.

Ok, last defense line of my wicked intuition:
what if a saucer is kept hovering in a gravitational field by an external, constant power laser from the ground. We have both const. power & force. How would you calculate power, given the force?


After reading all the posts, I am leaning toward a conclusion that the point is in defining the POWER. As K^2 explained:"If you read OP's post carefully, you'll notice that his confusion arises from the fact that constant power applied to propellant gives you constant thrust, while constant thrust applied to accelerating rocket requires increasing power." There is power applied to propellant and there is power of the whole system, right?

 

[norrrbit]

If you use light pressure to accelerate a craft, the force on the craft will be dropping. You can look at it from either of two perspective.

1) Power is constant, but force is applied to faster and faster traveling craft. So force must decrease.

2) The light source will get red-shifted as craft accelerates.

Either way, it's not a constant force.

Ok, last defense line of my wicked intuition:
what if a saucer is kept hovering in a gravitational field by an external, constant power laser from the ground. We have both const. power & force. How would you calculate power, given the force?


After reading all the posts, I am leaning toward a conclusion that the point is in defining the POWER. As K^2 explained:"If you read OP's post carefully, you'll notice that his confusion arises from the fact that constant power applied to propellant gives you constant thrust, while constant thrust applied to accelerating rocket requires increasing power." There is power applied to propellant and there is power of the whole system, right?

 

[K^2]

Ok, last defense line of my wicked intuition:
what if a saucer is kept hovering in a gravitational field by an external, constant power laser from the ground. We have both const. power & force. How would you calculate power, given the force?

Assuming a reflector at the bottom of the saucer, the force F= 2*dp/dt. And p for photon is E/c. So P = dE/dt = cdp/dt = (1/2)Fc.

Which is A LOT. So I would not recommend suspending saucers with laser beams. Use a jet, or something.

 

[norrrbit]

I'm getting over 5kW, which is a lot more reasonable. Check your numbers.

Ooops, you're right - that makes 5,3 kW.

If it was resting on the ground, it'd take zero power.

Obviously, that's why I have cut the text out, but you were faster :)

It doesn't take any energy to support something. It's only all the air you're moving. And with infinite area, you don't need blades. You can just make it a parachute, and it will be enough.

Then, what constrains engineers from making much bigger rotors, as it'd reduce power, i.e. energy consumption for hover flight - blades' mass and strength? New lighter materials will make rotors bigger and hover flight more economical?

 

[norrrbit]

Assuming a reflector at the bottom of the saucer, the force F= 2*dp/dt. And p for photon is E/c. So P = dE/dt = cdp/dt = (1/2)Fc.

Ok. So I understand (after small tutorial from wikipedia) that p is photons momentum and P=.5*c*mass*g. Thus finally I've got confirmed that Power is linearly dependent on Force... ;-). Just joking. I understand that it holds only for this specific experiment.

 

[norrrbit]

The thing that I learned most from this thread is that it seems one can't talk about power, force, etc. generally but only in specified context (as Gerenuk underlined).
Thank you K^2 and all you people for your effort. You've helped me a lot!

 

[Nagakumari]

Hi,
Can anyone tell me where to find solutions for the problems in Mechanics of Materials (Gere and Timoshenko) book.
That would be greatly appreciated.
Thank you.
Nagakumari

 

[rcgldr]

So you want to view the engine as doing negative work on the propellant?

It's only negative work (decrease in kinetic energy) if rocket speed is > .5 V_e (exit velocity relative to rocket). It's zero work on propellent if rocket speed = .5 V_e (just a sign change of exhaust velocity), and positive work (increase in kinetic energy) on propellent if rocket speed < .5 V_e.

In the case of the hovering object, no work is done on the object. In the case of a helicopter, work is done on the air, and the amount of work done depends on rotor efficiency. Rotor length is a big contributor to efficiency, but there are practical limits.

If you had a slug that snugly inside a friction free cylinder with an open top, then the slug would descend until the pressure of the trapped air supported the slug. Once a stable state is achieved, no work is done.

 

[K^2]

Then, what constrains engineers from making much bigger rotors, as it'd reduce power, i.e. energy consumption for hover flight - blades' mass and strength? New lighter materials will make rotors bigger and hover flight more economical?

