Gravitational Fields, Superposition, and Continuous Mass Distributions

(Topic)

Gravitational Fields, Superposition, and Continuous Mass Distributions

M07-01 described gravity as a force between masses.

The gravitational-field viewpoint separates the source of gravity from the test object used to probe it.

r,

(1)

define The Gravitational Field by

|-----------| | Fg- | g (r) = m , | -------------

(2)

where

m

(3)

is a test mass placed at that point.

Thus the gravitational force on a test mass is

|----------| -Fg-=-mg.--|

(4)

The field is a property of the source masses and their arrangement.

It does not depend on the mass of the test object when gravitational and inertial mass are identified in the Newtonian model.

Figure 1. Source masses create a gravitational field in the surrounding space. A test mass placed at a point experiences force F_g = mg.

1 Units and dimensions of the gravitational field

Because

g = F ∕m,

(5)

the SI units are

|------| -N∕kg.--

(6)

Using

1N = 1kg m ∕s2,

(7)

we obtain

|----------------| 1 N ∕kg = 1m ∕s2.| ------------------

(8)

Thus gravitational field and gravitational acceleration have the same dimensions.

For a freely falling test object,

|a =-g.| --------

(9)

2 Field of a single point mass

Place a source mass

M

(10)

at position

r′.

(11)

A test mass

m

(12)

r

(13)

experiences

′ F = GM m -r--−-r-. g |r′ − r|3

(14)

Divide by

m :

(15)

|--------------------| | r′ − r | |g(r) = GM --------.| ------------|r′ −-r|3-

(16)

If the source is at the origin,

′ r = 0,

(17)

then

|----------------| | r | |g(r) = − GM -3.| --------------r---

(18)

The magnitude is

|------------| | GM---| g (r) = r2 .| --------------

(19)

The direction is radially inward.

Figure 2. The gravitational field of a point mass is radial, inward, and inverse square in magnitude. The field is undefined at the idealized point source itself.

3 Field versus force

The distinction is important.

The field

g

(20)

is created by the source masses.

The force

Fg

(21)

depends additionally on the test mass:

|----------| |Fg = mg. | -----------

(22)

If the test mass doubles:

the gravitational field at that location is unchanged,
the gravitational force doubles.

4 Example 1: field of a small source mass

Let

M = 5.00kg

(23)

and

r = 2.00 m.

(24)

Then

g	=	(25)
	=	(26)
	= 8.34 × 10⁻¹¹ m∕s².	(27)

2.00 kg

(28)

test mass at that point would experience force magnitude

|--------------------| |F = 1.67 × 10 −10N. | ---------------------

(29)

5 Superposition of gravitational fields

Newtonian gravity is linear in the source masses.

If several point masses are present, the total field is the vector sum of the individual fields:

|-------∑--------| g (r) = gj(r).| | j | ------------------

(30)

For point masses

Mj

(31)

at positions

rj,

(32)

|--------------------------| | ∑ -rj-−-r- | |g(r) = G Mj |rj − r|3. | ------------j--------------|

(33)

Once

g

(34)

is known, the force on any test mass is

mg.

(35)

Figure 3. Gravitational fields obey vector superposition. The net field is the vector sum of the fields produced by all source masses.

6 Example 2: two equal masses at a midpoint

Let equal source masses

M

(36)

be placed at

x = +a

(37)

and

x = − a.

(38)

At the origin, the field from the right mass points in the

+x

(39)

direction.

The field from the left mass points in the

− x

(40)

direction.

The magnitudes are equal:

2 GM ∕a .

(41)

Therefore

|----------| -g(0)-=-0.-|

(42)

This is cancellation by symmetry.

The field is zero at the midpoint even though each source produces a nonzero field there.

7 Example 3: unequal collinear masses

Place

M1

(43)

x = 0

(44)

and

M 2

(45)

x = d.

(46)

At a point between them,

0 < x < d,

(47)

the fields point in opposite directions.

Taking

+x

(48)

to the right,

|------------------------| | GM1-- -GM2---- | |gx = − x2 + (d − x)2.| --------------------------

(49)

A zero-field point satisfies

M1--= --M2----. x2 (d − x )2

(50)

Taking positive square roots,

√M---- √M---- -----1= ----2. x d − x

(51)

Solving,

|---------√--------| | d M1 | |x = √-------√----.| -------M1--+---M2---

(52)

The balance point lies closer to the smaller mass.

8 From discrete masses to continuous matter

Real objects contain enormous numbers of particles.

