Celestial Mechanics: Newtonian Gravitation - Worked Problems and Complete Solutions

CM01 introduced Newton’s universal law of gravitation as the dynamical starting point for celestial mechanics. This companion article develops fluency with that law before gravitational potential, orbital energy, or conic-section dynamics are introduced.

The problems begin with direct force-law and field-strength calculations and then progress to vector components, inverse-square scaling, two-body acceleration, superposition, balance points, continuous spherical mass distributions, and a numerical implementation of the Newtonian field.

The central equations are

(1)

(2)

and, for The Gravitational Field produced by a point source mass M,

(3)

For several sources,

(4)

Unless otherwise stated, use

(5)

(6)

(7)

and

(8)

Part I: Exercises

Exercise 1: Earth–Moon gravitational force

Treat Earth and the Moon as point masses separated by their mean center-to-center distance.

1. Compute the magnitude of the gravitational force Earth exerts on the Moon.
2. State the magnitude of the force the Moon exerts on Earth.
3. Explain why the two bodies do not have equal accelerations even though the force magnitudes are equal.

Exercise 2: gravitational field strength at Earth’s surface and at 400 km altitude

Using the point-mass/spherical-Earth approximation, calculate

1. the gravitational field magnitude at Earth’s mean surface;
2. the gravitational field magnitude at altitude h = 400 km;
3. the ratio g(R_E + h)∕g(R_E).

Explain why astronauts in low Earth orbit are not beyond Earth’s gravity.

Exercise 3: inverse-square scaling without repeating the full calculation

At distance r from a point mass, the gravitational field magnitude is g.

Determine the field magnitude if

1. the distance is doubled;
2. the distance is tripled;
3. the source mass is doubled while the distance is tripled;
4. the source mass is reduced to one half while the distance is reduced to one quarter.

Do this using ratios rather than substituting G numerically.

Exercise 4: force versus field and cancellation of test mass

At altitude 1000 km above Earth’s mean radius:

1. calculate the gravitational field magnitude;
2. calculate the force on a 1.00 kg test mass;
3. calculate the force on a 1000 kg spacecraft;
4. calculate the acceleration of each object.

Use the result to explain the difference between gravitational force and gravitational field.

Exercise 5: vector gravitational force from Cartesian coordinates

Two point masses have

(9)

Their positions are

(10)

and

(11)

Find

1. the relative-position vector r = r₂ − r₁;
2. its magnitude and unit vector;
3. the gravitational force vector on m₂ due to m₁;
4. the force vector on m₁ due to m₂.

Exercise 6: equal force, unequal acceleration in the Earth–Moon system

Using the force from Exercise 1:

1. calculate Earth’s acceleration toward the Moon;
2. calculate the Moon’s acceleration toward Earth;
3. calculate the ratio a_M∕a_E;
4. show that this ratio is equal to M_E∕M_M.

Then calculate the distances of Earth and Moon from their common center of mass using

(12)

(13)

Exercise 7: altitude where Earth’s gravitational field is one half of its surface value

Find the altitude h above Earth’s mean surface where

(14)

Do not substitute numerical constants until after solving symbolically for h.

Exercise 8: recover the gravitational parameter and source mass from a field measurement

At distance

(15)

from the center of an approximately spherical body, a gravitational field magnitude

(16)

is measured.

1. Determine the body’s gravitational parameter μ = GM.
2. Determine the body’s mass M.
3. Explain why orbital mechanics often works directly with μ rather than with G and M separately.

Exercise 9: the Sun’s field at Earth and the force on Earth

At one astronomical unit:

1. calculate the magnitude of the Sun’s gravitational field;
2. calculate the gravitational force magnitude on Earth;
3. compare the solar field at Earth with Earth’s field at its own surface by computing the ratio g_⊙(1 AU)∕g_E(R_E).

Explain why a comparatively small acceleration can control an orbit over astronomical time scales.

Exercise 10: superposition above two equal masses

Two equal masses

(17)

are located at

(18)

with

(19)

An observation point is at

(20)

1. Find the field vector from each source at P.
2. Show explicitly that the horizontal components cancel.
3. Find the net gravitational field vector and magnitude.

Exercise 11: zero-field point between two unequal masses

Two masses M₁ and M₂ are separated by distance d, with M₁ > M₂. Consider only their Newtonian gravitational fields along the line joining them.

1. Derive the position x, measured from M₁, where the net gravitational field is zero between the masses.
2. Apply the result to the Earth–Moon pair.
3. Give the distance of this zero-field point from the Moon’s center.
4. Explain why this point is not the Earth–Moon L₁ Lagrange point.

Exercise 12: surface gravity of a uniform-density spherical body

A spherical body has uniform density ρ and radius R.

1. Use

   (21)

   to derive a formula for its surface gravitational field g_s in terms of ρ and R.

2. Evaluate the result for

   (22)

Exercise 13: dimensional analysis of G and μ

Starting from

(23)

derive the SI dimensions of G.

