Rodin's Rocket Problem

[rodin]

Calculus is a mathematical way of describing things which are not constant.

Now you might think a 100Kg moving rocket in deep space is not constant, but unless it is firing its engine, it moves with constant velocity. In fact, without measuring its velocity relative to something else, there is no way of telling if it is moving at all.

Now let's say it fires its engine for a tiny fraction of a second. It's velocity changes from x (unknown) to x +dx where dx is 1 mph. Meanwhile the rocket exhaust shoots off in the opposite direction at - 1000mph relative to the rocket's "frame of reference". Note the appearance of the word RELATIVE.

The "burn" is repeated. Because the rocket is moving the second burst exits at -999mph, and the rocket is now moving at 2mph relative to its starting velocity.

 

 I am ignoring small scale effects for now, such as the fact that after each burn the rocket is a teensy weensy bit lighter. I say I am justified in doing this, because THERE IS NO LOWER LIMIT to how LITTLE fuel I could have burned in each case. the ultimate case is where the amount trends to zero. This is EXACTLY the argument that is the foundation for calculus.

 

Now we will consider what has happened from a CONSERVATION OF ENERGY point of view.

In the first instance we convert a small amount of chemical energy into

1) Heat (wasted)
2) Thrust

At the end of each burst the resultant energy must be the same as went in, by conservation of energy. We will ignore heat for now, and look at the energy that was transferred into MOVEMENT.

After burning 1 UNIT of fuel, the spent chemicals and rocket are moving away from each other.

The rocket's kinetic (moving) energy is given by the formula e = 1/2mvr^2. The fuel's kinetic energy is given by the formula e = -1/2 mvf^2 (reaction is equal and opposite) 

where vr is the velocity of the rocket and vf is the velocity of the fuel

Now we burn the second unit of fuel.

The rocket's kinetic energy is given by the formula e = 1/2m(2vr)^2. The spent fuel's total kinetic energy is given by the formula e = -1/2mvf^2 - 1/2m(vf*0.999)^2

Check the Maths, and the arithmetic.

It appears that the second burst of chemical energy has resulted in

1) Approximately twice the amount of spent fuel kinetic energy
2) Approximately 4 times the amount of Rocket kinetic energy

This proves that either

1) Energy is not conserved and Newton's Law is wrong

OR

2) Kinetic energy does not equal energy

Discuss

Edited April 25, 2015 by rodin

[CraigD]

Your problem, Rodin, is due to incorrectly applying the principle of the conversation of energy

 

The rocket's kinetic (moving) energy is given by the formula e = 1/2mvr^2. The fuel's kinetic energy is given by the formula e = -1/2 mvf^2 (reaction is equal and opposite)

The kinetic energy of the fuel is not negative. In ordinary classical mechanics, kinetic energy is never negative.

 

I believe what you’re trying to do there is represent kinetic energy as a vector. A simple way to represent vectors is to limit your space to 1 dimension (a line), with quantities in one direction positive, in the other negative. But energy is a scalar, not a vector, so you can’t do this.

 

Momentum is a vector quantity, and like energy, is conserved, so you can use this simplifying technique to write

[math]m_r v + \sum_i m_e v_i =0[/math]

And solve (using simple algebra, steps not shown) it for your example, like this

t  m_r  v               m_e   v_1   v_2
0  999  0               1     0
1  999  1               1     -999
2  998  1995002/997002  1     -999  -997001/999

Notice here we’re avoiding the need to use calculus, by having the rocket exhaust happen in discrete, mass 1 pieces, rather than treating it as a continuous stream of infinitesimal particles. Though it’s impractical to do this for a realistic rocket exhaust, which consists of a huge number of particles of varying masses and velocities, it’s useful for an instructive example like this.

 

Also notice that it’s not necessary, and inaccurate, to ignore the change in the rocket’s mass, and that, given a constant exhaust speed, noe of the speeds after the first iteration are nice whole numbers, though they are nice, exact rational ones.

 

Energy is, or course, conserved, but that’s because the increase in kinetic energy and incidental energy, such as radiation, of the system is equal and opposite to the decrease in potential energy, the rocket’s energy supply, which could be from a chemical fuel, or something else.