Instead of summing individual masses, we replace the sum by an integral.

An infinitesimal source mass

dm ′

(53)

r′

(54)

produces

|--------------------| | r′ − r ′ | |dg = G |r′ −-r|3dm .| ---------------------

(55)

Therefore

|---------∫--------------| | -r′ −-r- ′| g (r) = G |r′ − r|3dm .| --------------------------

(56)

This is the basic gravitational-field integral for a continuous mass distribution.

9 Line mass density

For a thin wire or rod, define line mass density

|--------| | dm | λ = d-ℓ . ----------

(57)

Its units are

|------| -kg∕m.--

(58)

Then

|----------| dm--=-λ-dℓ.-

(59)

If the density varies along the object,

λ = λ (ℓ).

(60)

10 Surface mass density

For a thin sheet, disk, or shell, define surface mass density

|---------| Σ = dm-. | -----dA----

(61)

Its units are

|-------| | 2 | kg-∕m-.--

(62)

Then

|------------| -dm--=-Σ-dA.-|

(63)

11 Volume mass density

For a three-dimensional body, define volume mass density

|--------| | dm--| ρ-=--dV-.-

(64)

Its units are

|-------| kg-∕m3.--

(65)

Then

|-----------| dm--=-ρ-dV.--

(66)

Figure 4. Continuous mass models use line density λ, surface density Σ, or volume density ρ according to whether the source is idealized as one-, two-, or three-dimensional.

12 The three common gravitational integrals

For a line distribution,

|---------∫------------------| | ′ r ′ − r ′ | |g(r) = G λ(ℓ)|r′ −-r|3dℓ .| ------------------------------

(67)

For a surface distribution,

|------------------------------| | ∫ r′ − r | |g(r) = G Σ (r′)--′----3dA ′. | ------------------|r-−-r|------

(68)

For a volume distribution,

|---------∫------------------| | ′ -r′ −-r- ′| g (r) = G ρ(r )|r′ − r|3dV . ------------------------------

(69)

The structure is always the same.

The only change is the mass element.

13 A reliable setup procedure

For continuous gravitational-field problems:

Choose coordinates that match the geometry.
Identify the observation point.
Write the source position r′.
Write the vector from observation point to source element.
Write dm′ using the correct density.
Form dg.
Use symmetry before integrating.
Integrate only the components that survive.
Check units and limiting cases.

Symmetry often removes most of the vector algebra.

14 Symmetry before calculus

Suppose a mass distribution is symmetric about an axis.

At a point on that axis, source elements related by rotation produce transverse field components that cancel.

Only the axial component survives.

Similarly, for a distribution symmetric about a plane, some field components cancel in pairs.

Using symmetry first can turn a difficult vector integral into one scalar integral.

15 Field on the axis of a uniform ring

Consider a thin ring of:

radius a,
total mass M,
center at the origin,
plane in the xy plane.

Find the field at

(0,0,z).

(70)

Every ring element lies the same distance from the observation point:

|----√---------| |s = a2 + z2.| ---------------

(71)

By rotational symmetry, all transverse components cancel.

Only the

z

(72)

component survives.

Figure 5. For a point on the symmetry axis of a uniform ring, transverse field components cancel in pairs and only the axial component survives.

16 Derivation of the ring field

For a mass element

dm,

(73)

the field magnitude is

dg = G --dm--- . a2 + z2

(74)

The axial projection factor is

----z---- √a2-+--z2.

(75)

For

z > 0,

(76)

the field points toward the ring plane, so

dgz = − G ---zdm-----. (a2 + z2)3∕2

(77)

The geometric factor is constant around the ring.

Therefore

g_z	= −G ∫ dm	(78)
	= −GM.	(79)

Thus

|------------------------| | -----z------^ | |g = − GM (a2 + z2)3∕2z.| -------------------------

(80)

17 Checks on the ring result

At the center,

z = 0,

(81)

|------| |g = 0.| --------

(82)

Far from the ring,

z ≫ a.

(83)

Then

(a2 + z2)3∕2 ≈ z3,

(84)

and

|------------| | GM | |gz ≈ − --2-.| ---------z----

(85)

The ring looks like a point mass from far away.

Near the center along the axis,

|z | ≪ a.

(86)

Then

2 2 3∕2 3 (a + z ) ≈ a ,

(87)

|--------------| |gz ≈ − GM--z. | ---------a3----|

(88)

Along the symmetry axis, the field is approximately linear near the center.

