# Gravitational force problem: finding r

1. Dec 18, 2012

### Apollinaria

This is embarrassing but I need to understand it. I'm trying to understand how to solve a problem but I don't understand how one of the variables was isolated for and I was hoping someone here could explain it please. It's seemingly basic algebra. And it's either really late and I'm tired or I'm dense.

So, how the heck did r end up alone?

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2. Dec 18, 2012

### Sourabh N

Exchange LHS numerator and RHS denominator. Take square root. Do you see it now?

3. Dec 18, 2012

### Apollinaria

Then take the sqrt of the whole thing?.... Not yet.

Edit: I need a step by step..

4. Dec 18, 2012

### Sourabh N

Step 0: $G\frac{m_{earth}m_{spacecraft}}{r^2} = G\frac{m_{moon}m_{spacecraft}}{(r_{earth-moon}-r)^2}$

Step 1: $\frac{(r_{earth-moon}-r)^2}{r^2} = G\frac{m_{moon}m_{spacecraft}}{Gm_{earth}m_{spacecraft}}$

Step 2: $\frac{(r_{earth-moon}-r)}{r} = \frac{\sqrt{m_{moon}m_{spacecraft}}}{\sqrt{m_{earth}m_{spacecraft}}}$

Step 3: $\frac{r_{earth-moon}}{r} - 1 = \frac{\sqrt{m_{moon}}}{\sqrt{m_{earth}}}$

5. Dec 18, 2012

### Apollinaria

The -1 was a +1 on the other side when it was the spacecraft?... Is this a quadratic?

6. Dec 18, 2012

### I like Serena

Hi Apollinaria!

No, this is not a quadratic.
And I'm not sure what you mean by the -1 being a +1 on the other side?
Anyway, the next step would be to add +1 on both sides, which cancels the -1 on the left side, and which as a +1 on the right side.
Is that what you meant?

To make r end up alone, you need to apply a series of fraction manipulations.
In this particular case, one of the steps is that:
$$\frac{r_{earth-moon} - r}{r} = \frac{r_{earth-moon}}{r} - \frac r r = \frac{r_{earth-moon}}{r} -1$$

Last edited: Dec 18, 2012
7. Dec 18, 2012

### Apollinaria

Hi PF, this problem came up in my homework assignment 2 months ago and the assignment online shows a solution that includes the correct formula. However, I still don't understand how to do the problem or how the formula is derived. A similar question is also on one of my worksheets and the problem has come to haunt me.

1. The problem statement, all variables and given/known data

On a trip to the sun, in a straight line from earth, there will be a point where the gravitational pull on the spaceship due to the earth is balanced by the pull due to the sun. How far is this point from earth? Include an FBD and give a complete explanation

Sun-Earth = 1.5E11 m
Mass Sun = 1.99E30 kg
Mass Earth = 5.98E24 kg

2. Relevant equations

F = Gm1m2 / r2

3. The attempt at a solution

I know that...
FS= FE

FS= G mSmship / r2
FE= G mEmship / r2

This is where I get confused. I attached the formula given online. And I understand that both forces will equal but I don't understand the denominator values.
I don't know how to isolate for r.

I had this posted originally as a thread on how to isolate for r but now I wonder if there is a DIFFERENT way to solve the problem entirely. Any insight?

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8. Dec 18, 2012

### Apollinaria

Hey there :) I was browsing online for solutions. Google, yahoo, youtube, everywhere and I found alternatives of it solved as a quadratic but those don't help me either and I'd rather not go there :rofl:

I see what you did there now! But I still can't begin to imagine how r is isolated. I've never done such a complex manipulation in all of my years of HS and uni (sad, I know). Or maybe I don't recall. It's frustrating. I wonder if there is a different way of solving the problem and have reposted the thread...

9. Dec 18, 2012

### I like Serena

Effectively it's a linear equation.
But you can only see that after a number of algebraic manipulations.
Sourabh showed a couple of steps, but did not complete it.

If you want to (re)learn how to do it, I guess you should start with simpler equations.
Like: 2x+3=5
Or 2/x = 3/(5-x)
Or 4/x^2=9/(5-x)^2

Do you know how to solve those?

10. Dec 18, 2012

### tiny-tim

Hi Apollinaria!
(your formula is using the moon rather than the sun)

No, that's the easiest and correct way to solve it.

Try multiplying both sides by the two denominators, you should get a straightforward quadratic equation in r …

show us what you get

11. Dec 18, 2012

### Staff: Mentor

Each force expression must use the distance between the corresponding masses. One uses the Sun-ship distance; the other uses the Earth-ship distance. Those distances must add up to the total Sun-Earth distance. So if one is called "r", the other is "Total Distance - r".
Just multiply it out. Do what you need to do to get rid of those denominators.

12. Dec 18, 2012

### Apollinaria

Hi Tim! :)
I was hoping to avoid using the quadratic because we haven't used it this semester, nor does he encourage it. I will be a few moments...

13. Dec 18, 2012

### Apollinaria

I see what you're saying now. So, what I'm actually looking for is d and not r?
I will multiply it out in a few min.
Also, does it matter which is the r and which is the d-r? Or can I attach it to either expression? :P
And, will the denominator be (d-r)2 OR (d-r2)2? :uhh:

14. Dec 18, 2012

### Apollinaria

Can't I cross multiply to get rid of the denominators?...

15. Dec 18, 2012

### Staff: Mentor

No. "d" is the total distance from Sun-Earth, which is given.
Since they want the distance measured from earth, call that distance "r".
d-r is a distance, so (d-r)2 is a distance squared. That makes sense.

d-r2 is I don't know what. (You can't subtract a distance squared from a distance.)

16. Dec 18, 2012

### Staff: Mentor

Sure.

17. Dec 18, 2012

### Apollinaria

Sorry, office with scanner in it is occupado. Had to take a crappy photo.

I haven't learned how to use the forum sqrt thingies and such yet. Will get on that later.

Edit: WOW, worser quality than I thought.

d√(Gmsmship) - r√(Gmsmship) = r√(GmEmship)

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18. Dec 18, 2012

### Apollinaria

Oooooo, I think I have it..... Sec.

19. Dec 18, 2012

### Apollinaria

Negative. Yeah, I have no idea.

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20. Dec 18, 2012

### Staff: Mentor

Looks like you've got the right idea to me.

Note that certain factors cancel out, like G and the mass of the ship.