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Rocket and Gravitational Potential Energy

  1. Jun 29, 2007 #1
    1. The problem statement, all variables and given/known data
    a rocket of mass 1.5e6 kg is launched upward from the surface of the Earth (mass Me =5.97e24, radius Re = 6.38 X 10^6m) with a speed of 0.5 x 10^4 m/s. What is the maximum altitude the rocket will reach above the surface of the Earth?

    2. Relevant equations

    1/2 mv1^2 - (GmM)/R1 = 1/2 mv2^2 - (GmM)/R2
    where m is the mass of the rocket
    v1 is the velocity of the rocket (at rest?)
    G = 6.67e-11 (constant)
    m= mass of rocket
    M = mass of Earth
    v2 = velocity of rocket
    R1 = radius of Earth
    R2 = radius we are trying to find

    3. The attempt at a solution

    I plugged in numbers for the above equation and my answer was 3.18559e7. This is a very wrong answer but what did I do wrong?
  2. jcsd
  3. Jun 30, 2007 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    It looks like you've transposed v1 and v2.
  4. Mar 10, 2008 #3
    i have a question...what is that suppose to mean. i did the same thing as the above person, but what does transposed mean? thanks!
  5. Mar 10, 2008 #4
    I think that the problem lies within your interpretation of the speeds. Remember that the left side of your conservation of energy equation is meant to be the initial values of both kinetic and gravitational potiential energy while the right side is supposed to be your final values.

    I think you understand the mechanical energy approach of this question but are confused with the initial values... from what I can tell the problem states that the rocket is blasting-off with a non-zero initial speed (0.5e4 m/s) from the Earth's surface and you are supposed to find the height above Earth's surface at which it has lost all of its speed from a transfer of kinetic energy to gravitational potential energy, assuming it travels straight out into space.

    I'm expecting about 1.6e6 meters above the Earth's surface. Your first equation is all that is needed.

    btw transposed means switched the positions, i.e. you subbed in the wrong velocities on the wrong sides of the equation, due to your seeing the initial speed as 0. Don't try to aim for my value but try the equation again with what you now know. Message back if you get your desired answer or have an issue with anything I've written.
  6. Mar 10, 2008 #5

    D H

    Staff: Mentor

    You have to use R1 and v1 consistently. These statements are what made Janus think you transposed v1 and v2 (in other words, used v2 where you should have used v1, and vice-versa):
    Show us how you "plugged numbers in for the above equation".
  7. Mar 11, 2008 #6
    Hey, thanks so much for helping, but after much fustration and head scratching, i know what i did wrong.
    I still used 1/2mv**2 - GMm/R = 1/2m(vf)**2 - GMm/(Rmax)
    Initially, i thought that the initial height (R) was equal to zero, but it is not, R is the distance from the center of the earth, which is the radius of the earth.

    so in the end, i ended up with this equation:

    1/2mv**2 - GMm/R = -GMm/(Rmax)
    v = initial velocity
    G = constant
    M = mass of earth
    R = initial height (radius of the earth)
    Rmax = distance from the CENTER of the earth to the rocket
    and after you solve for Rmax, you can solve for the distance above the earth because you know that:

    Rmax = Rearth - h
    Rearth = radius of earth
    h = distance above earth


    Im not sure if this is what you were explaining, it could be, but i was not sure, BUT THANKS SO MUCH FOR THE HELP!! MUCH APPRECIATED!! ALSO, i hope this helps anyone else trying to figure this problem out!!
  8. Mar 11, 2008 #7
    hey, thanks so much for the help, my response is in another reply...i thought that i probably should not type it again, it is probably going to be above this post! :) thanks again!
  9. Mar 11, 2008 #8
    Oh, I was actually referring to your saying "v1 is the velocity of the rocket (at rest?)" when its actually the velocity they gave you, and v2 is the velocity of the rocket at rest. But if the problem was in the initial and final heights, then I think you've done okay. Hope you got the right answer!
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