Physics Question Gravitational pull?

Question

Need Help. I get 1.9949 * 10^-8 but it says its wrong. Can someone help me. Four 8.5 kg spheres are located at the corners of a square of side 0.68 m. Calculate the magnitude and direction of the gravitational force on one sphere due to the other three.

Answer 1

You need to use the equation F=(G*m1*m2)/r^2

Where G is the universal constant, m1 is the mass of one ball, m2 is the mass of the other, and r is the distance between them.

Draw a diagram to help you visualize...

O O

O O

Let's call the top left Ball 1, the top right Ball 2, bottom left Ball 3, and bottom right Ball 4

For two of the forces, the answer will be the same, since the distance between them will be 0.68m, but the direction will be at 90 degrees to each other. Using Ball 1 as our reference point, the force between Ball 1 and Ball 2, and Ball 1 and Ball 3 will be the same, just at a right angle to one another.

For the third force between Ball 1 and Ball 4, you need to solve for the hypotenuse of the square using the Pythagorean theorem. Plug this value into the equation for r for the third sphere.

Now use the hypotenuse as your x axis for simplification. Now you only need to solve for the x component of one of the two identical forces, which will be Fx=cos(45)*F (You could also use sin, because the angle is 45 degrees...) F is the force between one ball and another along the same edge of the square. Multiply this number by two and add it to the force between Ball 1 and Ball 4 and you have your answer.

You don't need to solve for the y components of the two identical forces because they will cancel each other out.

Answer 2

For the 2 closest spheres:
Fb,d = Gm²/d² = 6.7E-11*8.5²/.68² = 1.046875E-8 N

For the farthest sphere: Fc = Fb(1/√2)² = .5234375E-8

A diagram will show that The net force Fa will be toward c and equal to

Fa = Fc + (√2)Fb = 2.003942323E-8 N

Answer 3

the gravitational force is
Gm1m2/r^2

There G = 6.6742* 10^-11 N m^2/kg^2

So, if we label the corners a,b,c, and d, and want to find the pull on "a", then first look at the pull from B, which is in the direction of b:

pull towards b = g*8.5^2/.68^2, in the direction of b
pull towards d = g*8.5^2/.68^2 in the direction of d
The resulting pull is actually in the direction of c (vector addition), and has magnitude of sqrt(2) times the pull towards b (pythagorean theorem).
pull towards c (due to c) = g*8.5^2/(sqrt(2)*.68)^2

So putting it all together we get
total pull is towards c, and the magnitude is
g*.85^2*sqrt(2)/.68^2 + g*8.5^2/(2*.68^2)
simplifying, we get
g*.85^2/.68^2*(sqrt(2)+1/2)
= 5.33g = 3.56*10^-10

Which at least is different from your answer :)

Answer 4

R = -GM1M2/R^2
G = 6.6742*10-11N-m^2/kg^2
Let the masses be located @
1, (0,0)
2, (0, 0.68)
3, (0.68, 0.68)
4, (0.68, 0)
Calculating the force on the sphere @ (0,0),
F12 = G8.5^2/0.68^2 j
F13 = G8.5^2/(0.68^2 + 0.68^2) i + j
F14 = G8.5^2/0.68^2 i
F = G(8.5^2)[1/(2(0.68)^2) + √((1/0.68^2)^2 + (1/0.68^2)^2)] @ 45°
F = G[(8.5^2)/(0.68^2)][1/2 + √(1 + 1)] @ 45°
F = (6.6742*10^-11)[(8.5^2)/(0.68^2)][1/2 + √2] @ 45°
F = 1.9962*10^-8 N @ 45°

Answer 5

Gravitational stress operates under an inverse sq. regulation, it is, F=ok/r^2, the place ok is a gadget consistent. If r' = (a million/2)r, F'=ok/(r'^2)=ok/(a million/4)r^2 = 4k/r^2 F'/F = (r^2)/(r^2/4)=(a million/4) F'=(a million/4)F, it is, the stress reduces to (a million/2)^2, to one / 4 of the unique stress.