1. Limited time only! Sign up for a free 30min personal 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!

Acceleration on Rocket

  1. Feb 9, 2005 #1
    HELP! I'm trying to write a computer program (C) to plot the trajectory of the rocket as it flies from Earth to Moon. In the information sheet it says:

    ax = -GMe*(1/rr^2)*(xr/|rr|) - GMm*(1/[rr-rm]^2)*([xr - xm]/[|rr - rm|])

    ax = acceleration in x-direction (we're working in xy coords)
    rr = a vector from the Earth to the Rocket
    rm = a vectore from the Earth to the Rocket
    xr = Rocket's x coordinate
    xm = Moon's x coordinate
    Me = Earth's Mass
    Mm = Moon's Mass

    Now, in order to input this acceleration in the x direction into my program I have to resolve the vector's into their x-direction BUT also be aware that their y co-ords will have an effect on the gravitational force felt in the x direction.

    Could some-one explain to me what I actually have to compute then? Also, when it says "rr^2" does this mean the modulus squared?!

    Please help, I'm really stuck!!! Thanks
  2. jcsd
  3. Feb 9, 2005 #2


    User Avatar
    Science Advisor

    I'm guessing you meant to write "a vector from the Earth to the Moon" here, right?
    The rocket's position vector rr just has two components, the rocket's x-coordinate and its y-coordinate, or (xr, yr). So if you figure out rr^2 and |rr| as a function of the vector's components, you can see that they will depend on both the x-coordinate and the y-coordinate--that's all they mean when they say "be aware that their y co-ords will have an effect on the gravitational force felt in the x direction".
  4. Feb 9, 2005 #3
    Thanks, yes I did mean that sorry! Was very stressed and panicky as had been trying to write the program all day and was in desperate need to write the msg before I had a Physics tutorial and then a training session straight after.

    One more question...I asked someone else for advice on the same problem and they quoted the following:

    "Mathemactically, rr^2 = |rr|^2.cos(rr,rr) = |rr|^2.cos0 = |rr|^2. So you're right, it means modulus squared.

    xr/|rr| = cos(rr, Ox) where (rr, Ox) is the angle between rr and x positive coordinate.

    Therefore, if we consider the force acts on the Rocket by the Earth due to x axis
    Fe = [GMem/rr^2] .cos(rr, Ox) = [GMem/rr^2].xr/|rr|
    where m is mass of the Rocket"

    So obviously, they agree with you (and me!) that rr^2 is the same as the modulus squared. However, they suggest using angles and cosines when considering the force/acceleration! I didn't think this would really be necessay as the vector equations give the directions so surely I don't have to resolve them as well? :uhh:
  5. Feb 9, 2005 #4


    User Avatar
    Science Advisor

    Right, the equation already takes care of resolving the force into components. The idea of the equation is that the ratio (component of total force from earth along x-axis)/(total force from earth) will be equal to the ratio (distance from earth along the x-axis)/(total distance from earth)...so, just multiply both sides by the total force from earth, and you get (component of total force from earth along x-axis) = (total force from earth)*[(distance from earth along x-axis)/(total distance from earth)], or Fe*(xr/|rr|). You can see that those two ratios must be equal, because the total force vector and the total distance vector are parallel to each other, so if you make a right triange with the total force as the hypotenuse and the x-component and y-component of the force as the other two sides, it will look just like a right triangle with the total distance vector as the hypotenuse and the x-component and y-component of the distance as the other two sides...if two right triangles are similar, the ratio of the hypotenuse and the side along the x-axis will be the same for both. For a right triangle you could also find the length of the side along the x-axis by multiplying the length of the hypotenuse by the cosine of the angle the hypotenuse makes with the x-axis, but if you already know the ratio between xr and |rr| then you can just use the equal-ratio assumption to find the force along the x-axis.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook