1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Universal Gravitation and rockets

  1. Dec 15, 2007 #1
    [SOLVED] Universal Gravitation

    1. The problem statement, all variables and given/known data
    A small rocket is launched vertically, attaining a maximum speed at burnout of 1.0x10^2 m/s and thereafter coasting straight up to a maximum altitude of 1519 m. Assuming the rocket accelerated uniformly while the engine was on, how long did it fire and how high was it at engine cutoff? Ignore air friction

    2. Relevant equations
    Kinematics and Law of universal gravity equation

    3. The attempt at a solution
    I thought the time could be solved by using the 1st kinematics where Vf = 1.0 x 10^2m/s and Vi = 0m/s and a = 9.8m/s^2. The time I found was 10.2041s which isn't correct. Is that because gravity would change during the time the rocket reaches burnout? If so, how would I go about solving this problem.
  2. jcsd
  3. Dec 15, 2007 #2
    There are two parts to the rocket's motion. The time that was computed corresponds to which part?
  4. Dec 15, 2007 #3
    The time I found was the time of acceleration. It asks how long it "fired" so I assumed that the final velocity was what the question had and acceleration was 9.8m/s^2 and used the first kinematics Vf = vi +a(delta t)
  5. Dec 15, 2007 #4

    Shooting Star

    User Avatar
    Homework Helper

    Why should the accn of the rocket be 9.8 MKS when the engines are on? Find the accn 'a' directly by the formula relating Vi, Vf and assuming h for the hight gained. After that, it was in free fall, with initial velo Vf.

    Are you supposed to assume uniform g or an inverse square law?
  6. Dec 15, 2007 #5
    Using the 3rd kinematics accn = 13.166m/s^2 which I assume would be the acceleration of the rocket. So by subtracting 9.8m/s^2 for grav, the acceleration became 3.36656m/s^2.
    I then used the 3rd kinematics again but this time solving for delta y which gave me 14.852m but it was wrong.
  7. Dec 15, 2007 #6
    As posted here there are two parts in the rocket's motion. First the one with the engines on and the second one without them. In the first part the acceleration of the rocket must be used in the equations, however it is still unknown as is the height and the time after lift-off when the engines stop working. So it will become difficult to start from this. The second part of the rocket's motion contains however all the data you need for describing this part. The formula's you need are the ones of a linear motion under an acceleration, in this case g. Before starting to do the calculations, make a nice sketch and draw the axes system, X (the earth's surface) in the horizontal direction and Y pointing upwards. Now you know that the acceleration of the rocket is -g because it is pointing downwards and no other acceleration is at hand here because of the engines not working. This is equal to [tex]\frac{dv}{dt}[/tex] and to [tex]\frac{d^2y}{dt^2}[/tex]. Use this to obtain the necessary formula's. Take care of the boundary conditions, we are working on the second part and these are thus time t=0 (from here we start timing the second part) [tex]v_0=v_f=100\frac{m}{s}[/tex] and [tex]h_0=?[/tex] (unknown). From the formula of the velocities you can obtain the time (10.204s) from shutdown of the engines to it's final height. Knowing this and the formula for the height you can obtain the height (1008.8m) at which the engines stopped. The second part is now completely known.

    Secondly the lift-off part. In case you are not familiar with potential and kinetic energy, I will give you a tip on how to look for an equation without time, relating acceleration, velocity and height. Consider:


    Multiplying the left and right hand side with the same number is allowed, so we know that:


    Multiplying these gives:

    [tex]a \cdot \frac{dy}{dt}=v \cdot \frac{dv}{dt}[/tex]

    Leaving out the dt gives:

    [tex]a \cdot dy=v \cdot dv[/tex]

    Integrating this gives and using the symbols from above:

    [tex](a_{rocket}-g) \cdot y=\frac{1}{2}(v_f^2-v_0^2)[/tex]

    You know that [tex]v_0=0[/tex], and a (14.756m/s^2) is the only unknown. Solve for this. Then you should be able to obtain the time (20.176s) from lift-off to shutdown of the engines and the height (1008.8m) at which this occurs, which should be the same as the one from the beginning.
  8. Dec 15, 2007 #7
    Thanks a lot coomast, I got it. I started what you said to do in the first part to find the distance it traveled at 100m/s but I missed something in there. Thanks for all the help.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Universal Gravitation and rockets
  1. Universal Gravitation (Replies: 3)

  2. Universal Gravitation (Replies: 1)