Celestial Mechanics: Gravitational Potential and Spherical Gravity - Worked Problems and Complete Solutions

CM02 introduced gravitational work, potential energy, gravitational potential, and the relation

(1)

This companion article develops those ideas through complete worked problems. The first problems stay close to the point-mass potential

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and its corresponding potential energy

(3)

The later problems extend the same reasoning to spherical shells and uniform solid spheres. The escape-speed derivation is included explicitly because it is the first major celestial-mechanics application of gravitational potential energy.

Unless otherwise stated, use

(4)

(5)

(6)

and

(7)

Part I: Exercises

Exercise 1: gravitational work between two radii

A mass m moves radially outward from r₁ to r₂, where r₂ > r₁, in the field of a point mass M.

1. Starting from F_g = −GMmr∕r², derive the gravitational work W _g(r₁ → r₂).
2. Derive the corresponding change in potential energy ΔU.
3. Explain the signs of W_g and ΔU for outward motion.

Exercise 2: Earth’s surface gravitational potential

Using the point-mass exterior approximation for Earth:

1. calculate the gravitational potential Φ(R_E) at Earth’s mean surface;
2. calculate the potential energy of a 1000 kg spacecraft at that location;
3. explain why both values are negative when the reference Φ(∞) = 0 is chosen.

Exercise 3: exact altitude energy change versus mgh

Raise a 1.00 kg mass from Earth’s mean surface to altitude

(8)

1. calculate the exact change in gravitational potential energy using −μ_Em∕r;
2. estimate the change using mg₀h, where g₀ = μ_E∕R_E²;
3. compute the percentage by which the constant-g approximation differs from the exact result;
4. explain the source of the difference.

Exercise 4: recover the gravitational field from the potential

For a point mass,

(9)

Use

(10)

and spherical symmetry to recover the familiar point-mass gravitational field. Carefully track the sign.

Exercise 5: work along an equipotential surface

A test mass moves along a circular arc of constant radius r around a spherical gravitating body.

1. What is the change in gravitational potential ΔΦ?
2. What is the gravitational work W_g?
3. Show the same result directly from F_g ⋅ dr.
4. Explain why equipotential surfaces are perpendicular to The Gravitational Field.

Exercise 6: scalar superposition of gravitational potential

Two point masses M₁ and M₂ produce a gravitational potential at a point P.

1. Write the total potential in terms of distances r₁ and r₂ from P.
2. Evaluate the potential at the midpoint between Earth and Moon.
3. Explain why potentials add as scalars even though gravitational fields add as vectors.

Exercise 7: zero gravitational field does not mean zero potential

Use the Earth–Moon zero-field point derived in CM01E1,

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where x is measured from Earth’s center toward the Moon.

1. calculate x and the distance d − x from the Moon;
2. calculate the gravitational potential there;
3. explain how g = 0 can coexist with Φ≠0.

Exercise 8: derive and calculate Earth’s escape speed

A mass m is launched from Earth’s mean surface with just enough speed to reach infinity with zero remaining speed.

1. Write the initial and final mechanical energies.
2. Derive the escape-speed formula

   (12)

3. calculate Earth’s surface escape speed.
4. explain why the escaping object’s mass cancels.
5. explain why mgh is not the correct potential-energy model for escape to infinity.

Exercise 9: escape speed from low Earth orbit altitude

At altitude 400 km above Earth’s mean surface:

1. calculate the local escape speed;
2. compare it with the surface escape speed;
3. calculate the circular-orbit speed at the same radius using v_circ = ;
4. verify the relation v_esc = v_circ.

Exercise 10: potential of a thin spherical shell

A thin spherical shell has total mass M and radius R. Use the shell theorem and the convention Φ(∞) = 0.

1. determine Φ(r) for r ≥ R;
2. determine Φ(r) for r < R;
3. show that Φ is continuous at r = R;
4. use g = −∇Φ to recover the exterior and interior fields.

Exercise 11: field inside a uniform solid sphere

A solid sphere has radius R, total mass M, and constant density.

