What is the distance between star 1 and 2 if the net force of star 2 is zero?

[minimax]

Astronomy Que!

Homework Statement

Three stars lie on a line (1,2,3). The distance from star 1 to star 3 is labelled as D. If star 1 is four times that of star 3, and seven times of star 2, what is the distance between star 1 and 2 if the net force of star 2 is zero?

Homework Equations

F=\frac{Gm1m2}{r^{2}}

The Attempt at a Solution

Since the net force for star 2 is zero, I know that the force between star 1 and 2 is the same as the force between start 2 and 3

\frac{Gm1m2}{r^{2}}=\frac{Gm2m3}{r^{2}}

I tried to sub in values to find the force between star 1 and 3

F=\frac{Gm3(4m3)}{D}

but I'm not sure where to go from here...

Thank you

 

[Snazzy]

m_1=4m_3=7m_2

So then, the distance between star 1 and star 2 is x, and the distance between star 2 and star 3 is D-x. Can you go on from here?

 

[minimax]

errr..I think so.
Here's what I've 'tried' so far

\frac{Gm1m2}{x}=\frac{Gm2m3}{D-x}
\frac{G7m^{2}}{x}=\frac{Gm2m3}{D-x}

Cross multiply:
\frac{G7m2^{2}}{Gm2m3}=\frac{x}{D-x}
G's cancel, so do m2

\frac{7m2}{m3}=\frac{x}{D-x}

if 4m3=7m2
then m3=(7/4)m2

so,
\frac{1}{4}=\frac{x}{D-x}

now to find for x?

(D-x)=4x
D-x=4x
D=4x+x

x=\frac{D}{5}

not sure if it's the right way, or if I'm just messing around with variables, or if the force between 1 and 3 has any use (yikes!)
Anyone willing to verify?

but yeeeah...times like these when I wonder why I took physics as an option when I'm planning to do arts *sighs*
thanks for your help!

 

[Snazzy]

if 4m3=7m2
then m3=(7/4)m2

so,
\frac{G}{4}=\frac{x}{D-x}

1. Where did that G/4 come from?
2. F \propto \frac{1}{r^2}

 

[minimax]

whoops, the G's cancel out too

I got the 1/4 from dividing G7m2 by G(7/4)m2

so...\frac{1}{4}=\frac{x}{D-x}
to get x=D/5

 

[Snazzy]

\frac{7m2}{m3}=\frac{x}{D-x}

if 4m3=7m2
then m3=(7/4)m2

so,
\frac{1}{4}=\frac{x}{D-x}

Does \frac{7m_2}{m_3}=\frac{1}{4} when m_3=\frac{7m_2}{4}?

You also haven't addressed my 2nd point.

 

[minimax]

shizz..
no, it is not 1/4...but 4 instead
4=\frac{x}{D-x}

as for your second point: the inverse proportionality of distance to Force...
the larger the distance, the smaller the force...

 

[Snazzy]

But it's not distance, it's distance squared.

 

[minimax]

so what you're saying is that I have to square my x and D-x values right?

 

[Snazzy]

Well, if the formula is given by F=\frac{Gm_1m_2}{r^2}, should you?

 

[minimax]

merde. i feel like I am going around in circles...

ok, so if i squared it

4=\frac{x^{2}}{(D-x)^{2}}
square rooting it, i get

2=\frac{x}{D-x}
cross multiplying and expanding...
x=\frac{2D}{3}

 

[Snazzy]

Looks good to me.

 

[minimax]

yay! thanks for your help and patience Snazzy!