A 20 kg sphere is at the origin and a 10 kg sphere is at (x,y) = (20,0

In summary, the conversation discusses how to find the point or points where a small mass can be placed between two larger masses in such a way that the net gravitational force on it is zero. The solution involves setting the two forces equal to each other and using simple algebra to find the relationship between the two unknown radii. G is a constant that cancels out and is not necessary to know in order to solve the problem.
  • #1
NG12
5
0

Homework Statement



a 20 kg sphere is at the origin and a 10 kg sphere is at (x,y) = (20,0). at what point or points could you place a small mass such that the net gravitational force on it due to the spheres is zero?

Homework Equations



g=GM/r^2

The Attempt at a Solution



Can not figure out how to plug into the equation
 
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  • #2
$$F = G \frac {m_{1}m_{2}}{r^2}$$

Might be a better option.

##r## is the radius between the two masses.
 
  • #3
Thank you! How would I go about plugging my information into the equation?
I'm extremely new to physics and the textbook lacks information on solving this problem
 
  • #4
Well, you know that your forces have to cancel out.

This means you can set your forces equal. The book tells you that you can go ahead and ignore the mass of the particle that you want to set in between them, so you'll have two equations.

$$F = G \frac {m_{1}}{{r_{1}}^2}$$

and $$F = G \frac {m_{2}}{{r_{2}}^2}$$

So you can simplify that using simple algebra pretty easily if you set those equation equal to each other. This will give you a relationship between ##r_{1}## and ##r_{2}##.

You'll need one more equation to solve for the two unknowns. Just remember that your two radii have to add up to be the original radius between the two spheres. From there you've got 2 equations and 2 unknowns, and you're set!
 
  • #5
Thank you very much!
What does G equal?

And for the radius would I divide the distance by 2?
 
  • #6
NG12 said:
Thank you very much!
What does G equal?

And for the radius would I divide the distance by 2?

No. r1 and r2 are the quantities you're trying to find, remember? They are the unknowns of your problem. If you could just divide the distance by two there would be no need to solve the problem to begin with...
 
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  • #7
NG12 said:
Thank you very much!
What does G equal?

And for the radius would I divide the distance by 2?

G = 6.67 x 10^(-11) N m^2/kg^2, but you don't need to know that in order to solve that problem. Your approach ought to be 1st solve the algebraic equation and only then plug in the data. That should always be your approach. Algebra is your friend.
 
  • #8
Okay, now I'm lost again. If I don't need to know what G is then what equation should I be using?
 
  • #9
NG12 said:
Okay, now I'm lost again. If I don't need to know what G is then what equation should I be using?


G is a constant. It would end up cancelling out. Conceptually, there is nothing to this problem once you have the two force equations set equal to each other. It is simply algebra.

You need to examine this problem a little more closely and figure out how and what you are solving for.

You will have two unknowns, which means you will need at least two defining equations for your parameters. My earlier post gave you hints on how to find both.
 
  • #10
NG12 said:
Okay, now I'm lost again. If I don't need to know what G is then what equation should I be using?

Use the equation provided in post # 2 (twice - once for each force).

Solve the problem algebraically and then you will understand why you don't need to know the value of G.
 

FAQ: A 20 kg sphere is at the origin and a 10 kg sphere is at (x,y) = (20,0

1. What is the gravitational force acting on the 10 kg sphere?

The gravitational force acting on the 10 kg sphere is directly proportional to the mass of the sphere and inversely proportional to the square of the distance between the two spheres. In this case, the gravitational force can be calculated using the formula F = G(m1m2)/d^2, where G is the gravitational constant, m1 is the mass of the 20 kg sphere, m2 is the mass of the 10 kg sphere, and d is the distance between the two spheres.

2. What is the net force on the 10 kg sphere?

The net force on the 10 kg sphere is the sum of all the forces acting on it, which includes the gravitational force from the 20 kg sphere as well as any other external forces. This can be calculated using the formula Fnet = ma, where m is the mass of the 10 kg sphere and a is its acceleration.

3. What is the acceleration of the 10 kg sphere?

The acceleration of the 10 kg sphere can be calculated using the formula a = Fnet/m, where Fnet is the net force on the sphere and m is its mass. In this case, the net force would be the gravitational force from the 20 kg sphere.

4. How far apart are the two spheres?

The distance between the two spheres can be calculated using the distance formula d = sqrt((x2-x1)^2 + (y2-y1)^2), where (x1,y1) is the position of the 20 kg sphere at the origin and (x2,y2) is the position of the 10 kg sphere at (20,0). In this case, the distance between the two spheres would be 20 units.

5. What is the potential energy of the system?

The potential energy of the system is the energy that is stored in the interaction between the two spheres due to their positions. It can be calculated using the formula U = -G(m1m2)/d, where G is the gravitational constant, m1 is the mass of the 20 kg sphere, m2 is the mass of the 10 kg sphere, and d is the distance between the two spheres.

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