# Gravitational force problem: finding r

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?

#### Attachments

• Screen shot 2012-12-17 at 11.20.37 PM.png
15.4 KB · Views: 398

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

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

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

Edit: I need a step by step..

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

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

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

I like Serena
Homework Helper

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

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:
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.

## Homework Statement

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

F = Gm1m2 / r2

## 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?

#### Attachments

Last edited:

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:
$$\frac{r_{earth-moon} - r}{r} = \frac{r_{earth-moon}}{r} - \frac r r = \frac{r_{earth-moon}}{r} -1$$

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...

I like Serena
Homework Helper

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?

tiny-tim
Homework Helper
Hi Apollinaria! 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?

(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 Doc Al
Mentor
And I understand that both forces will equal but I don't understand the denominator values.
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".
I don't know how to isolate for r.
Just multiply it out. Do what you need to do to get rid of those denominators.

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 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...

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.

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:

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

Doc Al
Mentor
I see what you're saying now. So, what I'm actually looking for is d and not r?
No. "d" is the total distance from Sun-Earth, which is given.
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
Since they want the distance measured from earth, call that distance "r".
And, will the denominator be (d-r)2 OR (d-r2)2?
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.)

Doc Al
Mentor
Can't I cross multiply to get rid of the denominators?...
Sure.

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

Doc Al
Mentor
Negative. Yeah, I have no idea.
Looks like you've got the right idea to me.

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

No. "d" is the total distance from Sun-Earth, which is given.

Since they want the distance measured from earth, call that distance "r".

Doc, I think I misinterpreted what you said...

The r2 goes under the mE.
The (d-r)2 goes under the ms

I did it this way and got the correct answer.