18 Example 4: numerical ring field

Let

M = 10.0kg,

(89)

a = 0.200 m,

(90)

and

z = 0.100 m.

(91)

Then

\|g_z\|	= GM	(92)
	≈ 5.97 × 10⁻⁹ m∕s².	(93)

The direction is toward the ring plane.

19 Field of a finite uniform rod on its perpendicular bisector

Consider a uniform rod:

extending from x = −a to x = +a,
lying on the x axis,
line density λ.

Find the field at

(0,y ), y > 0.

(94)

A source element at

(x,0)

(95)

has mass

dm = λ dx.

(96)

The distance to the observation point is

∘ -------- s = x2 + y2.

(97)

By left-right symmetry, horizontal components cancel.

Only the vertical component remains.

Figure 6. On the perpendicular bisector of a uniform rod, horizontal field components cancel by symmetry and only the component directed toward the rod survives.

20 Derivation of the rod field

The field element magnitude is

dg = G --λdx---. x2 + y2

(98)

The vertical projection factor is

----y----- ∘ x2 + y2.

(99)

Thus

ydx dgy = − G λ--2----2-3∕2. (x + y )

(100)

Integrate:

g_y	= −Gλy ∫ _−a^a	(101)
	= −Gλy_−a^a.	(102)

Therefore

|------------------| | 2G λa | gy = − -∘--------.| --------y--a2-+-y2--

(103)

Since

M = 2a λ,

(104)

this can also be written

|------------------| | GM | gy = − -∘--------.| --------y--a2-+-y2--

(105)

21 Far-field check for the rod

y ≫ a,

(106)

then

∘ -2----2 a + y ≈ y.

(107)

Therefore

|------------| |g ≈ − GM--.| | y y2 | --------------

(108)

Again, a localized extended body looks like a point mass from sufficiently far away.

22 Infinite-line limit

Hold

λ

(109)

fixed and let

a → ∞.

(110)

From

2G λa gy = − -∘--2----2, y a + y

(111)

we obtain

|------------| | 2G-λ-| gy → − y .| --------------

(112)

An infinite line produces a

1∕y

(113)

field rather than an inverse-square field.

The geometry of the source changes how the field spreads.

23 Field on the axis of a uniform disk

Consider a thin disk:

radius R,
uniform surface density Σ,
lying in the xy plane.

Find the field at

(0,0,z ), z > 0.

(114)

Build the disk from concentric rings.

A ring of radius

s

(115)

and width

ds

(116)

has area

dA = 2πs ds.

(117)

Its mass is

|--------------| dm = 2 πΣs ds.| ----------------

(118)

Figure 7. A uniform disk can be decomposed into concentric rings. Each ring contributes only an axial field component at a point on the symmetry axis.

24 Derivation of the disk field

Using the ring result,

dg = − G ---zdm-----. z (s2 + z2)3∕2

(119)

Substitute

dm = 2π Σs ds :

(120)

dg = − 2πG Σz ----sds-----. z (s2 + z2)3∕2

(121)

Integrate from

s = 0

(122)

s = R :

(123)

g_z	= −2πGΣz ∫ ₀^R	(124)
	= −2πGΣz₀^R.	(125)

For

z > 0,

(126)

|-------------[-------------]--| | z | |gz = − 2πG Σ 1 − √--2----2- .| ---------------------z-+--R-----

(127)

The direction is toward the disk.

25 Infinite-sheet limit

Let

R → ∞.

(128)

Then

√---z-----→ 0. z2 + R2

(129)

Thus

|--------------| |g → − 2πG Σ. | --z-------------

(130)

The field of an infinite uniform sheet is independent of distance from the sheet.

This differs strikingly from:

point mass: g ∝ 1∕r²,
infinite line: g ∝ 1∕r,
infinite sheet: constant g.

26 Far-field check for the disk

For

z ≫ R,

(131)

expand

----z----- ------1----- R2-- √ -2----2-= ∘ -----2--2-≈ 1 − 2z2. z + R 1 + R ∕z

(132)

Then

g_z	≈−2πGΣ	(133)
	= −.	(134)

Since

M = πR2 Σ,

(135)

|------------| | GM---| |gz ≈ − z2 .| --------------

(136)

The finite disk also approaches point-mass behavior in the far field.

27 Example 5: numerical disk field

Let

2 Σ = 5.00 kg∕m ,

(137)

R = 0.400m,

(138)

and

z = 0.200 m.

