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Rocket Thrust in Vacuum

  1. Sep 2, 2013 #1
    Hello forum members and thank you for letting me post.

    My question refers to how rocket thrust works in a vacuum. I know this question has been posted before, but I am still confused about how it works.

    My understanding is that rockets do not move by pushing against air but move by pushing "against their own fuel," as is stated in Newton's Third Law according to which "for every action there is an equal and opposite reaction."

    The oft repeated analogy to explain how this works is throwing a ball off the stern of a small boat and the resulting forward movement of the boat, or the recoil energy resulting from firing a gun.

    Ok. So a rocket moves by recoil energy? Is this correct?

    Here is an analogy of the recoil generated by a a gun, in this case an M16, compared to the mass and force involved in a rocket, in this case a French rocket Ariane 5. (The analogy is not mine.)

    I would appreciate your comments.

    Weight of M16 RIFLE: 4kg https://en.wikipedia.org/wiki/M16_rifle

    Weight of an M16 bullet: 3.6g http://www.defencetalk.com/forums/army-security-forces/m16-vs-ak-47-a-4686-7/

    Weight of an Ariane 5 rocket: 760,000Kg http://en.wikipedia.org/wiki/Ariane_5

    Recoil Force involved in firing M16 bullet: "A bullet fired from an M16 rifle has approx 1763 Joules of kinetic energy as it leaves the muzzle, but the recoil energy exerted on the gun is less than 7 Joules." http://en.wikipedia.org/wiki/Recoil

    The kinetic energy exerted on the rifle is thus equal to 250 times less than the energy exerted on the bullet .

    If we multiply the weight of the M16 by 190,000, we get the weight of an Ariane 5 (760,000Kg).

    Continuing the analogy, if we multiply the weight of the M16 bullet by 190,000, we get 684Kg (3.6g x190,000).

    If the giant M16 were ejecting three giant bullets a seconds (2,054Kg), then this would approximately equal the fuel mass ejected by Ariane 5 every second (2,000Kg).

    Multiplying (by the 190.000 factor) the known kinetic energy exerted on an M16 rifle by a bullet exiting its muzzle (about 7 joules) we get (7X190.000= 1.330.000 joules).

    The giant M16 rocket will therefore produce a kinetic energy thrust equal to 7 x 190,000 x 3 = 3,990,000 joules, or approximately 4 million joules.

    Four million joules is approximately equal to 5,400 Horsepower.

    How can 5,400 Horsepower be sufficient to lift off a 760,000Kg rocket into Low Earth Orbit?
     
  2. jcsd
  3. Sep 2, 2013 #2
    Your statement that 4 million joules is equal to 5400 horsepower is incorrect. The units are wrong. horsepower can be converted to joules per second. The ariane rocket has a continuous flow of mass out its rockets, and the rate of mass outflow must be much larger than you assumed.
     
  4. Sep 2, 2013 #3
    Your calculations don't seem to be correct.

    Ariane 5 first stage is a powerful engine where just the fuel turbopump has 16 thousand horsepower (12 MW according to Wikipedia). That is just a pump and it already exceeds your calculation 3 times. And don't forget that Ariane 5 is using not only the first stage engine, but also the boosters which are even more powerful.
     
  5. Sep 2, 2013 #4
    Hi Chestermiller,

    In what way is the conversion from Joules to Horsepower incorrect? In what way are the units wrong? Could you please clarify any errors I have made? Thank you.

    You are right! The outflow of an Ariane rocket is 2,000Kg PER booster. Since the French fusée has two EAP (Étages d'Accélération à Poudre) boosters, the total fuel mass ejected will be double, which brings us to about 8 million Joules.

    According to Wikipedia the total thrust of an Ariane 5 at lift-off is powered 90% by the two lateral boosters, so the initial thrust produced by the main cryogenic stage is negligible compared to that produced by the two EAPs.

    Hi mpv_plate,

    The given explanation for how a rocket moves in space is that the rocket "pushes against its own fuel" just as when the firing of a gun produces a "recoil" force. This "recoil" force, as explained, is what produces the rocket "thrust" force. However using the analogy of recoil energy produced by a bullet exiting a gun, applied to the thrust of a rocket, appears to demonstrate this cannot possibly create the force required.
     
