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Astronomy Que

  • Thread starter minimax
  • Start date
minimax
Astronomy Que!!!!!

1. 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?

2. Homework Equations
F=[tex]\frac{Gm1m2}{r^{2}}[/tex]


3. 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

[tex]\frac{Gm1m2}{r^{2}}[/tex]=[tex]\frac{Gm2m3}{r^{2}}[/tex]

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

F=[tex]\frac{Gm3(4m3)}{D}[/tex]

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

Thank you
 

Answers and Replies

458
0
[tex]m_1=4m_3=7m_2[/tex]

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

[tex]\frac{Gm1m2}{x}[/tex]=[tex]\frac{Gm2m3}{D-x}[/tex]
[tex]\frac{G7m^{2}}{x}[/tex]=[tex]\frac{Gm2m3}{D-x}[/tex]

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

[tex]\frac{7m2}{m3}[/tex]=[tex]\frac{x}{D-x}[/tex]

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

so,
[tex]\frac{1}{4}[/tex]=[tex]\frac{x}{D-x}[/tex]

now to find for x?

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

x=[tex]\frac{D}{5}[/tex]

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!
 
Last edited by a moderator:
458
0
if 4m3=7m2
then m3=(7/4)m2

so,
[tex]\frac{G}{4}[/tex]=[tex]\frac{x}{D-x}[/tex]
1. Where did that G/4 come from?
2. [tex]F \propto \frac{1}{r^2}[/tex]
 
minimax
whoops, the G's cancel out too

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

so...[tex]\frac{1}{4}[/tex]=[tex]\frac{x}{D-x}[/tex]
to get x=D/5
 
458
0
[tex]\frac{7m2}{m3}[/tex]=[tex]\frac{x}{D-x}[/tex]

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

so,
[tex]\frac{1}{4}[/tex]=[tex]\frac{x}{D-x}[/tex]
Does [tex]\frac{7m_2}{m_3}=\frac{1}{4}[/tex] when [tex]m_3=\frac{7m_2}{4}[/tex]?

You also haven't addressed my 2nd point.
 
minimax
shizz..
no, it is not 1/4...but 4 instead
4=[tex]\frac{x}{D-x}[/tex]

as for your second point: the inverse proportionality of distance to Force...
the larger the distance, the smaller the force....
 
458
0
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?
 
458
0
Well, if the formula is given by [tex]F=\frac{Gm_1m_2}{r^2}[/tex], should you?
 
minimax
merde. i feel like I am going around in circles...

ok, so if i squared it

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

2=[tex]\frac{x}{D-x}[/tex]
cross multiplying and expanding...
x=[tex]\frac{2D}{3}[/tex]
 
458
0
Looks good to me.
 
minimax
yay! thanks for your help and patience Snazzy!
 

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