# Homework Help: Gravitational field between two planets.

1. Jan 23, 2008

### harelo

[SOLVED] Gravitational field between two planets.

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

Planet B and Planet A are separated by a distance of R.

2. Relevant equations

At what distance will the net gravitational field be 0, express R in terms of Planet A.

HINT: Use Gravitational Force.

3. The attempt at a solution

Since I don't need a numerical value for this, I had an algebraic approach to it, and since I know that if Fg = 0 then G would be = 0.

Fga = -Fgb

Gmamo/Ra^2 = -(Gmbmo/(Ra-R)^2

Where Ra is the ratio from the test mass to Planet A, and mo is the test mass

However, after I try all my algebraic arrangements to isolate Ra, I always end up square rooting a negative number, which is impossible.

What am I doing wrong?

PS: Anything that is not bolded is a subscript. Sorry about the messy report, the superscripts and subscripts aren't working properly for some reason :S

2. Jan 23, 2008

### rock.freak667

If the distance between the test mass and one planet is x then the other distance is R-x
If the mass of the test mass is B then the force between the test mass and one planet is the same as the test mass of the other planet and the test mass

so that

$$\frac{GMB}{x^2}=\frac{GmB}{(R-x)^2}$$

reducing to

$$\frac{M}{x^2}=\frac{m}{(R-x)^2}$$

$$\frac{M}{m}=\frac{x^2}{(R-x)^2}$$

$$\frac{M}{m}=(\frac{x}{R-x})^2$$

3. Jan 23, 2008

### G01

You should be able to get this answer using only magnitudes of the force vectors. You want to find the Ra, where the magnitude of both fields is zero, which you should be able to find with the negative problem. Try working with only magnitudes and not working with the negative sign.

4. Jan 23, 2008

### harelo

M being the mass of planet A and m being the test of Planet B I'm assuming?

and where did the negative that was there go to?

Sorry I just want to understand the problem before I do it blindly

G01, I went under the assumption that the forces should be equal in magnitude but different in direction, I don't think thats a wrong assumption, right?

5. Jan 23, 2008

### G01

How does doing almost all the calculations for the OP actually help them in the end?

6. Jan 23, 2008

### G01

That is definitely a correct assumption, but you shouldn't have to worry about direction here. The forces will ALWAYS be opposite in direction along the line between the planets, so the only thing changed by the position in this case is the magnitude of each force.

So, you can say that you want to find the position where the magnitudes are equal, since the direction will always be correct. Thus, this is why the minus sign will disappear, since we are setting the magnitudes of each force equal to each other.

Does this make sense?

7. Jan 23, 2008

### harelo

Yeah that makes perfect sense, so lets recap:

but by having:

$$\frac{GMB}{x^2}=\frac{GmB}{(R-x)^2}$$

I understand that you divide out the gravitational constant and the test mass, leaving us with

$$\frac{M}{x^2}=\frac{m}{(R-x)^2}$$

and by cross multiplication, you get

$$\frac{M}{m}=\frac{x^2}{(R-x)^2}$$

However, after this point, I am not sure how you would go ahead and isolate that X. perhaps:

$$\sqrt(\frac{M}{m})=\frac{x}{R-x}$$

But I'm somewhat lost after there :S

EDIT

After further factoring, I ended up with this:

$$\frac{R\sqrt(\frac{M}{m})}{1+\sqrt(\frac{M}{m})}=x$$

Am I right or am I completely out to lunch?

Last edited: Jan 23, 2008
8. Jan 23, 2008

### G01

Try this:

Take the inverse of both sides of your last line. Then, can you divide through by the lonely x that will be in the denominator on the one side? This should help isolate it.

9. Jan 23, 2008

### harelo

Not sure if my edit appeared on time or not, but:

After further factoring, I ended up with this:

$$\frac{R\sqrt(\frac{M}{m})}{1+\sqrt(\frac{M}{m})}=x$$

Am I right or am I completely out to lunch?

10. Jan 23, 2008

### G01

I don't get that answer. Can you show your factoring? Maybe my answer is just simplified more and we're both right, but I can't tell from your end result.

11. Jan 23, 2008

### harelo

Will do

$$\sqrt(\frac{M}{m})=\frac{x}{R-x}$$

I multiply the B radius to both sides

$$(R-x)\sqrt(\frac{M}{m})=x$$

I expand (R-x) out of the brackets

$$R\sqrt(\frac{M}{m})-x\sqrt(\frac{M}{m})=x$$

Move the expanded x to the other side

$$R\sqrt(\frac{M}{m})=x+x\sqrt(\frac{M}{m})$$

Factor out the x

$$R\sqrt(\frac{M}{m})=x(1+\sqrt(\frac{M}{m}))$$

Divide out to isolate x

$$\frac{R\sqrt(\frac{M}{m})}{1+\sqrt(\frac{M}{m})}=x$$

12. Jan 23, 2008

### G01

Ok. That work looks correct to me. I agree with your answer. Good job!

13. Jan 23, 2008

### harelo

Thank you :) would you mind sharing what your answer was with me tho? I'd like to know if you had another approach

14. Jan 23, 2008

### G01

No problem. Here is how I obtained my answer:

$$\sqrt{\frac{M}{m}}=\frac{x}{R-x}$$

Invert both sides:

$$\sqrt{\frac{m}{M}}=\frac{R-x}{x}$$

Split the right side into two fractions:

$$\sqrt{\frac{m}{M}}=\frac{R}{x}-\frac{x}{x}$$

$$\sqrt{\frac{m}{M}}=\frac{R}{x}-1$$

It should not be too hard to solve for x from here.

You should get the same answer either way.

Last edited: Jan 23, 2008
15. Jan 23, 2008

### harelo

yeah just tried with made up values and I got the same answer, thanks a lot for the help :)

16. Jan 23, 2008

### G01

Anytime! Good Work!