    Last edited: Sep 2, 2013
  6. Sep 2, 2013 #5
    According to Wikipedia: "the megajoule (MJ) is equal to one million joules, or approximately the kinetic energy of a one-ton vehicle moving at 160 km/h (100 mph)."

    http://en.wikipedia.org/wiki/Joule#Megajoule

    How can an 760,000Kg Ariane 5 rocket lift off its own weight and free itself from gravity into Low Earth Orbit using the thrust force equal to eight or nine one-ton vehicles moving at 160 Km/h (100mph)?
     
  7. Sep 2, 2013 #6

    Nugatory

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    Joules are a unit of energy and horsepower is a unit of power. Because energy and power are different things (not just different units for the same thing, like meters and miles) you can't just convert the one into the other.

    Energy and power are related to each other in the same way that distance and speed are related to one one another: Power is the amount of energy delivered in a given amount of time. A one-horsepower engine running for a day produces the same number of Joules as an 83,000 horsepower motor that runs for one second.
     
  8. Sep 2, 2013 #7

    A.T.

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    Joules measure energy. Horsepower measures power. Did you mean Joules/second (Watt)?

    You seem to be confusing kinetic energy and force here.
     
  9. Sep 2, 2013 #8
    Hi Nugatory,

    Thank you for this brilliant and simple explanation!

    CB
     
  10. Sep 2, 2013 #9
    Hi A.T.,

    Thank you for pointing my error. Is the sentence above, in which I replaced the terms "thrust force" with "kinetic energy," correct?
     
  11. Sep 2, 2013 #10

    Nugatory

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    The gun analogy is just that: an analogy to show that it's reasonable to expect that if you push something hard out the back of the rocket, the rocket will move forward. It's not, as you've just discovered, a good starting point for actual calculations that will give you quantitative, or even ballpark, numbers for how much the force will be.

    If you really want to do this right, you'll google for "Tsiolkovsky rocket equation" or "ideal rocket equation" - and be prepared to spend a few quality hours on the derivation.

    If you just want to get a quick sense of how the physics works here, you can try using the conservation of momentum: If you push something of mass ##m## off the back of the rocket with speed ##v##, its momentum will be ##mv##; and the momentum of the rocket in the forward direction will increase by the same amount so the rocket gains some forward speed. The mass of the rocket is likely larger than the mass thrown off the back, so the speed increase of the rocket is smaller to keep ##m_rv_r## equal to ##mv## - but the speed increase is still there, and it's cumulative so each push increases the speed a bit more.
     
  12. Sep 2, 2013 #11

    jim hardy

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    How I wish I were good with graphics.

    Here is a simple thought experiment that will make it intuitive.....

    Imagine a very un-streamlined rocket out in space. With no atmosphere there it need not be streamlined, does it?
    So in your mind make it a perfectly sealed and rigid cube one foot by one foot by one foot. (I use English units because i'm both old and in USA, use metric if you like and make it a 1 meter cube for same logic will apply).

    Now internally pressurize your rocket to 1 PSI.
    What are forces on each of the six sides? Clearly 144 pounds pushing outward on each side.
    Since the cube is rigid, the forces on opposite sides cancel out so there is no net force on the rocket. Up cancels down, left cancels right and forward cancels backward.

    Now open a 1 square inch hole(or valve) on any face - i'll pick the bottom.
    This is a simple thought experiment so we'll ignore refinements that would be dictated by proper fluid mechanics - entrance losses and vena contracta and all that.

    Bottom face of your cube is now only 143 square inches, but top face is still 144.
    So forces are no longer balanced. 143 pounds push down against bottom, but 144 still push up against top.
    So rocket will accelerate up.

    So - a rocket in a vacuum accelerates not because of propellant pushing against air, but because of propellant NOT pushing against anything!

    Last statement was shocking enough that I always remember this oversimplification.
    Rocket scientists add converging-diverging nozzles to refine the fluid mechanics.

    any help?


    Movie "October Sky" is a fun story about some youths interested in rocketry. It has interesting pictures of amateur rockets.


    old jim
     
    Last edited: Sep 2, 2013
  13. Sep 2, 2013 #12

    A.T.

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  14. Sep 2, 2013 #13
    M(rocket) x V(rocket) = M(Fuel) x V(fuel)

    I am assuming we are using meters, Kg, and Newtons?

