What is the magnitude of the net gravitational force

In summary, the conversation discusses the calculation of the net gravitational force on a third sphere with a given mass and position, due to two other spheres with known masses and positions. The net force is found to be 3.48 N, and the conversation then moves on to determining where the third sphere could be placed so that the net force is equal to zero. The correct position is determined to be at (0,1.43).
  • #1
QuarkCharmer
1,051
3
Gravitation Force (Solved, thanks)

Homework Statement


A sphere with mass of 65kg and center at origin. Another sphere with mass 79kg at point (0,3), also centered at origin.

A.) What is the magnitude of the net gravitational force due to these objects on a third uniform sphere with mass 0.490 kg placed at the point (4,0)

Homework Equations


[tex]F_{g} = \frac{GMm}{r^{2}}[/tex]
[tex]G = 6.67(10^{-11})[/tex]


The Attempt at a Solution


I made this image to help:
j91o3k.jpg


Force from the 79kg sphere:
[tex]FG_{1}=\frac{G(.490)(79)}{5^{2}}[/tex]

Force from the 65kg sphere
[tex]FG_{2}=\frac{G(.490)(65)}{4^{2}}[/tex]

So the net force on the .490 kg sphere is given by:
[tex]\sqrt{(FG_{2} + FG_{1}cos(tan^{-1}(\frac{3}{4})))^{2}+(FG_{1}sin(tan^{-1}(\frac{3}{4})))^{2}}[/tex]

I get the net force on the 0.490 kg sphere to be 3.48 N. Which is incorrect? I don't know what is wrong.
 
Last edited:
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  • #2
What did you use for G?

You can find the cosine and sine of the angle at the 0.49 kg mass directly from the right triangle.

ehild
 
  • #3
ehild said:
What did you use for G?

You can find the cosine and sine of the angle at the 0.49 kg mass directly from the right triangle.

ehild

I used 6.67 * 10^-11 for G.

What do you mean about the cos/sin ? arctan(3/4) is the angle near the .490kg sphere, so the equation should hold?
 
  • #4
QuarkCharmer said:
I used 6.67 * 10^-11 for G.

What do you mean about the cos/sin ? arctan(3/4) is the angle near the .490kg sphere, so the equation should hold?

The sine of the angle is opposite/hypotenuse=3/5. The cosine is adjacent/hypotenuse=4/5. Is not that simpler than taking the cosine and sine of arctan(3/4)? It is not wrong, just strange.

Check your calculation. I think you forgot the 10^-11.

ehild
 
  • #5
You are correct. I don't know why I was thinking that I needed to determine the angle via inverse trig.

I did the computations over and somehow it came out correct this time.
It was 2.24*10^5 N. I must have missed something somewhere.

Thanks

Now I am working through this part:

Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

But I'm still working it out.

I am thinking that:
[tex]\frac{G(.409)(79)}{d^{2}} = \frac{G(.409)(65)}{(3-d)^{2}}[/tex]
[tex]\frac{(79)}{d^{2}} = \frac{(65)}{(3-d)^{2}}[/tex]
[tex]9-6d+d^{2}=\frac{65}{79}d^}{2}[/tex]

so basically the third sphere needs to be at the point (0,1.43) ?

Edit: Yeah that's all correct!

Thanks for the help.
 
Last edited:

What is the magnitude of the net gravitational force?

The magnitude of the net gravitational force is the measure of the strength of the overall gravitational pull between two objects. It is a scalar quantity and is typically measured in Newtons (N).

How is the magnitude of the net gravitational force calculated?

The magnitude of the net gravitational force is calculated using the equation F = G * (m1 * m2)/d^2, where F is the net gravitational force, G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), m1 and m2 are the masses of the two objects, and d is the distance between them.

Does the magnitude of the net gravitational force depend on the masses of the objects?

Yes, the magnitude of the net gravitational force is directly proportional to the masses of the objects. This means that as the masses increase, the net gravitational force also increases.

How does the distance between objects affect the magnitude of the net gravitational force?

The magnitude of the net gravitational force is inversely proportional to the square of the distance between objects. This means that as the distance increases, the net gravitational force decreases.

Can the magnitude of the net gravitational force be negative?

Yes, the magnitude of the net gravitational force can be negative. This occurs when the two objects have opposite signs of mass, such as a positive and negative charge, resulting in an attractive force rather than a repulsive force.

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