Homework Help: 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.

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