    Let me recalculate the recoil involved in firing an M16:

    Weight of M16 = 4Kg; Weight of M16 bullet = 0.004Kg; speed of M16 bullet = 950m/s

    4Kg x V(M16 rifle) = 0.004Kg x 950m/s

    So using this formula the recoil, or the velocity of the M16 going in the opposite direction from the bullet is 0.95m/s (v = [(0.004Kgx950m/s)/4]

    The bullet goes 950m/s, the M16 rifle "goes" 1m/s in the opposite direction. Am I correct?

    Now for Ariane 5:

    Weight of Ariane 5= 760,000Kg; Fuel mass ejected = 3,925Kg/s; Average exhaust velocity (all 3 engines) =2,415.42m/s *

    If I plug the numbers in the formula:

    760,000Kg x V(rocket) = 4,000Kg x 2,400m/s

    V(rocket) = 12.6m/s

    The Ariane 5 is moving at 13m/s ? Is this correct?

    However:

    1. How do we compute the effects of gravity pulling down on the slowly rising rocket in this formula? (In opposition to the M16 example in which the bullet was fired parallel to the ground.); What about the slowing down of the rocket due to air drag?; and

    2. How does the speed increase from this initial velocity? You wrote "the speed is cumulative." How is it so if exhaust velocity is maximal at initial take-off (since that is when the rocket needs the most thrust)? How can the velocity of the rocket increase if all it has to accelerate is to "push on itself"?


    * Data from: https://campus.tum.de/tumonline/LV_TX.wbDisplayTerminDoc?pTerminDocNr=7357
     
    Last edited: Sep 2, 2013
  15. Sep 2, 2013 #14

    Nugatory

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    No. That's the amount by which the speed increases in the first second. In the each subsequent second, the engines are still working, they push another chunk of exhaust backwards and the momentum of the rocket increases again - and the speed increases by more than 13 m/sec because the rocket is lighter by the amount of fuel already exhausted. Give that Tsiolkovsky equation a try... Also the answer to your question 2 below.

    Gravity exerts a downwards acceleration of 9.8 m per second per second, and you've just shown that the rocket motor are good for 13 meters per second per second upwards (and it gets better as the rocket burns fuel and becomes lighter) so the motors win.

    The exhaust velocity is roughly constant relative to the rocket. If the rocket is moving at 3 km/sec relative to earth, and the exhaust velocity is 2.4 km/sec (your number) then the exhaust is moving at .6 km/sec relative to the earth, but the rocket still gets the full 2.4 km/sec of kick. To see why those should be so, imagine yourself sitting on the rocket watching the exhaust stream out of the combustion chamber through the nozzles - we're still throwing it out the stern at 2.4 km/sec, the location of the earth is irrelevant. This, every increment of mass that we push out the back with a speed v increases the momentum of the rocket by mv, no matter how fast the rocket is already moving.
     
  16. May 26, 2016 #15
    This is only valid if the AK47 doesn't move at the start. If you took one in a car and fired it backwards, the situation would be much better already. If the gun/rocket moves with the same speed as the bullets/exhaust, all of the energy will go to the rocket and the exhaust ends up stationary with no kinetic energy.
    If the rocket speed is higher than the exhaust speed, the rocket will get more than the energy from the combustion, wich gets taken from the initial energy of the exhaust.
     
  17. May 26, 2016 #16

    jim hardy

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    Here's an easier mental model. Sum of forces = MA .

    Take an unstreamlined rocket.
    Since you stipulated vacuum it needn't be sleek.
    So, Let's make it a cube.

    Rocket_Unsleek.jpg

    It's not where it pushes,
    it's where it DOESN't push that counts.

    Now any mechanical engineer will tell you that a converging-diverging nozzle will give more flow
    so we would add a nozzle to our square edged orifice in that right hand face.

    Any help ?


    old jim
     
  18. May 26, 2016 #17
    Yes but what is giving it the fright? :) Sorry, had to throw that one in....A frightful reply will no doubt be coming:)

    And I don't even KNOW how to pronounce Fleft:)

    Anyway, why can't you just say the rocket thrusts against it's own mass?
     
  19. May 26, 2016 #18

    russ_watters

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    Assuming your numbers are correct: it does that every second
     
  20. May 26, 2016 #19

    SammyS

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    This is a pretty old thread!

    Do you realize that the last post in this thread, prior to your post, was submitted more than two and one half years ago?
     
  21. May 26, 2016 #20

    jim hardy

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    oops, my bad too
     
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