Then define

(24)

and determine the dimensions of μ.

Finally show that

(25)

has dimensions of acceleration.

Exercise 14: Julia implementation of Newtonian gravitational superposition

Write a Julia function that evaluates

(26)

for an arbitrary list of point masses.

Use the two equal masses from Exercise 10 and verify numerically that

1. the field at the midpoint (0, 0) is zero;
2. the field at (0, 10⁶ m) has zero x component;
3. its y component agrees with Exercise 10.

Part II: Complete Worked Solutions

Solution 1: Earth–Moon gravitational force

Newton’s scalar force law gives

(27)

Substitute the numerical values:

F	= (6.67430 × 10−11)	(28)
	≈ 1.98 × 1020 N.	(29)

Therefore

(30)

Newton’s third law requires an equal and opposite force on Earth:

(31)

The accelerations differ because

(32)

The same force divided by the much smaller lunar mass produces a much larger acceleration of the Moon than of Earth.

Solution 2: gravitational field strength at Earth’s surface and at 400 km altitude

At Earth’s surface,

(33)

Thus

g(RE)	= (6.67430 × 10−11)	(34)
	≈ 9.8203 m/s2.	(35)

Hence

(36)

At altitude h = 400 km,

(37)

Therefore

g(RE + h)	= G	(38)
	≈ 8.6943 m/s2.	(39)

Thus

(40)

The ratio is

	= 2	(41)
	≈ 0.885.	(42)

So

(43)

Gravity at this altitude is still almost 89% of its surface value. Orbiting astronauts appear weightless because they and their spacecraft are in continuous free fall, not because gravity has become negligible.

Solution 3: inverse-square scaling without repeating the full calculation

Because

(44)

ratios can be formed directly.

If r′ = 2r,

(45)

so

(46)

If r′ = 3r,

(47)

If M′ = 2M and r′ = 3r,

(48)

so

(49)

Finally, if

(50)

then

	= 2	(51)
	= (16)	(52)
	= 8.	(53)

Therefore

(54)

The inverse-square dependence can overwhelm a moderate change in source mass because distance enters quadratically.

Solution 4: force versus field and cancellation of test mass

At altitude 1000 km,

(55)

The gravitational field is

g	= G	(56)
	≈ 7.3365 m/s2.	(57)

Thus

(58)

For a 1.00 kg mass,

(59)

so

(60)

For a 1000 kg spacecraft,

(61)

so

(62)

The acceleration of either body is

(63)

Therefore

(64)

The force scales with test mass; the gravitational field does not. Dividing the force by inertial mass removes the test mass from the acceleration.

Solution 5: vector gravitational force from Cartesian coordinates

The relative vector is

r	= r2 − r1	(65)
	= (3.00 × 106x + 4.00 × 106y) m.	(66)

Its magnitude is

r	=	(67)
	= 5.00 × 106 m.	(68)

Hence

(69)

The force magnitude is

F	= G	(70)
	= (6.67430 × 10−11)	(71)
	= 4.27155 × 1017 N.	(72)

The force on body 2 points opposite r:

F2←1	= −Fr	(73)
	= −(4.27155 × 1017)(0.600x + 0.800y).	(74)

Therefore

(75)

Newton’s third law gives

(76)

Solution 6: equal force, unequal acceleration in the Earth–Moon system

Using

(77)

Earth’s acceleration is

aE	=	(78)
	≈ 3.3163 × 10−5 m/s2.	(79)

Thus

(80)

The Moon’s acceleration is

aM	=	(81)
	≈ 2.6976 × 10−3 m/s2.	(82)

Thus

(83)

The acceleration ratio is

(84)

Numerically,

(85)

Now compute the center-of-mass distances:

rE	= rEM	(86)
	≈ 4.668 × 106 m,	(87)

and

rM	= rEM	(88)
	≈ 3.7973 × 108 m.	(89)

Therefore

(90)

(91)

Both bodies move about their common center of mass; the center lies much closer to Earth because Earth is much more massive.

Solution 7: altitude where Earth’s gravitational field is one half of its surface value

Write

(92)

Cancel GM_E:

(93)

Invert both sides:

(94)

Take the positive square root:

(95)

Therefore

(96)

Numerically,

h	= ( − 1)(6.371 × 106)	(97)
	≈ 2.639 × 106 m.	(98)

Thus

(99)

The gravitational field does not fall to one half until the radial distance has increased to times Earth’s radius.

Solution 8: recover the gravitational parameter and source mass from a field measurement

For a spherical source,

(100)

Therefore

(101)

Substitute the measurement:

μ	= (8.13475)(7.00 × 106)2	(102)
	≈ 3.98603 × 1014 m3∕s2.	(103)

Thus

(104)

The mass is

(105)

Hence

M	=	(106)
	≈ 5.9722 × 1024 kg.	(107)

Therefore

(108)

This is essentially Earth’s mass. Orbital dynamics often uses μ directly because the equations of motion depend on the product GM, and μ can be determined very accurately from orbital observations even when G and M separately are less precisely known.