1. find the enclosed mass M(r) for r < R;
2. use the shell theorem to derive the field inside the sphere;
3. show that the field magnitude grows linearly with r;
4. verify that the interior field matches the exterior surface field at r = R.

Exercise 12: potential inside a uniform solid sphere

Using the field from Exercise 11 and continuity with the exterior potential at r = R, derive the interior potential

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Then find:

1. the potential at the surface;
2. the potential at the center;
3. the potential difference between center and surface.

Exercise 13: escape speed from the center of a uniform sphere

Imagine a hypothetical nonrotating Earth with the same total mass M_E and radius R_E but uniform density. Ignore tunnels, pressure, and material interactions.

1. use the interior potential to derive the escape speed from the center;
2. calculate its numerical value;
3. compare it with the surface escape speed.

Exercise 14: continuity of potential and behavior of the field at a surface

Compare a thin spherical shell and a uniform solid sphere of equal M and R.

1. show that Φ is continuous across r = R for both objects;
2. determine whether the radial field g_r = −dΦ∕dr is continuous across r = R for the thin shell;
3. determine whether it is continuous for the uniform solid sphere;
4. explain physically why the answers differ.

Exercise 15: Julia check of spherical field and potential profiles

Write a Julia program that evaluates Φ(r) and g_r(r) for

1. a thin spherical shell;
2. a uniform solid sphere;

using normalized units G = M = R = 1.

Check numerically that

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away from the thin shell’s surface, and verify continuity of Φ at r = R for both models.

Part II: Complete Worked Solutions

Solution 1: gravitational work between two radii

For radial motion,

(15)

Therefore

Wg	= ∫ r1r2 Fg ⋅ dr	(16)
	= −GMm∫ r1r2 r−2 dr	(17)
	= −GMmr1r2	(18)
	= GMm.	(19)

Thus

(20)

For a conservative force,

(21)

Hence

(22)

If r₂ > r₁, then 1∕r₂ < 1∕r₁, so

(23)

Gravity points inward while the displacement is outward, so gravity removes kinetic energy unless some external agent supplies energy. The gravitational potential energy becomes less negative as the body is moved farther away.

Solution 2: Earth’s surface gravitational potential

The potential is

(24)

Substituting the given values,

Φ(RE)	= −	(25)
	≈−6.2565 × 107 J/kg.	(26)

Thus

(27)

For a 1000 kg spacecraft,

U	= mΦ	(28)
	= (1000)(−6.2565 × 107)	(29)
	≈−6.2565 × 1010 J.	(30)

Therefore

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The negative sign follows from the convention

(32)

At any finite distance from an attractive positive mass, energy must be added to move a test mass all the way to infinity. The finite-distance state therefore lies below the zero reference.

Solution 3: exact altitude energy change versus mgh

The initial radius is

(33)

and the final radius is

(34)

For m = 1.00 kg,

ΔU	= μEm	(35)
	≈ 3.6960 × 106 J.	(36)

So

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The surface gravitational acceleration from the same spherical model is

g0	=	(38)
	≈ 9.8203 m/s2.	(39)

The constant-g estimate is

mg0h	= (1)(9.8203)(4.00 × 105)	(40)
	≈ 3.9281 × 106 J.	(41)

Thus

(42)

The relative difference is approximately

(43)

Hence the constant-g model overestimates the energy change by about

(44)

The reason is that mgh assumes the surface value of g acts over the entire height. In reality,

(45)

decreases continuously as the mass rises.

Solution 4: recover the gravitational field from the potential

For spherical symmetry,

(46)

With

(47)

we have

	= −GM(r−1)	(48)
	= .	(49)

Therefore

g	= −∇Φ	(50)
	= −r.	(51)

Thus

(52)

The derivative dΦ∕dr is positive because the negative potential rises toward zero as r increases. The extra minus sign in g = −∇Φ makes the field point toward decreasing potential, which is inward.

