# Newton's argument, gravitational force inside a sphere

• PFuser1232
In summary: So, if you were to define the gravitational field as ##g(\vec{x})##, it would be discontinuous at ##r=R##, and its jump would be given by the mass contained in the sphere of radius R. In summary, the reason why gravitational force vanishes inside a spherical shell can be seen by a simple argument due to Newton. Consider the two small mass elements marked out by a conical surface with its apex at m. The amount of mass in each element is proportional to its surface area. The area increases as (distance)2. However, the strength of the force varies as 1/(distance)2, where the distance is measured from the apex to the shell. Thus the forces of the two mass
PFuser1232
This is an excerpt from "Introduction to Mechanics" by Kleppner and Kolenkow:

"The reason why gravitational force vanishes inside a spherical shell can be seen by a simple argument due to Newton. Consider the two small mass elements marked out by a conical surface with its apex at ##m##.

The amount of mass in each element is proportional to its surface area. The area increases as (distance)2. However, the strength of the force varies as 1/(distance)2, where the distance is measured from the apex to the shell. Thus the forces of the two mass elements are equal and opposite, and cancel. We can pair up all the elements of the shell this way and so the total force on ##m## is zero."

Are the cones similar because of the spherical symmetry of the shell?

Yes, in a word, it is due to symmetry.
In a few more words:
Consider a 1 Kg mass at point P some distance inside the spherical shell, not at the center.

Area A1 = θr1^2, for thickness t Volume = θr1^2t, Mass = θr1^2tp
The gravitational Force exerted on the 1 kg mass by this section of shell is
F = G(θr1^2tp) / r1^2 = Gθtp
Using the same method the mass at Area A2 = θr2^2tp
F = G(θr2^2tp) / r2^2 = Gθtp
The gravitational forces are exactly equal and cancel out and this holds true for all of the outer shell material taking all of the forces in pairs.
But be careful to note this only applies to a spherical shell. Inside a solid sphere there is also a solid inner sphere that will exert a gravitational force on the object that varies with the distance from the center. But if that point is exactly at the center so that there is no inner sphere then the total gravitational force is zero.

PFuser1232 and Delta2
Tom_K said:
Yes, in a word, it is due to symmetry.
In a few more words:
Consider a 1 Kg mass at point P some distance inside the spherical shell, not at the center.

Area A1 = θr1^2, for thickness t Volume = θr1^2t, Mass = θr1^2tp
The gravitational Force exerted on the 1 kg mass by this section of shell is
F = G(θr1^2tp) / r1^2 = Gθtp
Using the same method the mass at Area A2 = θr2^2tp
F = G(θr2^2tp) / r2^2 = Gθtp
The gravitational forces are exactly equal and cancel out and this holds true for all of the outer shell material taking all of the forces in pairs.
But be careful to note this only applies to a spherical shell. Inside a solid sphere there is also a solid inner sphere that will exert a gravitational force on the object that varies with the distance from the center. But if that point is exactly at the center so that there is no inner sphere then the total gravitational force is zero.

So the area of the spherical segment is ##r^2 \theta##?
Also, what is the gravitational force experienced at ##r = R## (for a thin shell of radius ##R##)?

So the area of the spherical segment is ##r^2 \theta##?
Yes but keep in mind that θ in this case is the solid angle, measured in steradians. Since the surface area of a sphere = 4πR^2 you can deduce there are exactly 4π steradians of solid angle in a sphere.
Also, what is the gravitational force experienced at ##r = R## (for a thin shell of radius ##R##)?

For all points inside the shell the gravitational force is zero, including the point at the center where r = R.

So the area of the spherical segment is ##r^2 \theta##?
Also, what is the gravitational force experienced at ##r = R## (for a thin shell of radius ##R##)?

At the shell, the force is not well defined, but this does not matter much for any type of application. In reality, you cannot have this situation, your assumption that the shell is thin means quite some idealisation, as does the assumption of a point like test particle.

Tom_K said:
For all points inside the shell the gravitational force is zero, including the point at the center where r = R.
The center is r=0, r=R is the shell itself.

Orodruin said:
At the shell, the force is not well defined, but this does not matter much for any type of application. In reality, you cannot have this situation, your assumption that the shell is thin means quite some idealisation, as does the assumption of a point like test particle.The center is r=0, r=R is the shell itself.

I was referring to the drawing I posted earlier, in which case the distances r1 and r2 are not measured from the center of the sphere but from the object inside the sphere to the outer shell. In that case, when r = R the object is at the center of the sphere. When stating any measurement, point of reference should be clearly stated also.

Tom_K said:
In that case, when r = R the object is at the center of the sphere.

Yes, but I do not believe this is what OP was referring to, there is nothing special about that point (other than being the centre of the mass distribution and having complete rotational symmetry around it, meaning the force there is always zero).

The quickest modern argument is that Newtonian gravitation is described by Poisson's equation
$$\Delta g=4 \pi \gamma \rho.$$
For a spherically symmetric mass distribution around the origin of the coordinate system, with ##\rho(\vec{x})=0## for ##r=|\vec{x}|<R## the unique solution in this region is $$g(\vec{x})=\text{const}$$.

Outside of the mass distribution, i.e., if ##\rho(\vec{x})=0## also for ##r \geq R'##, the unique solution there is
$$g(\vec{x})=-\frac{\gamma M}{r}, \quad M=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \rho(\vec{x}).$$

## What is Newton's argument about gravitational force inside a sphere?

Newton's argument states that the gravitational force inside a uniform spherical body with a consistent density is directly proportional to the mass of the object and inversely proportional to the square of the distance from the center of the sphere.

## How does Newton's argument apply to real-life objects?

This argument applies to real-life objects such as planets and stars, as they can be approximated as uniform spheres with consistent densities. It helps explain why objects in space have a consistent gravitational force, and why the force increases as you get closer to the center of the object.

## What are the implications of Newton's argument for objects inside the Earth?

According to Newton's argument, the gravitational force inside the Earth would decrease as you move closer to the center of the Earth. This means that objects at the center of the Earth would experience less gravitational force compared to objects on the surface.

## Does Newton's argument hold true for non-spherical objects?

No, Newton's argument is specifically for objects that can be approximated as uniform spheres. For non-spherical objects, the distribution of mass is not consistent, so the gravitational force would vary and cannot be accurately calculated using this argument.

## How does Newton's argument relate to the concept of escape velocity?

Newton's argument helps explain the concept of escape velocity, which is the minimum speed an object needs to escape the gravitational pull of a planet or star. This is because the gravitational force inside a sphere decreases as you move away from the center, so the closer you are to the surface, the stronger the gravitational force and the higher the escape velocity needs to be.

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