Question

If a point particle of a mass m is located inside, but not at the center of a uniform spherical shell of Mass M and radius R, which of the following is true?
Answer is The gravitational force on the point particle is zero. I know F= (G*m1*m2)/R^2 formula

Answer 1

Hi! What is your question?

Answer 2

I dont understand why the answer is correct

Answer 3

Do you know Gauss law for electrostatic field or gravitational field?

Answer 4

I remember a little bit but i dont understand it i mean if the particle is 1 mm inside of the surface does it still have no gravitational force action on it? Because on the surface there is a gravitational force right?

Answer 5

Since the mass distribution has a spherical symmetrical distribution, the gravitational field will be :
- radial (pointing towards the centre of the sphere)
- dependent only on the distance to the centre OM = r

Do you agree with this?

Answer 6

yes general formula being Fgravitation= (G*m1*m2)/R^2

Answer 7

True, but that formula will be of no real help here.

Answer 8

Now, Gauss law states that, on any closed surface S,

\(\Large \iint_S \vec g \cdot d \vec S = -4\pi \, G \sum M_{\text{int}}\)

Answer 9

Now, if you choose surface S as shown: sphere of centre O and radius r, then the integral (the flux) is equal to

\(-g(r)\times Area=-4\pi r^2\,g(r)\)

Since \(M_{\text { int}} = 0\) (no mass inside S), then g(r) = 0 as long as r < R

Answer 10

I'm sorry i don't quite follow, but as far as i understand, from complex calculations we end up proving that a particle inside earth's surface has no gravitational force acting on it? Am i right?

Answer 11

Well, inside that distribution (superficial mass), yes, but not inside Earth's surface.
Inside earth's surface, field is roughly proportional to radius r.

Answer 12

Ha, since earth's mass distribution is not homogeneous there exist a gravitational force inside earth. But if the sphere is UNIFORM we can say that gravitational force on a particle inside the sphere is zero?

Answer 13

No, it is because if the mass is superficial, then there is no mass below your feet => g = 0.

In the case of earth, even with a uniform symmetrical distribution, the distribution is in volume, and not just on the surface. So you will always have some mass below your feet, except at the very centre.

Answer 14

As I understand it - the mass of any object can be considered to act through its CoG. No matter if it is a shell, a solid sphere or a random shape.
The formula that you quote gives the attraction between 2 bodies due to their mass (i.e. the gravitational force).
It makes no difference that the point mass is inside the shell. If is not at the centre then there will be a resultant force acting towards the centre, given by the formula you quote. I therefore disagree with the given answer of zero force.

(Can Vincent explain why he discards the general equation so firmly?)

(I assume we are considering the 2 bodies in isolation - where other forces are negligible)

Answer 15

I do not discard the general equation. Actually, the structure of this relation is responsible for the validity of Gauss law. If the force was not an inverse-square one, then Gauss law would not hold for gravitational field.

It is just that proving the point that g =0 inside the hollow shell is very tiresome using that law; yet it can be proved.

Answer 16

Not wishing to start a flame!
I have not come across Gauss relating to gravity and understood that the F=G(m1*m2)/r^2 was a universal equation.

Are you saying that it is a special case - and does not apply in the circumstances given?
(I can see that it falls over when r=0)

If it does apply - then , since the CoG of the 2 masses are not co-incident then the resultant force is not zero.

In fact, rom the equation - the resultant force can only be zero if one of the masses is zero.

Maybe my understanding is simplistic.

Answer 17

As far as i know the equation is not valid inside the earth its valid as the object goes away from the surface of the earth. I found from the book that " When a point mass m is inside a uniform spherical shell of mass M, the potential energy is the same no matter where inside the shell the point is located. The force from the masses' mutual gravitational interaction is zero." university physics thirteenth international edition page 420.

Answer 18

F=G(m1*m2)/r^2 is a universal equation indeed, between point masses.

When you are far away from a mass, you can do as if it was concentrated at its CoM and reason with point masses.
But when you are within the massive object itself, you cannot reason using centres of mass anymore.
This is why the result g = 0 inside the hollow sphere seems counter intuitive.

Answer 19

Please excuse the persistence - it seems I have become the questioner - not the answerer.
At what point does the 'point mass' assumption break down? For instance - a satellite around the earth is very small compared to the earth, but conversely the earth is huge in contrast to the satellite. The circumference of the orbit can be relatively similar to the circumference of the earth - yet I understand that the simple equation applies in this instance.

I can see that intuitively the mass in the centre of the sphere has no resultant - since it is symmetrically attracted to all points on the sphere.
Once displaced from the centre it is subject to the integral of forces to all points - is it therefore a 'simple' matter that the integral of the above equation over the surface of a sphere sums to zero for all points within the sphere? (Wouldn't like to try to prove THAT one!)

Answer 20

It can be proved that the point-mass (m2) view holds for bodies (m2) with spherical symmetry (even with non-uniform density), as long as the test-mass is outside the distribution itself.

This is why Earth's gravitational field on the surface can be computed using the point-mass point of view.

If the test-mass (m1) is within the mass distribution, the formula holds with m2 being replaced by the internal mass m'2.