Help with Circular Motion Homework: Find Gravitation Force Point

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To find the point where a third object experiences no net gravitational force between two masses, A (mass m) and B (mass 4m), located 2m apart, one must analyze the gravitational forces acting on the third object. The gravitational force from mass A on the third object is given by F1 = Gm*m2/r1^2, while the force from mass B is F2 = G(4m)m2/r2^2. The distances r1 and r2 relate to the position of the third object along the line connecting A and B, where R = r1 + r2 equals 2m. Setting the forces equal (F1 - F2 = 0) allows for solving the position where the forces cancel out, resulting in no net gravitational force on the third object. Understanding these calculations is crucial for solving the homework question effectively.
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I'm having troubles with one of my homework questions, and I was wondering if someone could help me with it.

okay here it is...

Q: Two masses, object A and object B are located 2m apart from each other, the mass of object a is m and the mass of object b is 4m.

Showing your calculations find the point between these two objects where a third object would experience no gravitation force.

I'm really stuck on this question, any help would be very much appreciated.
 
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"a third object would experience no gravitation force."
what they mean is, no net gravitation force
imagine a particle C lying on the line connecting A and B - what is the force that C feels from A? what is the force that C feels from B?
What is the net force?
 
I don't understand that at all...

I would have thought I'd try and figure out what m was = to then, maybe do that... because from what I know I don't think I can figure out the net gravititional force from just what I was given...

A little more help? heh
 
Are you aware that the force for gravitation is
\vec{F}=-\frac{Gm_1m_2}{r^2}\hat{r}

You have two masses m and 4m and some unknown mass m_2, so the magnitude of the gravitational attraction between m and m_2 is
F_1=\frac{Gmm_2}{r_1^2}
where r_1 is the distance between m_2 and m. The strength of the gravitational attraction between 4m and m_2 is
F_2=\frac{G(4m)m_2}{r_2^2}
where r_2 is the distance between 4m and m_2. If m_2 lies along the line between 4m and m and the distance between m and 4m is R, then
R=r_1+r_2, so the second equation becomes
F_2=\frac{G(4m)m_2}{(R-r_1)^2}
If m_2 is right between A and B, then the forces act in opposite directions and so cancel each other out. When F_1-F_2=0, m_2 will feel no force.
 
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