Gravitational Potential Energy and Gravitational Potential

(Topic)

Gravitational Potential Energy and Gravitational Potential

Newtonian gravity is a conservative interaction.

That means gravitational work can be represented by a potential-energy function.

For two point masses

M

(1)

and

m

(2)

separated by distance

r,

(3)

the gravitational force magnitude is

GM m F = --r2--.

(4)

With the source mass

M

(5)

fixed at the origin, the vector force on

m

(6)

|----------------| | GM---m- | |Fg = − r2 ^r.| -----------------

(7)

We will derive the potential energy from the work done by this force.

Figure 1. Moving a test mass radially through a point-mass gravitational field. The force points inward while the radial displacement may point inward or outward.

1 Work done by gravity

For an infinitesimal displacement

dr,

(8)

the gravitational work is

dWg = Fg ⋅ dr.

(9)

For radial motion,

dr = dr ^r.

(10)

Therefore

dW_g	= −r ⋅ dr r	(11)
	= −dr.	(12)

Integrating from

r1

(13)

r2,

(14)

W_g	= −GMm∫ _r₁^r₂	(15)
	= GMm_r₁^r₂.	(16)

Thus

|------------(--------)--| | 1 1 | Wg = GM m --− -- .| ---------------r2---r1----

(17)

2 Potential-energy difference

For any conservative force,

|----------------------| -ΔU--=-−-Uwork-=-−-Wg.--

(18)

Therefore

( ) U (r2) − U (r1) = − GM m 1- − -1 . r2 r1

(19)

Rearranging,

|------------------------------------| | ( 1 1) | |U (r2) − U (r1) = GM m -- − -- .| --------------------------r1---r2----

(20)

Only differences in potential energy are physically required by mechanics.

The absolute zero may be chosen conveniently.

3 Choosing the zero at infinity

For an isolated gravitational system, the standard convention is

|----------| U-(∞-)-=-0.-

(21)

Let

r1 = ∞

(22)

and

r2 = r.

(23)

Then

-1 = 0. r1

(24)

Hence

|----------------| | GM--m--| U (r) = − r .| ------------------

(25)

The negative sign is essential.

Figure 2. With the zero chosen at infinity, point-mass gravitational potential energy is negative and approaches zero from below as separation increases.

4 Why gravitational potential energy is negative

With

U (∞ ) = 0,

(26)

a finite separation has

U (r) < 0.

(27)

This means energy must be supplied to separate the masses completely to infinity with zero final speed.

Conversely, allowing masses to fall together releases gravitational potential energy into kinetic energy or other forms.

Negative potential energy is therefore a statement about the chosen zero and the bound nature of the attractive interaction.

5 Force from potential energy

For a one-dimensional conservative force,

dU- Fr = − dr .

(28)

Using

GM m U (r) = − ------, r

(29)

(30)

Therefore

|--------------| | GM m | Fr = − ---2--.| ----------r-----

(31)

The negative sign recovers the inward gravitational force.

In vector form,

|------------| |Fg = − ∇U. | -------------

(32)

6 Gravitational potential

Potential energy depends on the test mass.

It is useful to divide by that mass and define the gravitational potential:

|--------| | U- | |Φ = m .| ---------

(33)

For a point source mass,

|--------------| | GM---| Φ (r) = − r .| ----------------

(34)

The SI units are

|------| -J∕kg.-|

(35)

Since

2 2 1 J∕kg = 1 m ∕s,

(36)

potential can also be written in units of

2 2 m ∕s .

(37)

7 Potential versus potential energy

The distinction is parallel to field versus force.

Gravitational potential:

|------------| | GM---| |Φ = − r | -------------

(38)

depends only on the source and location.

Potential energy:

|--------| U--=-m-Φ--

(39)

also depends on the test mass.

Doubling the test mass doubles

U

(40)

but leaves

Φ

(41)

unchanged.

8 Field from potential

Since

Fg = mg

(42)

and

Fg = − ∇U,

(43)

while

U = m Φ,

(44)

we obtain

g-=--−-∇Φ.-| ------------

(45)

For spherical symmetry,

|----------| | dΦ | gr = − ---.| --------dr--

(46)

Using

Φ = − GM ∕r,

(47)

|------------| |g = − GM--.| --r------r2---

(48)

Figure 3. The Gravitational Field points in the direction of decreasing potential. For a point mass, the negative radial slope of Φ = −GM∕r gives the inward inverse-square field.