Solution 9: the Sun’s field at Earth and the force on Earth

The solar field at one astronomical unit is

g⊙	= G	(109)
	= (6.67430 × 10−11)	(110)
	≈ 5.9303 × 10−3 m/s2.	(111)

Thus

(112)

The force on Earth is

F	= MEg⊙	(113)
	≈ (5.9722 × 1024)(5.9303 × 10−3)	(114)
	≈ 3.542 × 1022 N.	(115)

Therefore

(116)

The ratio to Earth’s surface field is

	=	(117)
	≈ 6.04 × 10−4.	(118)

So

(119)

The solar acceleration is small compared with surface gravity, but it acts continuously. Celestial motion is controlled by sustained acceleration over long times rather than by requiring a large instantaneous acceleration.

Solution 10: superposition above two equal masses

The two source positions are

(120)

and the observation point is

(121)

For the left source,

(122)

For the right source,

(123)

Each source is a distance

(124)

from P.

The two field contributions are

(125)

and

(126)

For the numerical values in the problem, each Cartesian component has magnitude

(127)

Thus

(128)

and

(129)

Add them:

gnet	= g1 + g2	(130)
	= −y.	(131)

The x components cancel exactly.

For

(132)

we obtain

|gnet|	=	(133)
	≈ 4.7194 × 10−3 m/s2.	(134)

Therefore

(135)

Symmetry removes one component before any difficult arithmetic is required.

Solution 11: zero-field point between two unequal masses

Let the balance point lie a distance x from M₁. Its distance from M₂ is d − x.

Between the masses the two gravitational fields point in opposite directions. Zero net field requires equal magnitudes:

(136)

Cancel G and take the positive square root:

(137)

Cross multiply:

(138)

Thus

(139)

so

(140)

For Earth and Moon,

x	=	(141)
	≈ 3.4603 × 108 m.	(142)

Therefore the zero-field point is

(143)

Its distance from the Moon is

rM,P	= rEM − x	(144)
	≈ 3.8367 × 107 m,	(145)

or

(146)

This point is not L₁. The L₁ point is an equilibrium in a frame rotating with the Earth–Moon system and therefore includes the rotating-frame inertial terms associated with the required orbital angular speed. Exercise 11 balances only the two Newtonian gravitational fields in an inertial description.

Solution 12: surface gravity of a uniform-density spherical body

For a uniform sphere,

(147)

The surface field is

(148)

Substitute the mass:

gs	=	(149)
	= πGρR.	(150)

Therefore

(151)

For

(152)

we obtain

gs	= π(6.67430 × 10−11)(5514)(106)	(153)
	≈ 1.5416 m/s2.	(154)

Thus

(155)

At fixed density, surface gravity scales linearly with radius because total mass grows as R³ while the inverse-square field divides by R².

Solution 13: dimensional analysis of G and μ

From

(156)

solve dimensionally for G:

(157)

Since

(158)

we obtain

[G]	=	(159)
	= m3kg−1s−2.	(160)

Therefore

(161)

For

(162)

we have

(163)

Thus

(164)

Finally,

(165)

which is exactly the dimension of acceleration.

Solution 14: Julia implementation of Newtonian gravitational superposition

One direct implementation using three-component vectors is

using LinearAlgebra
using Printf

const G = 6.67430e-11

function gravity_field(r, masses, positions)
g = zeros(3)
for i in eachindex(masses)
dr = r - positions[i]
d = \href{https://physicslibrary.org/encyclopedia/NormInducedByInnerProduct.html}{norm(}dr)
g .+= -G * masses[i] * dr / d^3
end
return g
end

M = 1.0e20
masses = [M, M]
positions = [
[-1.0e6, 0.0, 0.0],
[ 1.0e6, 0.0, 0.0]
]

r_mid = [0.0, 0.0, 0.0]
r_top = [0.0, 1.0e6, 0.0]

g_mid = gravity_field(r_mid, masses, positions)
g_top = gravity_field(r_top, masses, positions)

println("g(midpoint) = ", g_mid)
println("g(top) = ", g_top)
@printf("|g(top)| = %.8e m/s^2\n", norm(g_top))

At the midpoint, symmetry gives

(166)

At the point directly above the midpoint, the two x components cancel numerically. The result should be approximately

(167)

This agrees with the analytic superposition calculation in Exercise 10.

The implementation also exposes the mathematical structure that will later become the Newtonian N-body problem: every source contributes a vector term proportional to

(168)

1 What CM01E1 adds to the series

CM01 introduced the law. CM01E1 develops the ability to use it in several complementary forms:

(169)

The exercises also establish several scaling ideas that recur throughout celestial mechanics:

(170)

(171)

and

(172)

The next main theory article, CM02, develops gravitational potential and potential energy. That change of viewpoint converts the inverse-square vector field into a scalar potential and prepares the later derivation of orbital-energy conservation.

References