Homework Help: Simple Gravitation - 1 problem

1. Apr 19, 2007

Stuffy01

Simple Gravitation - With Answer - Help me work the middle steps

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 as in Figure P13.23. 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

Find the magnitude and direction of the gravitational field g.

3. The attempt at a solution

The direction is 45 degrees (measured counterclockwise from the positive x-axis)

G((4m+M)/(2l^2)) was the final simplified answer I got for the magnitude, but it was wrong.

Attached Files:

• p14-03alt.gif
File size:
2.4 KB
Views:
265
Last edited: Apr 19, 2007
2. Apr 19, 2007

Stuffy01

Associated work:

Base equation: Gravitational field force = GMm/(l^2)

I see three gravitational forces at work:
Gm/l^2 on the y axis

Gm/l^2 on the x axis

GM/((sqrt(2) times l)^2)

Add all three so 2Gm/2l^2 + 2Gm/2l^2 + GM/2l^2 = (G(4m+M))/(2((l)^2))

3. Apr 19, 2007

denverdoc

Welcome to PF.

Careful, as in Why do all the denominators look alike when you showed the diagonal object to have longer radius?

Never mind i see what you did.

4. Apr 19, 2007

Stuffy01

Thanks. See any easy mistakes or did I go the wrong direction?

5. Apr 19, 2007

denverdoc

No I found no mistakes so I was sort of head scratching on this one, hoping someone else might see a problem if there is one. Of course these are vector fields, which might be the issue.

Last edited: Apr 19, 2007
6. Apr 19, 2007

Stuffy01

I GOT THE ANSWER FROM A FRIEND:

(G/l^2)(M/2+sqrt(2)m)

Can anyone help me work it out now that the answer is known?

7. Apr 19, 2007

It is to do with the fact that the fields are vector fields.

You can't just arithmetically add the 3 fields up, you have to consider their direction. It's like if you push a block north with 1N and east with 1N the overall force is $$\sqrt{2}N$$ in the NE direction, not 2N. So I'd recommend drawing a diagram and doing a bit of pythagoras.

8. Apr 19, 2007

denverdoc

No problem, resolve the diagonal force into x and y components which can be added like scalars

cos and sin are both sqrt(2)/2 so mult your diag. expression and add to the x and y components from other forces, then calculate magnitude by taking the sqrt of sum of squares. Falls out like honey in a bucket. I suspect you'll never make this mistake again. good problem and post.

9. Sep 3, 2011

Saudi Arabia

Hi Guys, i guess this is the answer, i'm not sure! but? if someone has a comment reply pls.xxx
tanθ = gy / gx
= [ Gm / L2 + G M /2L2 sinθ ] / [ Gm /L2 + G M / 2L2 cosθ ]

θ = tan-1 { [ Gm / L2 + G M / 2L2 sinθ ] / [ Gm /L2 + G M / 2L2 cosθ ]}