9 Potential difference and gravitational work

Between two points,

|------------∫---------| | 2 | Φ2 − Φ1 = − g ⋅ dr. ---------------1--------

(49)

For a test mass

m,

(50)

|-----------------------| U2 − U1 = m (Φ2 − Φ1 ). | -------------------------

(51)

The gravitational work is

|------------------| |Wg = m (Φ1 − Φ2).| --------------------

(52)

10 Path independence

Because Newtonian gravity is conservative, the work between two fixed endpoints does not depend on the path.

For a point mass source,

GM g = − --2-^r. r

(53)

Tangential displacement contributes no work because it is perpendicular to the radial field.

Only the change in radius matters.

Therefore all paths connecting the same

r1

(54)

and

r2

(55)

produce the same gravitational work.

11 Equipotential surfaces

An equipotential surface has constant

Φ.

(56)

For a point mass,

Φ = − GM ∕r.

(57)

Thus constant potential means constant

r.

(58)

The equipotential surfaces are spheres centered on the mass.

Because

g = − ∇Φ,

(59)

the gravitational field is perpendicular to equipotential surfaces.

Figure 4. Point-mass equipotential surfaces are concentric spheres. The gravitational field is perpendicular to each equipotential surface and points toward decreasing potential.

12 Superposition of gravitational potential

Potential is a scalar.

For several point masses,

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

(60)

The corresponding field is

|----------| g-=--−-∇Φ.--

(61)

Scalar superposition is often easier than directly summing vector fields.

For a continuous mass distribution,

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

(62)

With volume density,

|--------------------------| | ∫ ρ(r′) | |Φ(r) = − G ------′d3r′.| ---------------|r −-r|-----

(63)

13 Potential of a thin spherical shell

Consider a thin spherical shell of radius

R

(64)

and total mass

M.

(65)

Outside the shell, the shell theorem gives the same field as a point mass.

With

Φ (∞ ) = 0,

(66)

the exterior potential is

|--------------------------| |Φ(r) = − GM---, r ≥ R. | ------------r---------------

(67)

Inside the shell,

g = 0.

(68)

Therefore

∇ Φ = 0,

(69)

so the potential is constant throughout the interior.

Continuity at the shell gives

|--------------------------| | GM--- | |Φ(r) = − R , r ≤ R. | ----------------------------

(70)

Figure 5. A thin spherical shell has constant gravitational potential throughout its interior even though the gravitational field there is zero. Outside, the potential follows −GM∕r.

14 Zero field does not mean zero potential

Inside a spherical shell,

g = 0.

(71)

But

|------------| |Φ = − GM---| ---------R---|

(72)

is generally not zero.

A zero field means the potential has zero spatial gradient.

It does not require the potential itself to vanish.

This distinction is important throughout field theory.

15 Potential of a spherically symmetric body

For a spherically symmetric density

ρ(s),

(73)

consider a field point at radius

r.

(74)

Interior shells with

s < r

(75)

act like central point masses for the field, while exterior shells produce zero field but still contribute constant potential.

The total potential can be written

|------------------------------------------------| | [1 ∫ r ∫ ∞ ] | Φ (r) = − G -- 4 πρ(s)s2ds + 4πρ(s)sds .| -------------r--0----------------r----------------

(76)

For a finite body of radius

R,

(77)

the second integral ends at

R.

(78)

16 Uniform solid sphere

For a uniform sphere of:

radius R,
total mass M,
constant density ρ,

the exterior potential is

|--------------------------| | GM | |Φ(r) = − -----, r ≥ R. | ------------r---------------

(79)

Inside, the gravitational field is

gr = − GM---r. R3

(80)

Use

gr = − dΦ-. dr

(81)

Thus

dΦ- GM--- dr = R3 r.

(82)

Integrate:

Φ (r) = GM--r2 + C. 2R3

(83)

Continuity at

r = R

(84)

requires

Φ (R) = − GM---. R

(85)

Therefore

|--------------------------------------| | GM ( ) | |Φ (r) = −----3 3R2 − r2 , r ≤ R. | -----------2R--------------------------

(86)

Figure 6. The potential of a uniform solid sphere is quadratic inside and joins continuously to the exterior −GM∕r curve at the surface.

17 Potential at the center of a uniform sphere

Set

r = 0.

(87)

Then

|----------------| |Φ(0) = − 3GM---.| ------------2R---|

(88)

At the surface,

|----------------| | GM | |Φ (R) = − -R---.| -----------------

(89)

The center therefore lies deeper in the gravitational potential well.

