# I Understanding point masses.

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1. Aug 20, 2016

### Faiq

"Understand that, for a point outside a uniform sphere, the mass of the sphere can be considered a point mass."
This is one of the "learning outcomes" in the A level Course. Here are my questions related to it.
1.) What is meant by point masses? (A definition will do)
2.) Why do we consider point masses? What's the problem with dealing a uniform spherical objects as normal spherical objects?
3.) What are the limitations for this consideration? When this consideration can't be used?
4.) They are dealing with "point outside a uniform sphere". Why is the consideration not suitable when the point is inside the uniform sphere?

2. Aug 20, 2016

### phinds

A point. That has mass.
Do you have any IDEA how complex the math would be if you did not consider it as a point mass?
It can be used exactly when the statement you quoted says it can be used.
Only the part of the mass inside the radius of the object under consideration can be counted as the point mass, not the entire sphere including the part that has a larger radius.

Last edited: Aug 20, 2016
3. Aug 20, 2016

### Staff: Mentor

There isn't a problem, it is just simpler math. For example, to calculate the gravitational acceleration a distance of $r$ away from a uniform spherical mass of radius $R$ (with $R<r$) and mass $M=\rho \frac{4}{3} \pi R^3$, you can either calculate:
$$a=GM/r^2$$
or you can calculate (assuming I even set up the math correctly)
$$a= \int_{-\sqrt{R^2-x^2-y^2}}^{\sqrt{R^2-x^2-y^2}} \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} \int_{-R}^R \rho G \frac{(r-x,y,z)}{\left( (r-x)^2+(y)^2+(z)^2 \right)^{3/2}} \; dx \, dy \, dz$$
The answer is the same.

4. Aug 21, 2016

### Faiq

Can you also answer the other questions I mentioned here?

5. Aug 21, 2016

### PeroK

You've got to work out the answer to 4) for yourself!

6. Aug 21, 2016

### phinds

What did you not find satisfactory about the answers I already gave you? Do you think they are wrong? Are they incomplete?

7. Aug 21, 2016

### Staff: Mentor

I don't have anything to add to the excellent response from @phinds

8. Aug 22, 2016

### Andrew Mason

A point mass in this context is the sphere treated as if all of its mass were located at a single point at its centre.
Someone has already done the math and it works out that, to a body outside the sphere, the contributions to gravity of all the matter in the uniform sphere are equivalent to the contribution to gravity of a point mass having the same mass as the sphere but located at the centre of the sphere. See Dale's post above.
I am not sure what this question is asking. The mass of the sphere must be uniformly distributed throughout (i.e. uniform density).
Consider a point located a distance s from the centre of the sphere but inside the sphere (ie. at a radius s < R, where R is the sphere radius). Newton, using calculus, showed that the mass that is located within the sphere at a radius greater than s (ie. a distance r from the centre where s < r ≤ R) does not affect the gravitational field at s. Only the mass within the sphere at a radius ≤ s contributes.

AM

9. Aug 22, 2016

### Faiq

No I dont consider them incomplete. I was expecting an answer which I can write in examination. If I wrote "a point that has mass" I would be deducted marks for incomplete definition or some sort of those things.

10. Aug 22, 2016

### Faiq

About your first question, I was asking when is this consideration not suitable. Example what are situations in which using point masses would give me wrong answer.
And can you please explain the sentences in quotation marks?

11. Aug 22, 2016

### phinds

Then use the more complete answer given by Andrew. It says the same thing really, but it's more elegant

12. Aug 22, 2016

### phinds

It says the same thing really, but it's more elegant but on the other hand it answers a more extensive question than what you asked. My answer is exactly the right answer for your exact question.

13. Aug 22, 2016

### Andrew Mason

The Wikipedia article gives a pretty thorough explanation of this.

AM