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!