# Acceleration on Rocket

1. Feb 9, 2005

### Hoofbeat

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|])

Where:
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?!

Chloé
xxx

2. Feb 9, 2005

### JesseM

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".

3. Feb 9, 2005

### Hoofbeat

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:

4. Feb 9, 2005

### JesseM

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.