In classical mechanics, the shell theorem gives gravitational simplifications that can be applied to objects inside or outside a spherically symmetric body. This theorem has particular application to astronomy. Isaac Newton proved the shell theorem and stated that:
A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its center. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell. A corollary is that inside a solid sphere of constant density, the gravitational force within the object varies linearly with distance from the center, becoming zero by symmetry at the center of mass. This can be seen as follows: take a point within such a sphere, at a distance r {\displaystyle r} from the center of the sphere. Then you can ignore all of the shells of greater radius, according to the shell theorem (2). But the point can be considered to be external to the remaining sphere of radius r, and according to (1) all of the mass of this sphere can be considered to be concentrated at its centre. The remaining mass m {\displaystyle m} is proportional to r 3 {\displaystyle r^{3}} (because it is based on volume). The gravitational force exerted on a body at radius r will be proportional to m / r 2 {\displaystyle m/r^{2}} (the inverse square law), so the overall gravitational effect is proportional to r 3 / r 2 = r {\displaystyle r^{3}/r^{2}=r} , so is linear in r {\displaystyle r} . These results were important to Newton's analysis of planetary motion; they are not immediately obvious, but they can be proven with calculus. (Gauss's law for gravity offers an alternative way to state the theorem.) In addition to gravity, the shell theorem can also be used to describe the electric field generated by a static spherically symmetric charge density, or similarly for any other interaction whose force follows an inverse square law. The derivations below focus on gravity, but the results can easily be generalized to the electrostatic force.
Derivation of gravitational field outside of a solid sphere There are three steps to proving Newton's shell theorem (1). First, the equation for a gravitational field due to a ring of mass will be derived. Arranging an infinite number of infinitely thin rings to make a disc, this equation involving a ring will be used to find the gravitational field due to a disk. Finally, arranging an infinite number of infinitely thin discs to make a sphere, this equation involving a disc will be used to find the gravitational field due to a sphere. The gravitational field E {\displaystyle E} at a position called P {\displaystyle P} at ( x , y ) = ( − p , 0 ) {\displaystyle (x,y)=(-p,0)} on the x-axis due to a point of mass M {\displaystyle M} at the origin is E point = G M p 2 {\displaystyle E_{\text{point}}={\frac {GM}{p^{2}}}}
Suppose that this mass is moved upwards along the y-axis to the point ( 0 , R ) {\displaystyle (0,R)} . The distance between P {\displaystyle P} and the point mass is now longer than before; It becomes the hypotenuse of the right triangle with legs p {\displaystyle p} and R {\displaystyle R} which is p 2 + R 2 {\textstyle {\sqrt {p^{2}+R^{2}}}} . Hence, the gravitational field of the elevated point is:
E elevated point = G M p 2 + R 2 {\displaystyle E_{\text{elevated point}}={\frac {GM}{p^{2}+R^{2}}}}
The magnitude of the gravitational field that would pull a particle at point P {\displaystyle P} in the x-direction is the gravitational field multiplied by cos ( θ ) {\displaystyle \cos(\theta )} where θ {\displaystyle \theta } is the angle adjacent to the x-axis. In this case, cos ( θ ) = p p 2 + R 2 {\displaystyle \cos(\theta )={\frac {p}{\sqrt {p^{2}+R^{2}}}}} . Hence, the magnitude of the gravitational field in the x-direction, E x {\displaystyle E_{x}} is:
