Gravitational attraction between 3 masses.


by lsatwd
Tags: attraction, gravitational, masses
lsatwd
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#1
Nov2-08, 06:23 PM
P: 5
1. The problem statement, all variables and given/known data
A small mass M and a small mass 3M are 3.60m apart. Where should you put a third small mass so that the net gravitational force on it due to the other two masses is zero?(From mass M)


2. Relevant equations

F = G(m1)(m2)/d^2

3. The attempt at a solution
I envisioned the problem to originally look like this: 
[1M]                                       [3M]
And adding in the 3rd mass I envisioned it to look something like this:
[1M]    (1M)                               [3M]
I then tried solving the problem by setting the two unknown variables (the distance between the 3rd mass and 1st mass and the 3rd mass and the 2nd mass) like this:
x = Distance between [1M] and (1M)
3.60-x = distance between (1M) and [3m]

I was thinking this would give me an equation looking like this:

G[1M](1M)/x^2 = G(1M)[3M]/(3.60-x)^2

I tried solving this but it's not giving me the right answer... Should the right equation above be G(2M)(3M)/(3.60-x)^2 because of combined mass?
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#2
Nov2-08, 06:45 PM
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P: 14,476
Quote Quote by lsatwd View Post
I was thinking this would give me an equation looking like this:

G[1M](1M)/x^2 = G(1M)[3M]/(3.60-x)^2

I tried solving this but it's not giving me the right answer.
As an aside, the test mass does not have to be of mass M; it would be better to write it's mass as m (little m), meaning it has very little mass.

You claimed the quoted equation is "not giving me the right answer." That implies that either you know what the right answer should be or that some agent knows your result is wrong. Show your work.
lsatwd
lsatwd is offline
#3
Nov2-08, 06:50 PM
P: 5
ahh never mind, i see.

It turns out i just suck at math. that equation gave me 1.32 which was the right answer. thanks for the help


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