Solution 5: work along an equipotential surface

At constant radius r,

(53)

is constant. Therefore

(54)

For a test mass m,

(55)

Hence

(56)

We can see the same result directly. Along a circular arc, dr is tangent to the circle while gravity is radial. Therefore

(57)

and

(58)

Thus gravity does no work along an equipotential surface. More generally, since

(59)

the field points normal to surfaces of constant Φ.

Solution 6: scalar superposition of gravitational potential

For point masses,

(60)

At the midpoint between Earth and Moon,

(61)

Therefore

Φmid	= −G	(62)
	= −.	(63)

Numerically,

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Potential is a scalar, so each source contributes one signed scalar value and the contributions are added directly. The gravitational field is a vector, so its source contributions must be added with direction.

Solution 7: zero gravitational field does not mean zero potential

The balance point satisfies

(65)

Using the stated masses and separation,

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Its distance from the Moon is

d − x	≈ 3.8380 × 107 m	(67)
	≈ 38380 km.	(68)

The potential is

(69)

Numerically,

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At this location the two field vectors are equal in magnitude and opposite in direction, so

(71)

But the potentials are both negative scalars:

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They therefore add rather than cancel. A zero gradient does not imply a zero value of the scalar function. It means only that the local slope vanishes.

Solution 8: derive and calculate Earth’s escape speed

The total mechanical energy of a mass m in Earth’s field is

(73)

For minimum escape from radius r₀, the object reaches infinity with speed approaching zero. Therefore

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Energy conservation gives

(75)

Thus

(76)

Cancel m:

(77)

Therefore

(78)

At Earth’s mean surface,

vesc	=	(79)
	≈ 1.1186 × 104 m/s.	(80)

Thus

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The escaping object’s mass cancels because both kinetic energy and gravitational potential energy are proportional to m.

The approximation mgh is not suitable for escape to infinity because it assumes nearly constant gravitational acceleration. Escape traverses distances comparable with and much larger than Earth’s radius, over which

(82)

changes drastically. The exact −μ_Em∕r potential must be used.

Solution 9: escape speed from low Earth orbit altitude

The radius at 400 km altitude is

(83)

The local escape speed is

vesc	=	(84)
	≈ 1.0851 × 104 m/s.	(85)

Therefore

(86)

This is smaller than the surface value 11.19 km/s because some gravitational potential energy has already been gained by climbing to the higher radius.

The circular speed at the same radius is

vcirc	=	(87)
	≈ 7.6726 km/s.	(88)

Then

vcirc	≈ 1.4142(7.6726)	(89)
	≈ 10.85 km/s,	(90)

so

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at the same radius in an inverse-square point-mass field.

Solution 10: potential of a thin spherical shell

For r ≥ R, the shell theorem says the gravitational field is the same as that of a point mass M at the center. With zero potential at infinity,

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At the surface,

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Inside a thin spherical shell, the shell theorem gives

(94)

Since

(95)

zero field means the potential is spatially constant throughout the interior. Continuity with the surface fixes that constant:

(96)

Thus the complete potential is

(97)

The potential is continuous at r = R.

Taking the radial derivative gives

(98)

and

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The field jumps at the idealized surface because the mass density is concentrated in an infinitesimally thin layer.

Solution 11: field inside a uniform solid sphere

The constant density is

(100)

At radius r < R, the enclosed mass is

M(r)	= ρπr3	(101)
	= M.	(102)

By the shell theorem, spherical shells outside radius r make no net contribution to the field at that interior point. Therefore

gr(r)	= −	(103)
	= −.	(104)

Thus

(105)

Its magnitude is

(106)

The field therefore grows linearly from zero at the center.

At r = R,

(107)

which matches the exterior point-mass field at the surface. Unlike the ideal thin shell, a uniform volume density does not create a jump in g at the outer boundary.

Solution 12: potential inside a uniform solid sphere

For r < R,

(108)

Since

(109)

we have

(110)

Integrate from the surface R to an interior radius r:

(111)

Therefore

Φ(r) −

= (r2 − R2).