E x = G M cos θ p 2 + R 2 {\displaystyle E_{x}={\frac {GM\cos {\theta }}{p^{2}+R^{2}}}}
Substituting in cos ( θ ) {\displaystyle \cos(\theta )} gives
E x = G M p ( p 2 + R 2 ) 3 / 2 {\displaystyle E_{x}={\frac {GMp}{\left(p^{2}+R^{2}\right)^{3/2}}}}
Suppose that this mass is evenly distributed in a ring centered at the origin and facing point P {\displaystyle P} with the same radius R {\displaystyle R} . Because all of the mass is located at the same angle with respect to the x-axis, and the distance between the points on the ring is the same distance as before, the gravitational field in the x-direction at point P {\displaystyle P} due to the ring is the same as a point mass located at a point R {\displaystyle R} units above the y-axis:
E ring = G M p ( p 2 + R 2 ) 3 / 2 {\displaystyle E_{\text{ring}}={\frac {GMp}{\left(p^{2}+R^{2}\right)^{3/2}}}}
To find the gravitational field at point P {\displaystyle P} due to a disc, an infinite number of infinitely thin rings facing P {\displaystyle P} , each with a radius y {\displaystyle y} , width of d y {\displaystyle dy} , and mass of d M {\displaystyle dM} may be placed inside one another to form a disc. The mass of any one of the rings d M {\displaystyle dM} is the mass of the disc multiplied by the ratio of the area of the ring 2 π y d y {\displaystyle 2\pi y\,dy} to the total area of the disc π R 2 {\displaystyle \pi R^{2}} . So, d M = M ⋅ 2 y d y R 2 {\textstyle dM={\frac {M\cdot 2y\,dy}{R^{2}}}} . Hence, a small change in the gravitational field, E {\displaystyle E} is:
d E = G p d M ( p 2 + y 2 ) 3 / 2 {\displaystyle dE={\frac {Gp\,dM}{(p^{2}+y^{2})^{3/2}}}}
Substituting in d M {\displaystyle dM} and integrating both sides gives the gravitational field of the disk:
E = ∫ G M p ⋅ 2 y d y R 2 ( p 2 + y 2 ) 3 / 2 {\displaystyle E=\int {\frac {GMp\cdot {\frac {2y\,dy}{R^{2}}}}{(p^{2}+y^{2})^{3/2}}}}
Adding up the contribution to the gravitational field from each of these rings will yield the expression for the gravitational field due to a disc. This is equivalent to integrating this above expression from y = 0 {\displaystyle y=0} to y = R {\displaystyle y=R} , resulting in:
E disc = 2 G M R 2 ( 1 − p p 2 + R 2 ) {\displaystyle E_{\text{disc}}={\frac {2GM}{R^{2}}}\left(1-{\frac {p}{\sqrt {p^{2}+R^{2}}}}\right)}
To find the gravitational field at point P {\displaystyle P} due to a sphere centered at the origin, an infinite amount of infinitely thin discs facing P {\displaystyle P} , each with a radius R {\displaystyle R} , width of d x {\displaystyle dx} , and mass of d M {\displaystyle dM} may be placed together. These discs' radii R {\displaystyle R} follow the height of the cross section of a sphere (with constant radius a {\displaystyle a} ) which is an equation of a semi-circle: R = a 2 − x 2 {\textstyle R={\sqrt {a^{2}-x^{2}}}} . x {\displaystyle x} varies from − a {\displaystyle -a} to a {\displaystyle a} . The mass of any of the discs d M {\displaystyle dM} is the mass of the sphere M {\displaystyle M} multiplied by the ratio of the volume of an infinitely thin disc divided by the volume of a sphere (with constant radius a {\displaystyle a} ). The volume of an infinitely thin disc is π R 2 d x {\displaystyle \pi R^{2}\,dx} , or π ( a 2 − x 2 ) d x {\textstyle \pi \left(a^{2}-x^{2}\right)dx} . So, d M = π M ( a 2 − x 2 ) d x 4 3 π a 3 {\textstyle dM={\frac {\pi M(a^{2}-x^{2})\,dx}{{\frac {4}{3}}\pi a^{3}}}} . Simplifying gives d M = 3 M ( a 2 − x 2 ) d x 4 a 3 {\textstyle dM={\frac {3M(a^{2}-x^{2})\,dx}{4a^{3}}}} . Each discs' position away from P {\displaystyle P} will vary with its position within the 'sphere' made of the discs, so p {\displaystyle p} must be replaced with p + x {\displaystyle p+x} .
