Help with a Gravitational physics problem?

Question

Four 6.5 kg spheres are located at the corners of a square of side 0.78 m. Calculate the magnitude and direction of the gravitational force on one sphere due to the other three.

I have tried using the formula with Cavendish's constat it didn't work. I did Gm1m2/r^2 which i got my answer as 9.26E-9 which was WRONG. There is an identical problem in my book with 7.5 kg mass and .6 length of a square side. I did that one the same way and got the right answer in the back. 2.0E-8, could someone tell me how to do my problem correctly? Show steps so i can understand it.

we are using 6.67E-11 as cavendish's constant

what did you guys use as the Radius, r?

That's the problem i'm having using Gm1m2/r^2 how do you find r correctly?

Answer 1

You can figure out Gm^2...It's constant for all three forces call it C = 2.82 e -10 N m^2

It's the 1/r^2 that's interesting...
12
34

Let's look at 1....For 41 force 1/r^2 = 1/(2*(0.78)^2) = 0.822 C
The forces from 2 and 3 net a pull toward 4... 1/r^2 along side 41 = 1/sqrt(2)*r^2 = 1.16C but there are two of them = 2.32C
Net = 3.15 C = 8.87 e-9 N
============================================
for forces in 12 and 13: 0.78 for force 14: 0.78*sqrt(2)

But here's what the force between 1 and 2 are

--- but it can be changed into:
/
\
and force betweent 1 and 3 are | but it can be changed into
/\
The two of forces cancel each other out leaving only \ which adds and is in the same direction as between 1 and 4.

Answer 2

You have the right formula there must be a problem in the input I used the same equation and I got 8.8721*10E-9 N for the first problem and 1.9962*10E-8 N for the second.

G = 6.642810^-11 Nm^2/kg^2, F = -GmM/r^2 = -Gm^2/r^2
Fr = (Fx^2 + Fy^2)^1/2 = F√2 = Gm^2/r^2)√2, F = Gm^2/(r√2)^2
F = (1/2)Gm^2/(r)^2, Ft = Gm^2/r^2(1/2+√2), Ft = (1/2+√2)(6.642810^-11)(6.5^2)/... Ft = 8.8721*10E-9 N along the diagonal.

Answer 3

Sounds like it might be an arithmetic error. I get 8.87x10^-9. If that's right, here's how I got it. The distance between opposite corners is .78*1.414, right? Then Gm1m2/r^2 gives 34.7G.

To either adjacent corner, the force is (Gm1m2/r^2). Multiply by .707 - that gives the component of the force that points along the diagonal. I get 49.1G. Add 2 of these (one for each corner) to the 34.7G towards the diagonal. Add em up and multiply by G. I get 8.87x10^-9.

Answer 4

G = 6.642810^-11 Nm^2/kg^2
F = -GmM/r^2 = -Gm^2/r^2
for the moment ignore the one on the other end of the diagonal.
Fr = (Fx^2 + Fy^2)^1/2 = F√2 = Gm^2/r^2)√2
force exerted by the 3rd is
F = Gm^2/(r√2)^2
F = (1/2)Gm^2/(r)^2
Ft = Gm^2/r^2(1/2+√2)
Ft = (1/2+√2)(6.642810^-11)(6.5^2)/(0.78^2)
Ft = 8.82931*10^-9 N along the diagonal.

Answer 5

You must have a math error then, using the same equation, I get 8.8721*10E-9 N for the first problem and 1.9962*10E-8 N for the second.

Since you got the second right, you should have the correct equations. There should one force from the one diagonally opposite and two forces summed from the left and right ones.

Answer 6

If 2.0E-8 is correct then 8.889E-9 is correct answer in your case. Use your calculator attentively & do what you did for 2.0E-8!