# Gravitational Field problem Question

1. Nov 14, 2008

### jessedevin

1. The problem statement, all variables and given/known data

Three objects -- two of mass m and one of mass M -- are located at three corners of a square of edge length l. Find the gravitational field g at the fourth corner due to these objects. (Express your answers in terms of the edge length l, the masses m and M, and the gravitational constant G).

2. Relevant equations

g=-GM/r2

3. The attempt at a solution

g= ga+gb+gc
g= Gm/l2 $$\hat{i}$$+ (GM/(l$$\sqrt{2}$$)2)(cos($$\pi/4$$)$$\hat{i}$$+sin($$\pi/4$$)$$\hat{j}$$)+Gm/l2 $$\hat{j}$$

I know you have to take the magnitude of this, but when I did that , I still get the wrong answer. Here's what I got:

||g||=$$\sqrt{2G^2/l^4(m^2+M^2)}$$

Did I start it right? Can someone help?

Last edited: Nov 14, 2008
2. Nov 14, 2008

### LowlyPion

Isn't the gravitational field given by GM/r ?

You have 3 vectors to add, but happily the 2 m's at right angles gives one lying in the direction of M

So ... √2Gm/L + GM/(√2*L) = √2*G*(m + M/2)/L ?

3. Nov 14, 2008

### jessedevin

So do you take the magnitude of that? Im still confused, because my book says otherwise. Can you go through your process?

4. Dec 2, 2008

### SonHa

I think what you did originally is correct, but it ask for the magnitude without the vector sign. So just put down the answer using c^2 = a^2 + b^2 and then I believe you have to indicate the degree according to the x-axis. I'm doing a similar problem. Wait, yeah you did that, never mind.

5. Dec 2, 2008

### SonHa

I got it!
Instead of converting M vectors into g forces of x and y, why don't you convert the other 2 m mass into direction of M which is Gm/l^2 * cos(45) * 2.
Then add it to the g force of M
My answer is (1.41Gm + 0.5GM)/l^2.
Hope it helps, the post was like half a month ago, lol.