Replacing M {\displaystyle M} with d M {\displaystyle dM} , R {\displaystyle R} with a 2 − x 2 {\displaystyle {\sqrt {a^{2}-x^{2}}}} , and p {\displaystyle p} with p + x {\displaystyle p+x} in the 'disc' equation yields:
d E = ( 2 G [ 3 M ( a 2 − x 2 ) ] 4 a 3 ) a 2 − x 2 2 ⋅ ( 1 − p + x ( p + x ) 2 + a 2 − x 2 2 ) d x {\displaystyle dE={\frac {\left({\frac {2G\left[3M\left(a^{2}-x^{2}\right)\right]}{4a^{3}}}\right)}{{\sqrt {a^{2}-x^{2}}}^{2}}}\cdot \left(1-{\frac {p+x}{\sqrt {(p+x)^{2}+{\sqrt {a^{2}-x^{2}}}^{2}}}}\right)\,dx}
Simplifying,
∫ d E = ∫ − a a 3 G M 2 a 3 ( 1 − p + x p 2 + a 2 + 2 p x ) d x {\displaystyle \int dE=\int _{-a}^{a}{\frac {3GM}{2a^{3}}}\left(1-{\frac {p+x}{\sqrt {p^{2}+a^{2}+2px}}}\right)dx}
Integrating the gravitational field of each thin disc from x = − a {\displaystyle x=-a} to x = + a {\displaystyle x=+a} with respect to x {\displaystyle x} , and doing some careful algebra, yields Newton's shell theorem:
E = G M p 2 {\displaystyle E={\frac {GM}{p^{2}}}}
where p {\displaystyle p} is the distance between the center of the spherical mass and an arbitrary point P {\displaystyle P} . The gravitational field of a spherical mass may be calculated by treating all the mass as a point particle at the center of the sphere.
Outside a shell A solid, spherically symmetric body can be modeled as an infinite number of concentric, infinitesimally thin spherical shells. If one of these shells can be treated as a point mass, then a system of shells (i.e. the sphere) can also be treated as a point mass. Consider one such shell (the diagram shows a cross-section):
(Note: the d θ {\displaystyle d\theta } in the diagram refers to the small angle, not the arc length. The arc length is R d θ {\textstyle R\,d\theta } .) Applying Newton's Universal Law of Gravitation, the sum of the forces due to the mass elements in the shaded band is
d F = G m s 2 d M . {\displaystyle dF={\frac {Gm}{s^{2}}}dM.}
However, since there is partial cancellation due to the vector nature of the force in conjunction with the circular band's symmetry, the leftover component (in the direction pointing towards m {\displaystyle m} ) is given by
d F r = G m s 2 cos ( φ ) d M {\displaystyle dF_{r}={\frac {Gm}{s^{2}}}\cos(\varphi )\,dM}
The total force on m {\displaystyle m} , then, is simply the sum of the force exerted by all the bands. By shrinking the width of each band, and increasing the number of bands, the sum becomes an integral expression:
F r = ∫ d F r {\displaystyle F_{r}=\int dF_{r}}
Since G {\displaystyle G} and m {\displaystyle m} are constants, they may be taken out of the integral:
F r = G m ∫ cos ( φ ) s 2 d M . {\displaystyle F_{r}=Gm\int {\frac {\cos(\varphi )}{s^{2}}}\,dM.}
To evaluate this integral, one must first express d M {\displaystyle dM} as a function of d θ {\displaystyle d\theta }
The total surface of a spherical shell is
4 π R 2 {\displaystyle 4\pi R^{2}}
while the surface area of the thin slice between θ {\displaystyle \theta } and θ + d θ {\displaystyle \theta +d\theta } is
2 π R sin ( θ ) R d θ = 2 π R 2 sin ( θ ) d θ {\displaystyle 2\pi R\sin(\theta )R\,d\theta =2\pi R^{2}\sin(\theta )\,d\theta }
If the mass of the shell is M {\displaystyle M} , one therefore has that
d M = 2 π R 2 sin ( θ ) 4 π R 2 M d θ = 1 2 M sin ( θ ) d θ {\displaystyle dM={\frac {2\pi R^{2}\sin(\theta )}{4\pi R^{2}}}M\,d\theta ={\frac {1}{2}}M\sin(\theta )\,d\theta }
and