18 Potential energy inside a uniform sphere

A test mass

m

(90)

has potential energy

U = m Φ.

(91)

Thus

|----------------(---------)-------------| |U (r) = − GM--m--3R2 − r2 , r ≤ R. | ------------2R3--------------------------|

(92)

The corresponding radial force is

F_r	= −	(93)
	= −r.	(94)

This agrees with the uniform-sphere interior field derived in M07-03.

19 Near-surface approximation

Near the surface of a spherical body, let

r = R + h,

(95)

where

|h| ≪ R.

(96)

The exact point-mass potential energy outside is

GM m U (R + h) = − R--+-h-.

(97)

The change from the surface is

ΔU	= U(R + h) − U(R)	(98)
	= GMm.	(99)

For

|h| ≪ R,

(100)

--1---≈ 1- − -h-. R + h R R2

(101)

Therefore

|----------------| | GM m | |ΔU ≈ ---2--h. | ----------R------

(102)

Since

2 g = GM ∕R ,

(103)

|------------| -ΔU--≈-mgh.--|

(104)

Figure 7. Near the surface, the exact gravitational potential curve is nearly linear over small height changes. The familiar mgh expression is the local first-order approximation.

20 Why mgh is not the universal gravitational potential energy

The expression

mgh

(105)

assumes:

a nearly constant gravitational field,
a relatively small vertical displacement,
a chosen reference height.

The inverse-square result

|---------------| U = − GM m ∕r | -----------------

(106)

is the appropriate point-mass or exterior-spherical-body expression over large changes in radius.

Thus

mgh

(107)

is a local approximation, not a replacement for

− GM m ∕r.

(108)

21 Example 1: potential at Earth’s surface

Using

μE = GME ≈ 3.9860 × 1014 m3 ∕s2,

(109)

and

RE = 6.3710 × 106 m,

(110)

Φ(R_E)	= −	(111)
	≈−6.26 × 10⁷ J∕kg.	(112)

1.00 kg

(113)

mass therefore has potential energy approximately

|--------------| -−-6.26 ×-107J-|

(114)

relative to the zero at infinity.

22 Example 2: exact versus mgh

Raise a

1.00 kg

(115)

mass by

h = 1000 m

(116)

near Earth’s surface.

The exact spherical change is

[ ] ΔU = μ -1--− ---1---- . exact E RE RE + h

(117)

Numerically,

|----------------------| ΔUexact ≈ 9.82 × 103 J.| ------------------------

(118)

Using

g ≈ 9.82m ∕s2,

(119)

ΔUmgh = mgh ≈ 9.82 × 103J.

(120)

The agreement is excellent because

h∕RE

(121)

is very small.

23 Mechanical energy in a gravitational field

For a test mass moving in a fixed gravitational potential,

|------------| -E-=--K-+-U.--

(122)

For point-mass gravity,

|--------------------| E = 1-mv2 − GM--m-.| -----2----------r-----

(123)

If no nongravitational work is done,

|--------------| |E = constant.| ---------------

(124)

This form is central to orbital mechanics.

24 Escape speed

Escape means reaching

r → ∞

(125)

with nonnegative final speed.

The minimum escape condition has

v∞ = 0.

(126)

At infinity,

U (∞ ) = 0,

(127)

so the minimum final mechanical energy is zero.

At launch radius

r,

(128)

set

E = 0 :

(129)

1 2 GM m 2-mv esc − --r----= 0.

(130)

Cancel

m

(131)

and solve:

|-----∘---------| | 2GM | vesc = -----. | -----------r-----

(132)

In terms of circular speed,

∘ ------- vcirc = GM ∕r,

(133)

|------√-------| |vesc = 2vcirc.| ----------------

(134)

Figure 8. Escape corresponds to total mechanical energy reaching the threshold E = 0. Bound motion has negative total energy, while positive total energy is unbound.

25 Example 3: escape speed from spherical Earth

Using

μ = 3.9860 × 1014 m3 ∕s2, E

(135)

and

R = 6.3710 × 106 m, E

(136)

v_esc	=	(137)
	≈ 1.12 × 10⁴ m∕s.	(138)

Thus

|----------------| vesc ≈ 11.2km ∕s.| ------------------

(139)

This ideal value neglects atmosphere, Earth’s rotation, oblateness, and other bodies.

26 Escape speed is independent of test mass

The test mass cancels from

1 -mv2 − GM m ∕r = 0. 2

(140)

Therefore the ideal Newtonian escape speed depends on:

source mass,
launch radius,

but not on the escaping object’s mass.