(112)

Rearranging,

Φ(r)	= − + (r2 − R2)	(113)
	= − + .	(114)

Thus

(115)

At the surface,

(116)

At the center,

(117)

The potential difference from center to surface is

Φ(R) − Φ(0)	= − +	(118)
	= .	(119)

Hence

(120)

The potential rises smoothly from its most negative value at the center to the surface value, then approaches zero as −GM∕r outside.

Solution 13: escape speed from the center of a uniform sphere

At the center of a uniform sphere,

(121)

Minimum escape again corresponds to zero total specific mechanical energy at infinity:

(122)

Thus

(123)

Therefore

(124)

For the hypothetical uniform Earth,

vesc,0	=	(125)
	≈ 1.3700 × 104 m/s.	(126)

Hence

(127)

At the surface,

(128)

Thus

(129)

The center requires a greater escape speed because the potential is deeper even though the local gravitational field is zero exactly at the center. Again, field strength and potential value are distinct concepts.

Solution 14: continuity of potential and behavior of the field at a surface

For the thin shell,

(130)

and

(131)

So Φ is continuous.

Inside the shell,

(132)

Just outside,

(133)

Therefore the field has a jump discontinuity across the ideal infinitesimally thin mass layer.

For the uniform solid sphere,

(134)

matches the exterior value, so the potential is again continuous.

The interior field approaches

(135)

while the exterior field approaches

(136)

Thus the field is also continuous for the uniform sphere.

The distinction comes from the mass distribution. A thin shell has finite surface mass density concentrated on a zero-thickness layer, producing a jump in the normal gravitational field. A uniform solid sphere has a finite volume density with no singular surface layer, so the field transitions continuously.

Solution 15: Julia check of spherical field and potential profiles

One implementation is:

using Printf

G = 1.0
M = 1.0
R = 1.0

function phi_shell(r)
    r < R ? -G*M/R : -G*M/r
end

function g_shell(r)
    r < R ? 0.0 : -G*M/r^2
end

function phi_solid(r)
    if r <= R
        return -G*M/(2R) * (3 - r^2/R^2)
    else
        return -G*M/r
    end
end

function g_solid(r)
    if r <= R
        return -G*M*r/R^3
    else
        return -G*M/r^2
    end
end

function numerical_g(phi, r; dr=1e-5)
    return -(phi(r + dr) - phi(r - dr))/(2dr)
end

println(" r       shell g: a/n       solid g: a/n")
for r in (0.25, 0.75, 0.99, 1.01, 1.5, 2.0)
    @printf("%.2f   % .5f/% .5f   % .5f/% .5f\n",
            r, g_shell(r), numerical_g(phi_shell, r),
                                                                                         
                                                                                         
            g_solid(r), numerical_g(phi_solid, r))
end

println("\nPotential continuity check near r = R")
for r in (0.999999, 1.0, 1.000001)
    @printf("r=%.6f  shell Phi=% .9f  solid Phi=% .9f\n",
            r, phi_shell(r), phi_solid(r))
end

Away from r = R, the centered finite difference should satisfy

(137)

For the thin shell, the derivative is not defined at the idealized surface because the field jumps there. The potential itself remains continuous.

For the uniform solid sphere, both Φ and its first radial derivative are continuous at r = R, so the numerical derivative passes smoothly through the surface when the resolution is adequate.

1 What CM02E1 adds to the series

CM02 introduced potential as an alternative description of Newtonian gravity. CM02E1 develops the practical consequences:

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The escape-speed derivation demonstrates why potential energy is so valuable in celestial mechanics: a problem that would require integrating a continuously changing acceleration can instead be solved from one scalar conservation equation.

The spherical examples also establish several ideas that recur throughout astrophysics:

• exterior spherical gravity behaves like a point mass;
• a field can vanish where the potential is nonzero;
• potential is usually smoother than the field;
• the interior potential depends on the mass distribution, not merely the total mass;
• escape is determined by the depth of the potential well.

The next main lesson can now return to dynamics and derive Newton’s orbital equation of motion from the gravitational field.

References