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Rocket Thrust, Acceleration, Velocity and Altitude

  1. Jun 6, 2014 #1
    Having some trouble with this problem.
    I'm not sure if the first part is correct, I just need some feedback on that.
    Also I'm not sure how to calculate velocity or altitude after 1 second.
    Any help would be awesome!
    1. The problem statement, all variables and given/known data
    A rocket has an initial mass of 4000kg, of which 3000kg is fuel. Total burn time after launch is 8 seconds. Exhaust velocity is 400m/s. The rocket lifts off vertically.

    (a) Evaluate “Thrust”, V(dM / dt)

    (b) Using Newton’s second law, evaluate acceleration of the rocket after 1 second of flight.

    (c) Calculate the rocket’s velocity after 1 second.

    (d) Calculate the altitude reached after 1 second.

    (e) Calculate the kinetic energy ½ (Mv2) of the rocket after 1 second.

    (f) Calculate the gravitational potential mgh of the rocket after 1 second.


    2. Relevant equations
    g = 10m/s2
    (Thrust / recoil Mass) gives acceleration
    Evaluate “Thrust”, V(dM / dt)


    3. The attempt at a solution

    (a) Evaluate “Thrust”, V(dM / dt)
    3000/8 = 375kg/s of fuel each seconds
    Table -
    Time(t): 0.............1..........2.........3
    Mass(M): 4000....3625.....3250....2875
    Thrust: ..............75,000
    A: ........18.75....20.69....23.08
    Thrust = 400 * (375/2) = 75,000N

    (b) Using Newton’s second law, evaluate acceleration of the rocket after 1 second of flight.
    (Thrust / recoil Mass) gives acceleration
    A = 75,000 / 3625
    Acceleration: 20.69m/s2
    (c) Calculate the rocket’s velocity after 1 second.
    Time(t): 0 1 2
    A: 18.75 20.69 23.08
    (d) Calculate the altitude reached after 1 second.

    (e) Calculate the kinetic energy ½ (Mv2) of the rocket after 1 second.

    (f) Calculate the gravitational potential mgh of the rocket after 1 second.
     
  2. jcsd
  3. Jun 6, 2014 #2

    gneill

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    Staff: Mentor

    In (a), why are you dividing the mass ejection rate (375 kg/s) by two when you evaluate the thrust?

    Since the rocket is being launched vertically from the surface of the Earth, you need to take into account the effect of g on the net acceleration of the rocket. Take a look at the rocket related equations on the Hyperphysics site:

    Rocket Principles
     
  4. Jun 6, 2014 #3
    I was trying to figure out (dM / dt) and I got confused. Should it be:
    dM / dt
    3250 - 4000 / 2
    -750 / 2 = -375
    400 * -375 =-150,000N
    I'm trying to work of the V(dM / dt) equation given. Could you correct me on what I've done?
     
  5. Jun 6, 2014 #4

    gneill

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    Staff: Mentor

    The thrust formula involves the velocity of the mass being ejected and the rate at which the mass is being ejected. That is, V is the 400 m/s quantity and dM/dt is the rate 375 kg/s that you determined earlier. No need to go back to the actual mass values of the rocket at a particular time, you have the velocity and rate which remain constant over the burn.
     
  6. Jun 6, 2014 #5
    Ahh okay so its just 400 * 375 = 150,000N

    Is my calculation for Acceleration after 1 seconds correct?
    (Thrust / recoil Mass) gives acceleration -
    150,000 / 3625(mass at 1 second)
    Acceleration = 41.38 m/s2
     
  7. Jun 6, 2014 #6

    gneill

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    You haven't accounted for gravity's effect on the net acceleration of the rocket.
     
  8. Jun 6, 2014 #7
    Ahh of course, hows this?
    150,000 / 3625 - 9.8066 = 31.57m/s2
    Acceleration = 31.57m/s2

    Thank you for your help so far! I have to sleep now.
    If your not busy tomorrow I'll continue to post on this thread.
     
  9. Jun 6, 2014 #8

    gneill

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    Staff: Mentor

    Yes, that looks better :smile:

     
  10. Jun 7, 2014 #9
    I change g to = 10m/s2 instead of the usual 9.8, ask the question gave g = 10m/s2
    (c) Calculate the rocket’s velocity after 1 second.
    This is my calculation for velocity:
    v = v0 + u * ln(M0 / M)
    v = 0 + 400 * log(4000/3625)
    v = 17.1m/s-1

    (d) Calculate the altitude reached after 1 second.
    I'm not sure how to do this. Please help!

