Gravitational potential of a sphere

APPENDIX

GRAVITATIONAL POTENTIAL OF A SPHERE

E

Fig. E.1 shows a point mass m with Cartesian coordinates (x, y, z) as well a system of N point masses m1, m2, m3, ⋯, mN. The ith one of these particles has mass mi and coordinates (xi, yi, zi). The total mass of the N particles is M, M¼

N X

mi

(E.1)

i¼1

The position vector drawn from mi to m is ri and the unit vector in the direction of ri is ^i ¼ u

ri ri

FIG. E.1 A system of point masses and a neighboring test mass m.

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APPENDIX E GRAVITATIONAL POTENTIAL OF A SPHERE

The gravitational force exerted on m by mi is opposite in direction to ri, and is given by Fi ¼ 

Gmmi Gmmi ^i ¼  3 ri u ri 2 ri

The potential energy of this force is Vi ¼ G

mmi ri

(E.2)

The total gravitational potential energy of the system due to the gravitational attraction of all the N particles is V¼

N X

Vi

(E.3)

i¼1

Therefore, the total force of gravity F on the mass m is F ¼ rV ¼ 

  ∂V ^ ∂V ^ ∂V ^ i+ j+ k ∂x ∂y ∂z

(E.4)

where r is the gradient operator. Consider the solid sphere of mass M and radius R0 illustrated in Fig. E.2. Instead of a discrete system as above, we have a continuum with mass density ρ. Each “particle” is a differential element dM ¼ ρdv of the total mass M. Eq. (E.1) becomes ððð

M¼

ρ dv v

FIG. E.2 Sphere with a spherically symmetric mass distribution.

(E.5)

APPENDIX E GRAVITATIONAL POTENTIAL OF A SPHERE

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where dv is the volume element, and v is the total volume of the sphere. In this case, Eq. (E.2) becomes dV ¼ G

m dM ρ dv ¼ Gm r r

where r is the distance from the differential mass dM to the finite point mass m. Eq. (E.3) is replaced by ððð V ¼ Gm v

ρ dv r

(E.6)

Let the mass of the sphere have a spherically symmetric distribution, which means that the mass density ρ depends only on r0 , the distance from the center C of the sphere. An element of mass dM has spherical coordinates (r0 , θ, ϕ), where the angle θ is measured in the xy plane of a Cartesian coordinate system with origin at C, as shown in Fig. E.2. In spherical coordinates the volume element is dv ¼ r0 sinϕ dϕ dr 0 dθ 2

(E.7)

Therefore Eq. (E.5) becomes 2ðπ

R ð0

ðπ

M¼

1 0R 1 0 2π 10 π 10 R ð0 ð ð ð0 2 2 ρr sin ϕ dϕ dr 0 dθ ¼ @ dθA@ sin ϕ dϕA@ ρr 0 dr0 A ¼ ð2π Þð2Þ@ ρr 0 dr 0 A 02

θ¼0 r 0 ¼0 ϕ¼0

0

0

0

0

so that the mass of the sphere is given by R ð0

ρr 0 dr 0 2

M ¼ 4π

(E.8)

r 0 ¼0

Substituting Eq. (E.7) into Eq. (E.6) yields 2π ð

R ð0

ðπ

V ¼ Gm θ¼0 r 0 ¼0 ϕ¼0

2R 0 π 3 1 ð0 ð ρr 0 2 sin ϕ dϕ dr 0 dθ sinϕ dϕ Aρr 0 2 dr 0 5 ¼ 2πGm4 @ r r 0

(E.9)

0

The distance r is found by using the law of cosines,  1=2 2 r ¼ R2 + r 0  2r 0 R cos ϕ

where R is the distance from the center of the sphere to the mass m. Differentiating this equation with respect to ϕ, holding r0 constant, yields 1=2 dr 1  2 0 2 r 0 R sinϕ ð2r0 R sinϕ dϕÞ ¼ ¼ R + r  2r0 Rcos ϕ dϕ 2 r

so that sin ϕ dϕ ¼

r dr r0 R

It follows that ðπ ϕ¼0

sinϕ dϕ 1 ¼ 0 r rR

Rð + r0

dr ¼ Rr 0

2 R

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APPENDIX E GRAVITATIONAL POTENTIAL OF A SPHERE

Substituting this result along with Eq. (E.8) into Eq. (E.9) yields V¼

GMm R

(E.10)

We conclude that the gravitational potential energy, and hence (from Eq. E.4) the gravitational force, of a sphere with a spherically symmetric mass distribution M is the same as that of a point mass M located at the center of the sphere.