*Notes to a video lecture on http://www.unizor.com*

__Gravity Integration 3 -__

Thin Spherical Shell

Thin Spherical Shell

*1. Determine the potential of the gravitational field of a uniform infinitesimally thin spherical shell at any point outside it.*

Let's establish a system of coordinates with a spherical shell's center at the origin of coordinates and X-axis going through a point of interest

*, where we have to determine the gravitational potential.*

**P**Assume that the sphere's radius is

*and the mass is*

**R***. Then its surface is*

**M***and the mass density per unit of surface area is*

**4πR²***.*

**ρ=M/(4πR²)**Assume further that X-coordinate of a point

*, where we want to calculate the gravitational potential, is*

**P***, which is greater than the radius of a spherical shell*

**H***.*

**R**If, instead of a spherical shell, we had a point mass

*concentrated in its center at point*

**M***, its gravitational potential at a point*

**O(0,0,0)***would be*

**P**

**V**_{0}= −G·M/H(remember that the gravitational potential is negative, as it equals to amount of work needed to bring a unit of mass from infinity to point

*, and the field performs this work for us, so we perform negative work).*

**P**As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.

To calculate a gravitational potential of an infinitesimally thin spherical shell at point

*on the X-axis, let's divide a spherical shell into infinite number of infinitesimally thin rings that are parallel to the YZ-plane and, therefore, perpendicular to X-axis, that goes through a center of each ring.*

**P**The angle

*from X-axis (that is, from*

**φ***) to a radius from an origin of coordinates to any point on a ring will be our variable of integration.*

**OP**Then the radius of a ring will be

**r(φ) = R·sin(φ)**The distance from the origin of coordinates to a center of a ring is

*.*

**R·cos(φ)**The area of a ring between angles

*and*

**φ***will be equal to the product of infinitesimal width of a ring*

**φ+**d**φ***and its circumference*

**R·**d**φ**

**2πR·sin(φ)**Therefore, the mass of a ring will be

*d*

**m(φ) = ρ·2πR²·sin(φ)·**d**φ =**

= M·2πsin(φ)·d= M·2πsin(φ)·

**φ/(4π) =**

= M·sin(φ)·d= M·sin(φ)·

**φ/2**Knowing the mass of a ring

*d*, its radius

**m(φ)***and the distance from the ring's center to point of interest*

**r(φ)***, that is equal to*

**P***, we can use the formula of the ring's potential from a previous lecture*

**H−R·cos(φ)***V = −G·M /√R²+H²*

substituting

*d*instead of

**V(φ)***V*

*d*instead of

**m(φ)***M*

*instead of*

**H−R·cos(φ)***H*

*instead of*

**r(φ)***R*

Therefore,

*d*

**V(φ) = −G·**d**m(φ) /√r²(φ)+[H−R·cos(φ)]² =**

= −G·M·sin(φ)·d= −G·M·sin(φ)·

**φ /2√R²+H²−2R·H·cos(φ)**Now all we need is to integrate this by

*in limits from*

**φ***0*to

*π*.

Substitute

**y = √R²+H²−2R·H·cos(φ)**Incidentally, the geometric meaning of this value is the distance from point of interest

*to any point on a ring for a particular angle*

**P***.*

**φ**Then

*d*

**y = R·H·sin(φ)·**d**φ /√R²+H²−2R·H·cos(φ)**The limits of integration for

*from*

**φ***0*to

*π*in terms of

*are from |*

**y***| (which, for our case of point*

**H−R***being outside the sphere, equals to*

**P***) to*

**H−R***.*

**H+R**In terms of

**y***d*

**V(y) = −G·M·**d**y /(2R·H)**which we have to integrate by

*from*

**y***to*

**H−R***.*

**H+R**Simple integration of this function by

*on a segment [*

**y***] produces*

**H−R;H+R***in limits from*

**−G·M·y/(2R·H)***to*

**H−R***:*

**H+R**

**V = −G·M·(H+R)/(2R·H) +**

+ G·M·(H−R)/(2R·H) =

= −G·M/H+ G·M·(H−R)/(2R·H) =

= −G·M/H

**Remarkably, it's exactly the same gravitational potential, as if the whole mass was concentrated in a center of a spherical shell, as noted above as**.

**V**_{0}It's a justification for considering gravitational potential of space objects in many cases to be approximately equivalent to a potential of point-masses of the same mass concentrated in one point, the center of mass of an object.

*2. Determine the potential of the gravitational field of a uniform infinitesimally thin spherical shell at any point inside it.*

Using the same notation as in the previous case, this problem requires the distance from a point of interest

*to a center of a spherical shell*

**P***to be less than the radius*

**O***of a spherical shell.*

**R**Doing exactly the same manipulation and substitution

**y = √R²+H²−2R·H·cos(φ)**we see that the only difference from the previous case is in the limits of integration in terms of

*.*

**y**The limits of integration for

*from*

**φ***0*to

*π*in terms of

*are from |*

**y***| (which, in this case of point*

**H−R***being inside the sphere, equals to*

**P***) to*

**R−H***.*

**H+R**Integration by

*on a segment [*

**y***] produces*

**R−H;H+R***in limits from*

**−G·M·y/(2R·H)***to*

**R−H***:*

**H+R**

**V = −G·M·(H+R)/(2R·H) + G·M·(R−H)/(2R·H) =**

= −G·M/R= −G·M/R

Remarkably, it's constant and is independent of the position of point

*inside a spherical shell.*

**P**We have mentioned in the earlier lecture on gravitational field that in one dimensional case the gravitational force is a derivative of gravitational potential by distance from the source of gravity times mass of a probe object:

*F(r)=G·M·m /r²=m·**d*

**V(r)/**d**r**The fact that the gravitational potential is constant and, therefore, its derivative is zero, signifies that there is no force of gravity inside a spherical shell. The forces of gravity from all directions nullify each other.

An intuitive explanation of this is in the fact that, if you consider any conical surface with a vertex at point

*inside a sphere, cutting pieces of spherical shell's surface in both directions, the areas of the pieces will be proportional to a square of a distance from point*

**P***, while the gravitational forces produced by these pieces of surface are inversely proportional to a square of a distance from point*

**P***, thus both forces from opposite ends of a cone are equal in magnitude and opposite in direction, thus nullify each other.*

**P**