    (e) Calculate the kinetic energy ½ (Mv2) of the rocket after 1 second.
    KE = 0.5 * ( 3625 * 17.12) = 30993.75J

    (f) Calculate the gravitational potential mgh of the rocket after 1 second.
    I understand that i simply times mass by gravity and then by height, but I'm not sure how to work out height or altitude.
     
  11. Jun 7, 2014 #10

    gneill

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    Staff: Mentor

    Note that log is not the same thing as ln. One is a base 10 logarithm and the other is base e. You switched the function to log between your first and second equation lines.

    Also, once again you've lost the affect of gravity on the acceleration. Hint: Note that if you establish a general formula for the net acceleration with respect to time then you can integrate to find velocity, and integrate again to find distance, right?
     
  12. Jun 7, 2014 #11
    Not really sure what, or how to establish a general formula for 'net acceleration'.

    I reworked my equation including gravity and fixing the log/ln problem:
    v = v0 + u * ln(M0 / M) - g
    v = 0 + 400 * ln(4000/3625) - 10
    v = 400 * 0.09844007281 - 10
    v = 29.38 m/s

    How's this?
     
  13. Jun 7, 2014 #12

    gneill

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    Staff: Mentor

    Better, but not perfect. Numerically the answer you've found for v is correct for time = 1 second. Note though that strictly speaking that "lone" g on the right hand side is not a velocity and doesn't match the units of the other terms. The g should be multiplied by the time, (t = 1s) to obtain its contribution to the velocity after the given time. Remember the kinematic formula v = at. g is an acceleration...

    If you rewrite your formula replacing M with a suitable expression giving M at any time t, and fix the 'g' term to be gt, then you will have made the formula more general, applicable to any time t (within reason since g is only taken to be a constant for regions close to the Earth's surface).
     
  14. Jun 8, 2014 #13
    v = v0 + u * ln(M0 / Mt) - gt

    I put the subscript on the first M, 0 for the original mass, then the subscript t on the second for mass at the given time; M1 = 3625

    (d) Calculate the altitude reached after 1 second.
    x = v0t + 0.5 * at2 - gt
    a = 150,000 / 3625 - 10 = 31.38
    x = 0 + 0.5 * 31.38 * 1 - 10 * 1
    x = 5.69m
     
  15. Jun 8, 2014 #14

    gneill

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    Staff: Mentor

    Okay, but introducing subscripts is just a renaming of the variables, M0 and M. What you want is an expression for M(t) which will yield the mass for time t.

    Where does the above come from? It sort of looks like a standard kinematic formula with a "-gt" tacked on. But that won't work for two reasons: 1) The usual kinematic formulae assume that acceleration is a constant, but acceleration is not a constant in this problem; 2) the units of the term "gt" is velocity. You can't add velocity to distance.

    May I ask what level of course this problem comes from and what is the focus of the relevant chapter in your text? Does the course material assume that you've studied calculus?

    It could be that I've made an incorrect assumption about the level and intention behind the problem.
     
  16. Jun 8, 2014 #15
    Oh it came from https://www.google.com.au/webhp?sou...=UTF-8#q=distance calculator physics&safe=off
    except I tried to put gravity on the end of it. The course unit is Physics of Games (seconds year uni subject), there were no prerequisites for studying physics or any maths above General Maths in VCE (Last year of highschool in Victoria, Ausralia), so I don't know much about calculus or physics as I didn't study them prior to this course.
     
  17. Jun 8, 2014 #16

    gneill

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    Staff: Mentor

    That's a link to a whole page of Google search results. Is there a particular website that you are referring to?

    I see. So that may change things. While you could derive general formulas from basic principles to solve the problem "exactly" using calculus, they may be looking for simple numerical approximations. That is, simple strategies to simulate the movement of game objects so that they imitate real-world physics closely enough to fool the player, but simple enough to be computationally "lightweight" for implementation purposes. This would involve breaking the motions into small timesteps and using average values of force, mass, acceleration, etc., that apply to the individual timesteps. During each small timestep the assumption is made that these quantities are constant and so the basic kinematic formulas will hold.

    For example. If you were to suppose that the timestep size is 1 second, then you would evaluate the acceleration for the beginning of the timestep and the end of the timestep (which you can do here because you have a fixed thrust and you know how the mass changes over the timestep) and assume that the acceleration is the average of the two over the whole timestep. Then use Newton's laws and basic kinematic formulas to work out how the velocity and distance change over that 1s timestep.

    I don't know for certain that this is what they want you to do for this question. I'm not familiar with the particulars of the course you're taking.

    It could also be, given there are no particular physics or math prerequisites for the course, that they give you a handout with a list of relevant formulas to use and you are not expected to derive your own from scratch.
     
  18. Jun 8, 2014 #17
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