# Gravitational force between objects

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1. Feb 23, 2015

### SteveS

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

Three solid spheres of lead, each of mass 9.8 kg, are located at three corners of a square with side lengths of 50 cm. A small object is released at the forth corner. Considering only the gravitational forces among the four objects, determine the magnitude and direction of the acceleration of the smaller object when it is released.

2. Relevant equations
F = Gm1m2/r^2
a = F/m

mo = mass of object

3. The attempt at a solution

Ok so I drew a square and added the three 9.8 kg balls. Top right of the square is Ball #1, Top left of the square is ball #2, bottom left is Ball #3. The smaller object is in the bottom right corner of the square.

∑F = F1 Cos 45 + F2 + F3 Cos 45

Substituting into this the formula for force one gets

ΣF = G*mo*m1 / r^2 * Cos 45 + G*mo*m2 / 2r^2 + G*mo*m3 / r^2 Cos 45

This simplifies to:

ΣF = G*mo /r^2 ( Cos 45 * m1 + 0.5m2 + Cos 45*m3)

If we then equate this to F = mo *a, you get the mo to cancel and are left with:

a = G / r^2 (Cos 45*(9.8kg) + 0.5(9.8kg) + Cos 45 * (9.8kg))

substitue in G = 6.67x10^-11 and 0.05m for r and you get an a of: 5.0 x 10^-7 m/s^2 in the direction of the centre of the square.

Have I done this right?

2. Feb 23, 2015

### Suraj M

Are you sure about that?? the question says $50cm$! Do you still think your final answer is right?

3. Feb 23, 2015

### SteveS

lol d'oh. failure to move decimal places for the fail... let me fix that. r = 0.5m...

a = 5.0 x10^-9 m/s^2...

look better now?

4. Feb 23, 2015

### Suraj M

Looks right to me.
There isn't any mention of where the square is located, on earth? then horizontal or vertical? if in space then fine!
If you get time go through this, for the equations you typed.

5. Feb 23, 2015

### SteveS

The question doesn't specify a location, or weather it is vertical or horizontal. It only specifies to only consider the gravitational forces the objects exert on each other. So it seems like in space. That being said in space it shouldn't matter then if the square was vertical or horizontal relative to the observer then would it?

6. Feb 23, 2015

### Suraj M

thats why i said

7. May 8, 2015

### Brendan Webb

Why is the radius 0.5 meters? It says that the side of the square is 0.5m, meaning that the square is just points on a circle. Using Pythagoras I calculated the radius to be 0.35m approx. Where am I going wrong? (working on the question myself)

8. May 8, 2015

### SteamKing

Staff Emeritus
Make a sketch and check the geometry.

9. May 8, 2015

### jbriggs444

The "r" in the formula for gravitational attraction ($F=\frac{Gm_0m_1}{r^2}$) is not the radius of an arbitrarily selected circle that seems to be relevant to the problem. It denotes the distance between the [centers of masses of the] two objects whose masses are $m_0$ and $m_1$.

10. May 8, 2015

### Brendan Webb

Ok, now I understand the radius, thank you

11. May 8, 2015

### Brendan Webb

Hi again,

For the radius of F2, I don't understand why it would be 2r. It is shown in my textbook as SQRT(r^2+r^2) = SQRT(2r). This equals 2r as explained in the above problem. However, if we are using the Pythagorean Theorem to figure out the diagonal of the square isn't it r^2 + r^2 = r^2 (distance between m0 & m2) which in this problem equals 0.5^2 + 0.5^2 = Sqrt(0.5) which equals 0.707 meters. I made a square of equal size and measured it and yes the diagonal between a square of sides 0.5 meters equals 0.707 meters and not 1 meter as suggested by 2r. I am clearly not understanding something here but I don't know what, any help again would be much appreciated.

Thanks :)

12. May 9, 2015

### SteamKing

Staff Emeritus
If indeed your text has printed SQRT(r2+r2) = SQRT (2r), then this is in error, as simple algebra will show that:

SQRT(r2+r2) = SQRT(2*r2) = SQRT(2) * r {Note the placement of the parenthesis}