G- Field of a sphere

Classical Mechanics Level 2

A solid sphere of uniform density with mass $M$ and radius $R$ has its center at $(0,0,0)$ which is the origin of the coordinate system. A spherical cavity of radius $\frac { R }{ 2 }$ having center at $C$ is made in the sphere by removing part of the material as shown in the figure above. Then the gravitational field intensity at $C$ is

0

\frac { 7 }{ 16 } \frac { GM }{ { R }^{ 2 } }

\frac { 2GM }{ { R }^{ 2 } }

\frac { GM }{ 2{ R }^{ 2 } }

2 solutions

Josh Silverman Staff
May 8, 2014

Before the cavity is carved out, the field at $C$ arises from the matter inside the sphere of radius $R/2$ about the origin. The strength of this field is given by

$g_C=GM_{R/2}\frac{1}{\left(R/2\right)^2} = G\frac{M}{2^3}\frac{2^2}{R^2} = \frac{GM}{2R^2}$

Since the gravitational field is additive, we can write this field as a sum of the field $g_{\mathcal{C}av}$ from the matter within the surface of the cavity $\mathcal{C}av$ , and the field $g_{\mathcal{S}\setminus\mathcal{C}av}$ from the rest of the sphere $\mathcal{S}\setminus\mathcal{C}av$ .

$g_C = g_{\mathcal{C}av} + g_{\mathcal{S}\setminus\mathcal{C}av}$

Clearly, the field at $C$ due to the cavity is zero, due to spherical symmetry. Therefore $g_{\mathcal{C}av}$ is zero and it doesn't matter if we remove it.

This shows that the field at $C$ is unchanged upon forming the cavity, and is the same as in the case where the cavity is filed in.

$\boxed{g_C = g_{\mathcal{S}\setminus\mathcal{C}av} = \frac{GM}{2R^2}}$

Sorry... the given answer is correct but josh your reasoning is wrong.. We don't account gravitational field from the outer part of the surface on which the point of calculation lies only when there is symmetry because inside a spherical shell everywhere the field is zero and inside solid sphere outside the point of consideration there are infinite shells of infinitesimal thickness up to the radius of whole sphere and these infinite shells contribute nothing coz the point lies inside them but here we see that there is no symmetry like that so we can't assume just like that...

Apoorva Asthana - 7 years ago

@Apoorva Asthana my argument is laid out in fairly clear steps. Why don't you just identify which of the following steps you disagree with:

Consider $\partial\mathcal{B}$ to be the boundary surface of the cavity.

the field at $\mathcal{C}$ due to the whole sphere (without cavity removed) is $GM/2R^2$
the field at $\mathcal{C}$ due to the whole sphere, is the superposition of the field due to mass in inside $\partial\mathcal{B}$ , and the field due to the mass outside $\partial\mathcal{B}$ .
the contribution to the field at $\mathcal{C}$ due to the mass inside $\partial\mathcal{B}$ is zero.

Josh Silverman Staff - 7 years ago

In your 2nd point... The field due to the whole sphere (without cavity removed) can be the superposition of the field due to cavity and the outer part of the cavity only when cavity and the outer part superimpose to give whole sphere but on one hand having empty cavity which means nothing and on the second hand having the outer part you can never get a full solid original sphere. The superposition of the field of cavity and the outer part will give you the field of this structure "having" cavity and not what you assumed to calculate the outer part's field coz u knew that cavity would contribute nothing...

Apoorva Asthana - 7 years ago

@Apoorva Asthana –

The field due to the whole sphere (without cavity removed) can be the superposition of the field due to cavity and the outer part of the cavity only when cavity and the outer part superimpose to give whole sphere

That's exactly what happens in this problem. By definition, if you put the material that was removed from the cavity, back into the cavity, you again have the whole sphere.

coz u knew that cavity would contribute nothing...

That's the trick.

Josh Silverman Staff - 7 years ago

When the cavity is made, shouldn't we replace $M$ by $\frac{7M}{8}$ ? Then the answer will be $\frac{7GM}{16r^{2}}$ !

Avineil Jain - 7 years ago

Are you suggesting we make your replacement in the final equation of my solution?

Josh Silverman Staff - 7 years ago

Actually, I made a mistake, the mass to be taken is $M$ only.

Avineil Jain - 7 years ago

@Avineil Jain – Could you explain this to me, then? The sphere originally has mass M, it is of uniform density, and we remove material amounting to 1/8 of its volume. Shouldn't this reduce its mass to 7/8 of M? And wouldn't that reduce the strength of the gravitational field by the same factor?

Thomas Raffill - 1 year ago

I also did the same way but I couldn't understand one point and I think you can explain me that

you did

GM R/2(R/2 in subscript) = G M/8 * 4/R^2

But why is mass getting divided by 8 or 2^3?

Kartik Sharma - 7 years ago

It's because the volume of the sphere of radius $R/2$ is $\frac18^\text{th}$ the volume of a sphere of radius $R$ , ( $V \sim R^3$ ), so it has $\frac18^\text{th}$ the mass. Is that more clear?

Josh Silverman Staff - 7 years ago

Okay, i got it.

Thanks!

Kartik Sharma - 7 years ago

Cant we apply gravitational guass law.?

Prem Sagar - 2 years, 2 months ago

I am still confused. We took a sphere of mass M of uniform density, and removed 1/8 of its material. Is not 7M/8 the mass remaining? How is that reflected in the solution?

Thomas Raffill - 1 year ago

Aditya Kelkar
Sep 16, 2014

Consider the sphere to be intact. Hence find the gravitational field at a distance (R/2) from the center. This will come out to be[GM/2(R^2)]. Note this down. Now consider the hollow part. By the shell theorem we know that inside a hollow sphere, the gravitational field is zero. Hence, the net field at the point c will be same as what would have been if the sphere would have been left intact. Hence, the answer.

G- Field of a sphere

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