# Superposition of gravitational forces

1. Dec 10, 2015

### struggles

1. The problem statement, all variables and given/known data
Three uniform spheres are fixed at positions shown in the figure.

i) What is the magnitude of the force on a 0.0150-kg particle placed at P?
ii) What is the direction of this force?
iii) If the spheres are in deep outer space and a 0.0150-kg particle is released from rest 300 m from the origin along a line 45∘ below the -x-axis, what will the particle's speed be when it reaches the origin?
2. Relevant equations
Fg = -GMm/r2

3. The attempt at a solution
I think I've got the answers for part i) F = 9.67×10-12 N
ii) 45°.
I got i) by simply working out the force due to gravity on P due to each of the spheres and then resolving them to see the component of the force along the 45° angle.

However part iii) i'm struggling with. Obviously the force will increase as P gets closer to the origin. This will mean that the acceleration will increase.
I think Fg ∝ 1/r2 and the acceleration a = Fg/m.
Because the acceleration is non constant I think I have to integrate a with respect to t to find v. However there is no mention of t anywhere in this question! From here I'm not quite sure where to go and any help would be greatly appreciated!

2. Dec 10, 2015

### TSny

Are you familiar with the formula for the gravitational potential energy of two point masses separated by a distance r?

3. Dec 10, 2015

### RUber

You are right to think you should integrate.
If you simplify the 3 spheres to one representative mass M, how big would it be and where would it be?

Next, you have
$F_g = - \frac{GMm}{r^2}$
and
$a = F_g/m$ which gives you a form for acceleration in terms of distance.
Integrating over r instead of t should work just fine.

4. Dec 10, 2015

### struggles

Do you mean U = -GMm/r Tsny? Could I work out the potential energy at the distance of 300m and then using conservation of energy set it equal to U + KE of the origin?
How would I simplify the masses into 1 mass? I can see that it would be on the 45° line connecting the 2kg mass and the origin but not sure how I could work out the mass and exact location. We have done this in class but I'm struggling to make sense of both my notes and the textbook!

5. Dec 10, 2015

### TSny

That's the formula for U that I was suggesting. Yes, you can use conservation of energy.

300 m is so far away that you can just lump together the three spheres and think of the lumped mass as located at the origin. This will be accurate enough since your data has at most 3 significant figures. In fact, you could probably consider 300 m as "infinitely far" and still get a good answer.

But, of course, you cannot lump together the three masses of the spheres when the particle is at the origin.

6. Dec 10, 2015

### azizlwl

I think you can find COM of the three masses. Then you can find the energy(KE) to take the particle away from origin to 300m away, using integral here since force not constant

Just like throwing an object upward to a point, on earth, and return to the point of launching.

Last edited: Dec 10, 2015
7. Dec 10, 2015

### haruspex

No, that doesn't work. Consider two point masses. The potential half way between them is not the minimum.
You need to consider the potential due to each sphere separately and add them.

8. Dec 10, 2015

### azizlwl

If we take a line 45° line above x-axis, and consider 2 masses 1kg, the pull at any point is along that line will along that line only 45°, since two perpendicular force along the line pull by the 2 masses will be cancelled.

2kg mass is on the 45° line.

9. Dec 10, 2015

### haruspex

That's a different argument, relying on symmetry, not on CoM.
If you were to take some arbitrary point, such as (0, 1m), the net force from the three spheres would not be towards their CoM.
You probably understand that, but your post #6 could mislead a student into thinking that reducing the three spheres to their CoM works generally.

10. Dec 10, 2015

### azizlwl

Its a matter of solving part(iii)
To me that's how to solve that particular question.
Its all there as clue to the answer.
1. It is place 45° below - x axis, will it hard for student if placed elsewhere.
2. It passes the origin, will it passes origin if put somewhere else.

So we can deduce that it moves along that line passing try origin.
Why it moves it that manner, there be pulling it along that line.
Do we have calculate how heavy is Everest mountain, but we know object falling to
Earth with its COM at center of earth.

11. Dec 10, 2015

### haruspex

All true, but how does the CoM matter there? You haven't said so specifically, but I feel you are suggesting that the distance to the common CoM can be used to calculate the force. That is not the case. You would have to calculate the force of attraction to each sphere and add them vectorially. This will make the whole process somewhat harder than just using addition of potentials (as TSny suggests).
Bodies with radial symmetry are rather special. The field due to a uniform shell, outside that shell, is the same as if the mass were concentrated at the shell's centre.
To a first approximation, the Earth consists of nested uniform shells, so we can make that simplification. More accurately, it is an oblate spheroid. At mid latitudes, the gravitational force is not quite directed towards the Earth's centre. (The net force in the observer's frame is also redirected a little because of Earth's spin, but that's another story.)

12. Dec 10, 2015

### azizlwl

Thanks, I rest my case.

13. Dec 10, 2015

### haruspex

Not so fast.
In your post #6 you made a reference to CoM that may be highly misleading to the student. Please put this right either by satisfactorily explaining what you meant or withdrawing it.

14. Dec 12, 2015

### struggles

Thank you for all of your responses! I understand how to use conservation of energy but am still a bit confused over how to
. As far as I can see only a component of the 1kg would act as the other component would cancel each other out (although mass isn't a vector force so maybe I'm just getting myself very muddled!) How would I work out what the mass at the origin would be?

15. Dec 12, 2015

### TSny

When the particle is 300 m away from the origin, the potential energy between the particle and the three spheres is going to be very small. I suspect that you are meant to treat 300 m as essentially the same as "infinitely far away".

However, if you don't want to totally neglect the potential energy at 300 m away, you may consider all three spheres as being combined as one total mass of 4 kg located at the origin. The error that results from this simplification is very small because the distance of 300 m is so much larger than the roughly 1 meter separation between the spheres.

I think you will find that treating 300 m as "infinitely far" will give you an answer for the particle's speed at the origin that is accurate to at least three significant figures.

[Potential energy is not a vector quantity, so it doesn't have components that cancel out.]

16. Dec 13, 2015

### struggles

Thank you so much! I have finally managed to arrive at the correct answer.
Merry Christmas!