27 Potential of several masses

For point masses

Mj,

(141)

|------------------------| | ∑ --Mj---- | |Φ (r) = − G |r − r |.| --------------j-------j--|

(142)

The potential energy of a test mass is

|--------------| U (r) = m Φ(r).| ----------------

(143)

Then

|----------| g = − ∇Φ. | ------------

(144)

This scalar-first approach can simplify complicated many-source problems.

28 Potential energy of a system of point masses

For an isolated system of point masses, the gravitational potential energy is the sum over distinct pairs:

|------------------------| | ∑ mimj-- | |Usystem = − G rij .| ---------------i<j--------

(145)

The restriction

i < j

(146)

counts each pair once.

If one instead sums over all

i ⁄= j,

(147)

a factor of

1∕2

(148)

is required to avoid double counting.

29 Example 4: three point masses

For three masses,

m1, m2, m3,

(149)

the system potential energy is

--------------------------------------- | ( ) | |U = − G m1m2--+ m1m3--+ m2m3-- .| -------------r12------r13------r23-----|

(150)

There are three distinct interacting pairs.

30 Common mistakes

Forgetting the negative sign in U = −GMm∕r.
Confusing gravitational potential Φ with potential energy U.
Assuming potential must be zero where the field is zero.
Forgetting that the zero of potential energy is a convention.
Using mgh over distances where g changes substantially.
Treating mgh as the exact inverse-square potential energy.
Forgetting that g = −∇Φ.
Using g = +∇Φ.
Assuming equipotential surfaces are field lines.
Forgetting that field lines are perpendicular to equipotential surfaces.
Adding gravitational potentials as vectors instead of scalars.
Forgetting that a spherical shell has constant, nonzero interior potential.
Assuming the uniform-sphere interior potential is −GM∕r.
Forgetting to count each interacting pair only once in system potential energy.
Using escape speed without recognizing the E = 0 threshold.

31 Practice exercises

Starting from the inverse-square force, derive the gravitational work between radii r₁ and r₂.
Derive U = −GMm∕r using U(∞) = 0.
Show that F_r = −dU∕dr reproduces the gravitational force.
Define gravitational potential and derive Φ = −GM∕r.
Show that g = −∇Φ.
Derive the potential difference between two radii.
Show that point-mass equipotential surfaces are spheres.
Derive the scalar superposition formula for several point masses.
Derive the potential of a thin spherical shell inside and outside the shell.
Derive the interior potential of a uniform solid sphere.
Show that the uniform-sphere potential and its first radial derivative are continuous at the surface.
Derive the near-surface approximation ΔU ≈ mgh.
Compare exact spherical ΔU with mgh for a chosen small altitude.
Derive the escape-speed formula from Conservation of Mechanical Energy.
Write the total gravitational potential energy of four point masses without double counting any pair.

32 Summary

For two point masses,

|----------------| | GM--m--| -U(r)-=-−----r----

(151)

when

U (∞ ) = 0.

(152)

The gravitational potential is

|--------------| Φ (r) = − GM--.| -----------r----

(153)

Potential energy and potential are related by

|----------| |U = m Φ. | -----------

(154)

The field follows from the potential:

|----------| g-=--−-∇Φ.--

(155)

For several point masses,

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

(156)

A thin spherical shell has

|-------(------------------| | | GM--- | | { − r , r ≥ R, | |Φ (r) = | GM | | ( − -----, r ≤ R. | --------------R------------|

(157)

A uniform solid sphere has

|--------------------------------------| |Φ (r) = − GM--(3R2 − r2), r ≤ R. | -----------2R3-------------------------|

(158)

Near a spherical body’s surface,

|------------| -ΔU--≈-mgh.--|

(159)

The ideal escape speed from radius

r

(160)

|-----∘---------| | 2GM | vesc = -----. | -----------r-----

(161)

Potential energy turns Newtonian gravitation into an energy problem and prepares the way for orbital-energy analysis.

References

[1] PhysicsLibrary, M07-03, Newton’s Shell Theorem and Spherical Mass Distributions.

[2] PhysicsLibrary, M07-01, Newton’s Law of Universal Gravitation.

[3] PhysicsLibrary, M03-05, Conservative Forces and Potential Energy.

[4] J. R. Taylor, Classical Mechanics, University Science Books, 2005.

[5] D. Kleppner and R. Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge University Press, 2014.

"Gravitational Potential Energy and Gravitational Potential" is owned by bloftin.
(view preamble)

Gravitational Potential Energy and Gravitational Potential