(139)

Then

\|g_z\|	= 2πGΣ	(140)
	≈ 1.16 × 10⁻⁹ m∕s².	(141)

28 Why source dimension changes distance scaling

The field of one isolated point source spreads through three-dimensional space.

Its magnitude decreases as

1∕r2.

(142)

An infinite line has translational symmetry along one direction.

The remaining transverse spreading gives

1∕r.

(143)

An infinite sheet has translational symmetry in two directions.

Its field does not weaken with perpendicular distance.

These results can be understood geometrically even before introducing Gauss’s Law for gravity.

Figure 8. Source geometry strongly affects distance scaling: point-like sources give inverse-square behavior, an infinite line gives inverse-distance behavior, and an infinite sheet gives a constant field magnitude.

29 Far-field point-mass behavior

For any localized mass distribution with total mass

M,

(144)

the leading far-field behavior is typically

|-------------| g ≈ − GM r- | -----------r3--

(145)

when

r

(146)

is much larger than the source size.

The detailed shape of the source enters through smaller corrections.

A systematic treatment of those corrections leads to multipole expansions, which belong later in the gravitation sequence.

30 Field at symmetry points

Symmetry can make the field vanish.

Examples include:

the center of a uniform ring,
the midpoint between two equal point masses,
the center of many symmetric mass configurations.

A zero field does not imply a zero gravitational environment.

Nearby points may experience nonzero fields.

The spatial variation of the field determines tidal effects.

31 The field of a spherical body

A spherically symmetric body requires special treatment.

Outside such a body, the field turns out to be exactly the same as that of a point mass containing the same total mass:

r g = − GM --3. r

(147)

Inside a spherical shell, the net gravitational field vanishes.

These are Newton’s shell theorems.

They are important enough to deserve a separate derivation.

M07-03 develops those results from geometry and superposition.

32 Field singularities and idealizations

The point-mass field

g = GM ∕r2

(148)

diverges as

r → 0.

(149)

This singularity is a consequence of the point-particle idealization.

Real extended bodies have finite size and a distributed density.

Their fields must be computed using the actual mass distribution when the observation point lies within or very near the body.

Similarly:

an infinitely thin line,
an infinitely thin sheet,
an infinite spatial extent

are ideal mathematical models.

They are useful when the physical geometry is well approximated by those limits.

33 Gravitational field lines

A gravitational field line is drawn so that its tangent points in the direction of

g.

(150)

For positive mass in Newtonian gravity:

field lines point toward mass,
field-line density can be used qualitatively to indicate field strength,
field lines do not represent physical strings or trajectories.

A freely falling particle initially at rest begins accelerating tangent to the local field, but its later path need not remain on one field line because of inertia.

34 Common mistakes

Confusing gravitational field with gravitational force.
Multiplying by the test mass when only the field is requested.
Forgetting that the field points toward positive source mass.
Adding field magnitudes instead of vectors.
Using 1∕r² with the wrong separation vector.
Forgetting that a zero field at a symmetry point can result from cancellation.
Mixing line, surface, and volume density units.
Writing dm = ρdA or dm = Σ dV .
Integrating vector components that symmetry guarantees will cancel.
Forgetting the projection factor in ring, rod, or disk problems.
Applying point-mass behavior at distances comparable to the source size without justification.
Counting the fixed geometry incorrectly when taking an infinite-line or infinite-sheet limit.
Assuming every extended source retains an inverse-square dependence at all distances.
Applying the spherical point-mass result before establishing the shell theorem.
Treating ideal singularities as literal predictions for finite physical matter.

35 Practice exercises

Derive the point-mass field from Newton’s gravitational-force law.
Show that N∕kg is equivalent to m∕s².
Reproduce Example 1 for a 5 kg source mass at 2 m.
For two equal point masses, prove that the field vanishes at the midpoint.
Derive the zero-field location between two unequal collinear point masses.
Write the gravitational-field integral for line, surface, and volume densities.
Derive the field on the symmetry axis of a uniform ring.
Show that the ring field approaches point-mass behavior for |z|≫ a.
Show that the ring field is linear in z near the center along the symmetry axis.
Derive the field on the perpendicular bisector of a finite uniform rod.
Take the infinite-line limit of the finite-rod result.
Derive the on-axis field of a uniform disk by integrating rings.
Take the infinite-sheet limit of the disk result.
Show that a finite disk approaches point-mass behavior in the far field.
Explain geometrically why point, infinite-line, and infinite-sheet sources have different distance scalings.