Yes, it's mostly strength and added weight. And yes, with new materials, all sorts of aircraft have been becoming more efficient. Look at the modern sail planes.

It's only negative work (decrease in kinetic energy) if rocket speed is > .5 Ve

Well, yeah. Naturally. Alright, that works. But I still wouldn't call it the most intuitive explanation.

 

[Anamitra]

We turn on the engines which work with constant power: P - i.e. energy intake from the batteries is constant through time. To my intuition the ship should be now accelerating with constant accelaration: A. Then force acting on the ship should also be constant: F=ship_mass*A.

It would be convenient to use the formula:

F=ship_mass*A + (d/dt of ship_mass)*u
F=mA+(dm/dt)*u, [m is the mass of the spacecraft ]
P/u =mA+(dm/dt)*u ; [since P=F*u]
P= muA +(dm/dt)*u^2
[dm/dt is a negative quantity and we take it to be constant]
If P is constant(as assumed by norrbit)

We have,
dP/dt =0

Therefore,
m{(du/dt)*A+u*(dA/dt)}+uA*(dm/dt)+2u du/dt *dm/dt=0
[Noting that dm/dt is a constant ,we write dm/dt=-k;k>0]
mA^2 + mu (dA/dt)-uAk -2uAk=0
If A is constant as assumed by norrbit,

We have,
mA^2=3uAk
mA=3uk
A=3uk/m

Now m decreases with time and hence u has to decrease if A is to be a constant[and if k is constant]. Then again if u decreases A cannot be a constant
[if k is assumed to decrease we have to repeat the differentiation right from the beginning( since dm/dt= constant has simplified the differentiation that I have done). We cannot apply the simple formula A=3uk/m in this case]]
Therefore A cannot be a constant in this case if power is taken to be a constant[and if k is constant].

 

[rcgldr]

It's only negative work (decrease in kinetic energy) if rocket speed is > .5 V_e

Well, yeah. Naturally. Alright, that works. But I still wouldn't call it the most intuitive explanation.

Still switch this case to a rocket "hovering" above the moon (no atmosphere), the rocket isn't moving, so no work done on the rocket, only on the propellent expelled from the rocket. This is a similar case to the hovering helicopter (the heli doesn't move, only the air). I think it's a reasonable comparason.

 

[norrrbit]

After 4 pages of posts, I will try to recapitulate what I understood. I will use as simple and widely understood laws of physics as possible.

1. spaceship is propelled with constant drive power: P. That's because
2. it uses constant mass of propellant per second: m_p
3. which has final velocity of v_p (which I measure relatively to the ship, this velocity is constant & specific for propellant and engine construction)
4. thus propellant with m_p accelarates from 0 velocity to v_p in 1 sec which produces constant force F_p=m_p*v_p
5. F_p is counter-balanced (according to Newton's III law) by thrust F_s
6. which equals F_s=m_s*a_s (ship's mass*ship's accelaration)
7. thus F_p=F_s => m_p*v_p=m_s*a_s
8. Power is kinetic energy of a propellant's mass flow in a second: P=1/2*m_p*v_p^2
9. from 7&8 we have: P=1/2*m_s*a_s*v_p
10. because we use up the propellant, m_s drops and a_s increasesConclusions:
1. With constant power drive spaceship will accelerate faster
2. With constant power drive, thrust is also constant P=1/2*Thrust*v_p

So I was wrong in neglecting changes in ship's mass (and expecting constant acceleration of the ship - but that's why I used ionic drive - propellant's mass is realtively less important as it is more propellant-efficient), but my intuition about relation between power & thrust was not that absurd!Please correct me if I am wrong.

 

[rcgldr]

2. it uses constant mass of propellant per second: m_p

That should be "m-dot": ṁ_p, the mass flow per unit time (not mass flow per second, not important here, but the time it takes to reach terminal velocity is not needed to be known).

4. thus propellant with m_p accelarates from 0 velocity to v_p in 1 sec which produces constant force F_p=m_p*v_p

v_p is terminal velocity, and the time it takes to reach terminal velocity is not stated. The mass flow rate takes care of the time factor, so the thrust F = ṁ_p v_p.

8. Power is kinetic energy of a propellant's mass flow in a second: P=1/2*mp*vp^2

Again replace m_p with "m-dot": P=1/2 ṁ_p v_p² = 1/2 F v_p