# Homework Help: Mutual attraction force - did I do something wrong?

1. Jan 21, 2012

### pyroknife

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

I attached the problem.

2. Relevant equations
F=G*m1*m2 / r^2

3. The attempt at a solution
Fs=mutual attraction force between particle and sun
Fe=mutual attraction force between particle and earth
mass of sun(ms)=330,000*mass of earth(me)
mp(mass of particle)
x=distance between centers of earth and sun=149.6*10^9m
Fs=G*mp*ms / d^2 = G*mp*me / (x-d)^2
I cancel out G and mp and cross multiply to get
ms*(x-d)^2 = me*d^2
330000*me*(x-d)^2=me*d^2
330000 (x^2-2dx+d^2) = d^2
330000(149.6*10^6)^2 - 2*(149.6*10^6)*d+329999*d^2=0

I then applied the quadratic formula and got d=0, 2.992*10^11 m. Do those look right? Also the last part wants me to "justify the 2 solutions physically." I'm having some trouble on how to do that. Can someone give me a hint?

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2. Jan 22, 2012

### Simon Bridge

Where would you expect to find the equilibrium point? About?

3. Jan 22, 2012

### PeterO

Something must be wrong here?

You say the centre of the Earth is 149.6 x 109m = 1.496 x 1011m

At the point where the attraction from the sun and Earth match, the body must be between the Earth and Sun.

Your answer of 2.992 x 1011m is twice as far from the sun, as the earth is; not less distant from the sun ??

4. Jan 22, 2012

### Simon Bridge

Yeah that's what I was asking about ... the other solution is 0 too - suggesting there is a second equilibrium point at the center of the Sun. That would be the case with no Earth. (Note: this is an example of justifying the solutions physically.)

OP needs to redo the dervition. Putting x=1 (so distance is in AU and convert as needed) will help - and don't substitute until the end - all those numbers just obscure what is going on.

5. Jan 22, 2012

### PeterO

I am not sure that both these figures can be correct. it is either 149.6 x 106 or 149.6 x 109

6. Jan 22, 2012

### ehild

You should draw a picture first. You would see that there are two position for the particle where the force from Earth is the same as that from the Sun - one position between the Sun and Earth, the other one outside. Try to show.

When you have the equation ms*(x-d)^2 = me*d^2 you can rewrite it as (√ms(x-d))^2-(√med)^2=(√ms(x-d)-√med) (√ms(x-d)-√med)=0 You get two solutions at once. Substitute the numbers at the end.

ehild

7. Jan 22, 2012

### pyroknife

Sorry I typo'ed, I had it right on my paper, it's 149.6*10^9m=149.6*10^6km. I checked my answer and still got the same answer.

8. Jan 22, 2012

### ehild

As I suggested already, you get the solution easier and more accurately by using the identity a^2-b^2=(a-b)(a+b) instead of applying the quadratic formula.
The equation 330000*(x-d)^2-d^2=0 can be written as

$$(\sqrt{330000}(x-d)-d)(\sqrt{330000}(x-d)+d)=0$$

One factor has to be zero, either $\sqrt{330000}(x-d)-d=0$, which involves the solution $d=\frac{x}{1+1/\sqrt{330000}}$, when the particle is between the Sun and Earth

or $\sqrt{330000}(x-d)+d=0$, that corresponds to $d=\frac{x}{1-1/\sqrt{330000}}$: the particle is on the opposite side as the Sun.

ehild

9. Jan 26, 2012

### pyroknife

I used your equations and got 1.49*10^11 m and 1.50*10^11 m.

Since the distance between sun and earth =1.496*10^11m. Would a good justification be that the two distances are really close to the earth because for the equation of mutual attraction, mass is in the numerator and since the mass of the sun is 330000 bigger than earth, the distance between the particle and earth has to be closer than the distance between the particle and sun?

That to me doesn't seem to justify both solutions. Like why is one on the opposite side of the earth ?

10. Jan 26, 2012

### PeterO

The origonal question/diagram certainly implied that the answer sought was between the Sun and Earth.
In the context of the original diagram that means the Sun pulling left, and the Earth pulling right with equal sized forces. One of your solutions is probably that.

It has previously mentioned that another possibility was a position to the right of the Earth [again referencing the original diagram] where the Sun and Earth both pull left - with a force of equal size.

11. Jan 26, 2012

### pyroknife

oh thanks a bunch, I was thinking too hard for the solution.

12. Jan 26, 2012

### SammyS

Staff Emeritus
I would expect one solution to be between the Sun and the Earth, much closer to Earth than it is to the Sun. The other solution should be beyond the Earths orbit.

13. Jan 26, 2012

### ehild

Yes, the distance must be much smaller from the Earth to make the forces of equal magnitude, as its mass is very much smaller that the mass of Sun.
There is one place at distance d1<X from the Sun and distance X-d1 from the Earth where the attraction forces are equal, (and point in opposite directions) and there is a place at distance d2>X from the Sun and d2-X distance from the Earth where the attraction forces are equal and both point towards the Earth.

As it is difficult to handle big numbers, try to solve the problem pretending that G=1, M(Sun)=9 and m(Earth)=1 and x=4. Use both your way of solution, solving the quadratic, and mine one. Calculate the forces at both positions.

ehild

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14. Jan 27, 2012

### Simon Bridge

Looks like that answer suffers from rounding errors doesn't it.
Did you, perhaps, do the calculator part is several stages or just round off to 2dp?

If you make the variable you want to find the distance of the equilibrium point from the Earth rather than from the Sun your calculator will throw up more useful numbers. Also helps to use scaled units as ehild points out.

Which is why I suggested AU for distance.

So if R is the earth-sun distance = 1AU and r= distance to the equilibrium point, the your equation becomes:$$\frac{GM}{(R-r)^2}=\frac{Gm}{r^2} \Rightarrow \left ( \frac{M}{m} \right ) r^2 = (1-r)^2$$... (M/m)=330000 ... but leave that number until after you've finished the algebra before putting it in.

I'm getting a little over 1000th of an AU each side so you really needed an extra couple of dp in your answers. (Knowing where to round off to takes practice.)

Last edited: Jan 27, 2012