Maximum kinetic energy of a rocket in free space.

In summary, the student is returning to grad school for physics this fall, but has been out of the classroom for thirteen years. They are currently working their way through practice problems for the placement exam, but they do not have any way to verify correct answers for the problems without finding something similar online. To be fair, the student usually does find a similar problem online, but this one surprisingly they cannot find a direct analog for. The homework statement is that a rocket starts from rest in free space by emitting mass at a constant rate and with ejection velocity relative to the rocket. At what fraction of the original rocket mass is the kinetic energy of the rocket a maximum. Using conservation of momentum, reach the basic rocket equation: M dv
  • #1
Pianodan
6
0
I'm returning to grad school for physics this fall, but I've been out of the classroom for thirteen years. (teaching music, of all things) I'm working my way through practice problems for the placement exam, but I don't have any way to verify correct answers for the problems unless I find something similar online. To be fair, I usually do. But this one, surprisingly, I can't find a direct analog for.

Homework Statement



A rocket starts from rest in free space by emitting mass at a constant rate [itex]\alpha[/itex] and with ejection velocity u relative to the rocket. At what fraction of the original rocket mass is the kinetic energy of the rocket a maximum. (Assume the mass of the empty rocket to be negligible)

Homework Equations



Just kinetic energy and momentum, as far as I can tell. Forgive me, I'm not great with LaTex yet, so I'm trying to minimize typing it here.

The Attempt at a Solution



Using conservation of momentum, reach the basic rocket equation:

M dv = u dm

where M is the inital mass of the rocket. Separate variables and integrate:

v(t) = u ln (M / m(t))

where m(t) is the mass of the rocket at time (t)

The Kinetic Energy, T(t), of the rocket is:

1/2 (m(t) ) v(t)^2

Substitute m(t) = M - [itex]\alpha[/itex]t and the value of v(t) from above to get:

T = M u^2 ln (M/m(t)) - [itex]\alpha[/itex] t u^2 ln (M/m(t))

The kinetic energy should be at a maximum when the time derivative of the KE is zero, so differentiate (and this is the most likely place I screwed up) the above equation with respect to time to get:

[itex]\frac{dT}{dt}=ln(\frac{M}{m(t)})[/itex]

Setting that equal to zero, you find only one extremum in T at m(t) = M, or at the point of total fuel consumption. Given that the KE of a massless rocket is zero, I'm at a loss.

Just thinking about this in the abstract, it seems that the velocity over the entire period of thrust, the mass should decrease, and the KE, as the product of the mass and the square of the velocity (and 1/2), should increase to a maximum and then decrease to zero as all the mass moves to the ejecta.

What am I missing here?
 
Physics news on Phys.org
  • #2
I think that one of the mistakes you made is when you substitute the expression for v(t):
you should have [itex](ln (m/M))^2[/itex] instead of [itex]ln(m^2/M^2)=2ln(m/M)[/itex]
 
  • #3
Welcome to PF!

You may also want to find maximum kinetic energy using kinetic energy expressed as function of mass, Tm, since dTm/dm = 0 is a bit easier to solve by hand than dTt/dt = 0. Once you have a mass for maximum T you can easily insert that in your mass-time equation to find the corresponding time.
 
  • #4
Thanks to both of you for the help. It's clear that my facility with logarithms has eroded greatly. After making the correct substitution into the time differential, I discovered that it was indeed just as nasty as Filip had suggested. The mass differential was simpler, but still a bit of a chain rule slog.

T = 1/2 m(t) v^2 = 1/2 m(t) u (ln (M/m(t)) )^2

dT/dm = 1/2 u [ ln^2(M/m) - 2 ln (M/m) ]

Setting this equal to 0 yields the relatively simple looking:

ln(z)^2 = - ln^2(z) where z = (M/m(t))

Plugged it into Alpha and got a solution right away, but it took me a good half an hour of head scratching to understand it. I just about kicked myself when I got it - but that's always the way, isn't it? I hope my speed increases with practice - this was one of twelve questions on a three hour test.

Here's the final answer.

ln^2(z) + ln(z) + ln(z) = 0
ln(z)(ln(z) + 2) = 0

So there are two extrema - one at z = 1, and the other at z = e^-2

Looking at this, I think I lost a sign somewhere, since it implies that m / M at max KE is greater than one (e^2).
 
  • #5
I concur with your 2nd equation, but from that you should get

ln2(z) - 2 ln(z) = ln(z) (ln(z)-2) = 0

which implies that z = e2 and thus m = M/e2 is a solution for dT/dm = 0. I guess you are correct about loosing a sign along the way :smile:.
 
  • #6
Oh... right. (bangs head on desk)
 

Related to Maximum kinetic energy of a rocket in free space.

1. What is maximum kinetic energy of a rocket in free space?

The maximum kinetic energy of a rocket in free space is the maximum amount of energy that the rocket can possess while in motion. This energy is due to the rocket's mass and velocity.

2. How is the maximum kinetic energy of a rocket calculated?

The maximum kinetic energy of a rocket can be calculated using the equation KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass of the rocket, and v is the velocity of the rocket.

3. What factors affect the maximum kinetic energy of a rocket in free space?

The maximum kinetic energy of a rocket is affected by the mass of the rocket, the velocity of the rocket, and the direction of the rocket's motion.

4. Why is maximum kinetic energy important in rocket launches?

The maximum kinetic energy of a rocket is important in rocket launches because it determines the amount of thrust needed to overcome the force of gravity and propel the rocket into space. It also impacts the rocket's trajectory and the amount of fuel needed for the journey.

5. Can the maximum kinetic energy of a rocket be increased?

Yes, the maximum kinetic energy of a rocket can be increased by increasing its mass or velocity. However, this requires more fuel and can also have an impact on the rocket's stability and control during the launch.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
372
  • Introductory Physics Homework Help
2
Replies
42
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
919
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
30
Views
2K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
288
  • Introductory Physics Homework Help
Replies
19
Views
773
Back
Top