Inequalities in Force of Gravitation between Three Bodies

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1. Nov 19, 2014

jantdroid

1. The problem statement, all variables and given/known data
Given that the force of gravitation between Planet A (the one in the left side of the drawing), Fa=3000/da2 and the force of gravitation between Planet B and the rocket, Fb= 6000/ db2. Assuming that the three bodies involved is in stationary. What are the distances (ranges) from Planet A where the gravitational force is above 0.001 Newtons?

2. Relevant equations

3. The attempt at a solution
First, I solved if the Fa>0.001,

Fa=3000/da2
0.001<3000/da2
1732.054< da2

Therefore: 0 < da2 < 1732.054

Then I calculated if Fb>0.001
0.001<6000/db2
2442.75<db2

I subtracted the 2442.75 km from 6 000 000 km because the question asks what distances from planet A where the gravitational force is above 0.001 N.
5997557.25<da

0<da<5997557.25

Not sure if my answer was correct though, I'm still concerned with the middle body which is the rocket.

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2. Nov 20, 2014

Simon Bridge

That won't work - no.
Do you need the distance along the line joining planets A and B?

Planets A and B are separated by distance D. At a distance x along the line from A to B, the magnitude of the force of gravity towards A is given by: $F= F_A-F_B = \cdots$ ... you finish it.
Note: you'll need to express $d_A$ and $d_B$ in terms of x.

3. Nov 20, 2014

jantdroid

I think so, the question kinda confuses me.

"What are the distances (ranges) from Planet A to B where the gravitational force is above 6 000 000 and I'm concerned about the rocket in the middle too.

4. Nov 20, 2014

Simon Bridge

Which question is confusing you, there are several?
The trick with this question is to work out the equation for the sum of the forces - just do that part.
Don't use any of the given values - use letters for them instead.

5. Nov 20, 2014

jantdroid

The one that says find the distance(s) where gravitational force is above 0.001 N

6. Nov 20, 2014

Simon Bridge

OK - so did you follow the suggestion?

7. Nov 21, 2014

jantdroid

Yah, the equation is on the 4th degree. Not sure how to proceed with it by picking out the critical points and do some test points. :3

8. Nov 21, 2014

BiGyElLoWhAt

Are you saying the sum of the forces is a degree 4 polynomial? Because if it is, I think you made a mistake somewhere.

9. Nov 21, 2014

jantdroid

Yup, my classmate tried to solved it too still get the 4th degree equation. lol

10. Nov 21, 2014

Simon Bridge

I'm getting a 4th order polynomial that can be reduced to a 2nd order one.

Last edited: Nov 21, 2014
11. Nov 21, 2014

jantdroid

12. Nov 21, 2014

jantdroid

F=FA−FB

Since the force needed is greater than 0.001N

0.001< 3000/da2 - 6000/db2

transposing the 0.001 value to the right side of the equation,
0< 3000/da2 - 6000/db2 - 0.001

Finding the LCD of the solution

0< ( 3000db2-6000da2-0.001da2db2 ) db2 da2

Given that
db= 6 000 000 - da

then,

db2= (6 000 000)2 - 12 000 000 da + da2

substitute all to db and b2

0 < 3000( 3x1013 - 12 000 000 da + da2) - 6000 da2 - 0.001 (3x1013 - 12 000 000 da + da2) da2 ) / da2 (3.6x10 13 - 12 000 000 da + da2 )

13. Nov 21, 2014

Simon Bridge

Checking - I made a mistake in my quick once over :)

It's easier to handle big number symbolically:

putting x=d_A, (makes it easier to type) then d_B=D-x: D=6,000,000km
putting G=3000N.km2 means I can write the net force in the direction of planet A as:

$$F=\frac{G}{x^2}-\frac{2G}{(D-x)^2}$$ ... put H=F/G and multiply through by the common denominator:
$\implies H(D-x)^2x^2 = D^2-2Dx-x^2$
$\implies HD^2x^2-2HDx^3+Hx^4= D^2-2Dx-x^2$
$\implies Hx^4-2HDx^3 +(HD^2+1)x^2 + 2Dx - D^2 = 0$

You should double-check my working...
But you get the idea?

The solution we want lies between the planets, which corresponds to the only positive value of x < D.

$F=0.001\text{N} \implies H=(1/3000000) \text{km}^{-2} = 2/D$
... which simplified the equation a bit.

It is possible to solve a 4th order polynomial analytically, or you may prefer a numerical solution via Newton/Raphson.

Last edited: Nov 21, 2014
14. Nov 21, 2014

jantdroid

I don't quite get where the "H" came from and what exactly is the answer (distance) from planet a to b

15. Nov 21, 2014

jantdroid

Here, me trying to apply your method,

Let F> 0.001

0.001 < G (D-x)2 - 2G (x2) / (D-x)2 (x2)

0.001 < G (D2- 2Dx +x2) - 2Gx / (D2-2Dx+x2) (x2)

0.001 < GD2 - 2GDx + Gx2 -2Gx2/ X2D2 - 2Dx3 + x4

0 < GD2- 2GDx - Gx2 / x2D2- 2Dx3 + x4 - 0.001

Finding the LCD:

0 < ( GD2- 2GDx -Gx2 - 0.001 D2x2 - 0.002 Dx3 + 0.001 x4 / x2D2 - 2x3 +x4 ) x2D2 - 2x3 + x4

and now, I am lost. lol

16. Nov 22, 2014

Simon Bridge

H is what you get by dividing the first equation through by the common factor of G.
The distance from planet A to planet B is given in the problem statement - I gave this distance the label D, so D=6000000km.

The problem want you to find the distance from planet A to the place where the gravity experienced by the ship is above 0.001N strong.

Put F = 0.001N ... in order for the force to be stronger than this, the ship must be closer to planet A.

There are two places between A and B where the force will have this magnitude, but only one where the direction is towards planet A.

... I didn't bother to check the algebra ... I had 0.001/G = H because it's tidier at the outset.
I chose "H" because it's next in the alphabet from F and G.

Now you have to decide how to solve for x.
You can either use the link I gave you, or find x where F=0 as your starting point for a numerical calculation.

17. Nov 22, 2014

jantdroid

∠⊆⊆⊕⊕→ℝℝβ
Nice! Thanks for giving me this technique on making equations neat, I'll study how to get that quartic equation right away. I'll get back to you once I've confirmed the answer. Thank you! :D