F r = G M m 2 ∫ sin ( θ ) cos ( φ ) s 2 d θ {\displaystyle F_{r}={\frac {GMm}{2}}\int {\frac {\sin(\theta )\cos(\varphi )}{s^{2}}}\,d\theta }
By the law of cosines,
cos ( φ ) = r 2 + s 2 − R 2 2 r s {\displaystyle \cos(\varphi )={\frac {r^{2}+s^{2}-R^{2}}{2rs}}}
and
cos ( θ ) = r 2 + R 2 − s 2 2 r R . {\displaystyle \cos(\theta )={\frac {r^{2}+R^{2}-s^{2}}{2rR}}.}
These two relations link the three parameters θ {\displaystyle \theta } , φ {\displaystyle \varphi } and s {\displaystyle s} that appear in the integral together. As θ {\displaystyle \theta } increases from 0 {\displaystyle 0} to π {\displaystyle \pi } radians, φ {\displaystyle \varphi } varies from the initial value 0 to a maximal value before finally returning to zero at θ = π {\displaystyle \theta =\pi } . At the same time, s {\displaystyle s} increases from the initial value r − R {\displaystyle r-R} to the final value r + R {\displaystyle r+R} as θ {\displaystyle \theta } increases from 0 to π {\displaystyle \pi } radians. This is illustrated in the following animation:
(Note: As viewed from m {\displaystyle m} , the shaded blue band appears as a thin annulus whose inner and outer radii converge to R sin ( θ ) {\displaystyle R\sin(\theta )} as d θ {\displaystyle d\theta } vanishes.) To find a primitive function to the integrand, one has to make s {\displaystyle s} the independent integration variable instead of θ {\displaystyle \theta } . Performing an implicit differentiation of the second of the "cosine law" expressions above yields
− sin ( θ ) d θ = − 2 s 2 r R d s {\displaystyle -\sin(\theta )\,d\theta ={\frac {-2s}{2rR}}\,ds}
and thus
sin ( θ ) d θ = s r R d s . {\displaystyle \sin(\theta )\,d\theta ={\frac {s}{rR}}\,ds.}
It follows that
F r = G M m 2 1 r R ∫ s cos ( φ ) s 2 d s = G M m 2 r R ∫ cos ( φ ) s d s {\displaystyle F_{r}={\frac {GMm}{2}}{\frac {1}{rR}}\int {\frac {s\cos(\varphi )}{s^{2}}}\,ds={\frac {GMm}{2rR}}\int {\frac {\cos(\varphi )}{s}}\,ds}
where the new integration variable s {\displaystyle s} increases from r − R {\displaystyle r-R} to r + R {\displaystyle r+R} . Inserting the expression for cos ( φ ) {\displaystyle \cos(\varphi )} using the first of the "cosine law" expressions above, one finally gets that
F r = G M m 4 r 2 R ∫ ( 1 + r 2 − R 2 s 2 ) d s . {\displaystyle F_{r}={\frac {GMm}{4r^{2}R}}\int \left(1+{\frac {r^{2}-R^{2}}{s^{2}}}\right)\ ds\ .}
A primitive function to the integrand is
s − r 2 − R 2 s , {\displaystyle s-{\frac {r^{2}-R^{2}}{s}}\ ,}
and inserting the bounds r − R {\displaystyle r-R} and r + R {\displaystyle r+R} for the integration variable s {\displaystyle s} in this primitive function, one gets that
F r = G M m r 2 , {\displaystyle F_{r}={\frac {GMm}{r^{2}}},}
saying that the gravitational force is the same as that of a point mass in the center of the shell with the same mass.
Spherical shell to solid sphere It is possible to use this spherical shell result to re-derive the solid sphere result from earlier. This is done by integrating an infinitesimally thin spherical shell with mass of d M {\displaystyle dM} , and we can obtain the total gravity contribution of a solid ball to the object outside the ball
F total = ∫ d F r = G m r 2 ∫ d M . {\displaystyle F_{\text{total}}=\int dF_{r}={\frac {Gm}{r^{2}}}\int dM.}
Uniform density means between the radius of x {\displaystyle x} to x + d x {\displaystyle x+dx} , d M {\displaystyle dM} can be expressed as a function of x {\displaystyle x} , i.e.,
d M = 4 π x 2 d x