 

Finally, keep in mind that all of the physics here is classical, not relativistic, so isn’t exactly accurate. So long as the speeds involved are much less than the speed of light, however, classical physics gives close enough approximations for practical purposes.

[rodin]

Yes of course you are correct, energy having the velocity squared term loses its sign. And momentum is conserved. But... it still seems to me that energy as defined by E = 1/2 mv^2 is not.

 

I think we can totally ignore relativistic effects at 2 mph!

 

I have described the rocket with 1:1000 velocity gearing between the rocket and exhaust, I could just as easily have used 1:1,000,000 or any ratio you like. Obviously the greater the ratio, the smaller the discrepancies between the burns, tending AT THE LIMIT, as we get smaller, for each rocket burn to become identical, and for the rocket mass to lose next to no weight. Furthermore, since this is a thought experiment for now, we can say that the engine is 100% efficient, and the magical chemical reaction produces no heat, only expansion of mass from solid to gas. in other words, the only output energy is mechanical. At this limit

 

1) Each chemical input provides same energy, since same mass is expanded

 

2) However, the system output energy is not the same in each case, being more second time around by about a factor of 2

Edited April 26, 2015 by rodin

[CraigD]

I have described the rocket with 1:1000 velocity gearing between the rocket and exhaust, I could just as easily have used 1:1,000,000 or any ratio you like. Obviously the greater the ratio, the smaller the discrepancies between the burns, tending AT THE LIMIT, as we get smaller, for each rocket burn to become identical, and for the rocket mass to lose next to no weight.

The appropriate equation for this thought experiment is the rocket equation,

[math]\Delta v = \frac{ \left( \frac{m_0}{m_1} \right)^{\frac{2 v_e}{c}} -1}{\left( \frac{m_0}{m_1} \right)^{\frac{2 v_e}{c}} +1} c[/math]

where [math]m_0[/math] is the mass of the rocket without propellant, [math]m_1[/math] its mass with propellant, [math]v_e[/math] is the exhaust speed, and [math]c[/math] is the speed of light.

 

From this, we can see that as the mass of the exhaust (propellant, or reaction mass) approaches the limit of zero, [math]\Delta v[/math] approaches zero. [math]\Delta v[/math] increases as [math]v_e[/math], which you describe as a ratio in miles/hour, increases, but can’t be increased to any value we like, because it must be less than [math]c[/math].

 

Notice that, since we’re considering all possible exhaust speeds, we used the relativistic version of the rocket equation, so the only obstacle to realism remaining is that we’ve not accounted for a reduction in the rocket’s mass due to the energy used to accelerate its propellant. We can circumvent this obstacle by assuming the rocket is powered externally, such as via a “beamed power” scheme.

[rodin]

I used the non relativistic equation

 

ΔV = ve ln m0/m1

 

http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation

 

ve = 1000 m/s

m0 = 1000 Kg

m1 = 999 Kg (after 1 burst) ln m0/m1 = 0.00100

m1 = 998 Kg (after 2 bursts) ln m0/m1 = 0.00200

 

Velocity of rocket after first burst is 1.00 m/s

Velocity of exhaust after first burst is between -1000 and -999 m/s average -999.5 m/s

Mass of rocket after first burst is 999 Kg

Mass of Exhaust after first burst is 1 Kg

 

Velocity of rocket after second burst is 2.00 m/s

Velocity of exhaust after second burst is between -1000 and -998 Km/h average -999 m/s

Mass of rocket after second burst is 998 Kg

Mass of Exhaust after first burst is 2 Kg

 

Total Kinetic Energy after first burst

 

Rocket KE = ½ mv^2 = 499.5 Joules

Exhaust KE = 499500.1 J

Total KE = 499999.5 J

 

Total Kinetic Energy after second burst

 

Rocket KE = ½ mv^2 = 1996 J

Exhaust KE = 998001 J

Total KE = 999997 J

 

Unfortunately it seems that burning double the fuel does result in double the total kinetic energy.

 

I knew the KE equations worked for gravity etc. I guess I just did not visualise the immensity of the KE of the rocket exhaust.

 

Thanks for bearing with me.