Newton's law of universal gravitation

Newton’s Law of Universal Gravitation

Newtonian gravitation begins with a remarkably compact statement.

Any two point masses attract one another with a force whose magnitude is proportional to the product of their masses and inversely proportional to the square of their separation.

If two point masses

(1)

and

(2)

are separated by distance

(3)

the magnitude of the gravitational force is

(4)

The constant

(5)

is Newton’s gravitational constant.

A commonly used numerical value is

(6)

The force acts along the line joining the two masses and is always attractive in Newtonian gravitation.

1 What the law says

The scalar formula

(7)

contains three important scalings.

First,

(8)

Second,

(9)

Third,

(10)

Doubling either mass doubles the force.

Doubling both masses multiplies the force by four.

Doubling the separation reduces the force to one fourth.

Tripling the separation reduces the force to one ninth.

2 Inverse-square behavior

The inverse-square factor is central:

(11)

Thus

(12)

and

(13)

3 Dimensions of the gravitational constant

From

(14)

solve for

(15)

(16)

Since

(17)

[G]	=	(18)
	= m3 kg−1 s−2.	(19)

Equivalent units are

(20)

4 Point masses and separation vector

Let the masses occupy positions

(21)

and

(22)

Define the displacement from mass 1 to mass 2:

(23)

Its magnitude is

(24)

The corresponding unit vector is

(25)

5 Vector form of the force

The force on mass 1 due to mass 2 points from mass 1 toward mass 2.

Therefore

(26)

Equivalently,

(27)

The force on mass 2 due to mass 1 points in the opposite direction:

(28)

Thus

(29)

6 An equally common vector convention

Some texts define

(30)

which points from source mass 2 toward test mass 1.

Then the force on mass 1 is written

(31)

The minus sign expresses attraction.

Both conventions are correct when used consistently.

The safest procedure is:

1. define the separation vector explicitly,
2. determine which direction attraction should point,
3. then write the vector force.

7 Newton’s third law

The gravitational interaction satisfies Newton’s third law:

(32)

The forces have:

• equal magnitude,
• opposite direction,
• different points of application.

The forces are equal even when the masses are very different.

Their accelerations are not generally equal because

(33)

8 Example 1: unequal masses

Suppose

(34)

The force magnitudes remain equal:

(35)

But

(36)

while

(37)

Therefore

(38)

9 Gravitational mass and inertial mass

Newtonian mechanics distinguishes two concepts.

Inertial mass

(39)

appears in Newton’s second law:

(40)

Gravitational mass

(41)

measures coupling to gravity.

For a test body near a source mass

(42)

(43)

Thus

(44)

so

(45)

Experiment shows that gravitational and inertial mass are proportional to extremely high precision.

With the conventional choice of units,

(46)

Then the test body’s mass cancels:

(47)

This is why objects in the same gravitational environment have the same gravitational acceleration when nongravitational effects such as air resistance are negligible.

10 Mass independence of free-fall acceleration

Let a source mass

(48)

attract a test mass

(49)

Newton’s law gives

(50)

Newton’s second law gives

(51)

Equating them,

(52)

Cancel

(53)

(54)

The cancellation is one of the central features of Newtonian gravitation.

11 Example 2: force between laboratory masses

Let

(55)

and

(56)

Then

F	= G	(57)
	= 6.67430 × 10−11 N.	(58)

Gravity between ordinary laboratory-scale masses is extremely weak.

This small magnitude is one reason direct laboratory measurement of

(59)

is difficult.

12 Superposition

If several masses attract a test mass, the total gravitational force is the vector sum of the pairwise forces.

For masses

(60)

at positions

(61)

acting on test mass

(62)

at

(63)

(64)

Gravity in Newtonian mechanics is pairwise additive.

13 Example 3: two equal source masses

Place two equal source masses

(65)

symmetrically at

(66)

and

(67)

Place a test mass at the origin.

The two gravitational forces have equal magnitude and opposite direction.

Therefore

(68)

Zero net force does not mean there is no gravitational interaction.

It means the individual vector forces cancel.

14 Continuous mass distributions

An extended body can be treated as a collection of infinitesimal mass elements.

For source element

(69)

at position

(70)

the force on test mass

(71)

at

(72)

is

(73)

The total force is

(74)

If the mass density is

(75)

then

(76)

Thus

(77)

M07-02 develops this superposition viewpoint in terms of The Gravitational Field.

15 Spherically symmetric bodies

A point-mass law applies directly to point masses.

For extended bodies, one must in principle integrate over the mass distribution.

A particularly important result is that a spherically symmetric body produces the same external gravitational force as if all of its mass were concentrated at its center.

That statement follows from Newton’s shell theorem.

We will derive it later rather than assume it here.

Once that theorem is established, the point-mass formula can be applied outside spherical planets and stars:

(78)

where

(79)

is measured from the center.

16 Gravity near Earth’s surface

Using the spherical result for Earth,

(80)

At Earth’s surface,

(81)

Using approximate values

(82)

and

(83)

g	=	(84)
	≈ 9.82 m∕s2.	(85)

This simple spherical model is close to familiar terrestrial gravitational acceleration.

Real local gravity also depends on Earth’s rotation, shape, altitude, and mass distribution.

17 Weight

The gravitational force on an object of mass

(86)

is its gravitational Weight in the simple Newtonian model:

(87)

Near Earth’s surface this is often approximated as

(88)

The approximation treats

(89)

as nearly constant over the vertical distance of interest.

Mass and weight are not the same quantity:

• mass is measured in kilograms,
• weight is a force measured in newtons.

18 Altitude dependence

At altitude

(90)

above the spherical Earth’s surface,

(91)

Therefore

(92)

Relative to surface gravity,

(93)

19 Example 4: gravity one Earth radius above the surface

Let

(94)

Then

(95)

Therefore

(96)

Thus

(97)

An object one Earth radius above the surface is not beyond Earth’s gravity.

Its gravitational acceleration is simply reduced to one fourth of the surface value in the spherical model.

20 Two-body equations of motion

For two isolated masses,

(98)

and

(99)

the force on mass 1 is

(100)

Hence

(101)

Similarly,

(102)

Subtract the equations.

Define

(103)

Then

r	= r2 −r1	(104)
	= −G(m1 + m2).	(105)

Thus

(106)

The relative two-body motion is therefore another inverse-square problem.

21 One-body approximation

If

(107)

the acceleration of

(108)

is much smaller than that of

(109)

One often approximates the larger mass as fixed.

Then

(110)

This approximation is excellent for many satellite problems but is not exact.

The exact two-body treatment will be developed later in the M07 sequence.

22 Circular orbit as a first application

Suppose a small mass

(111)

moves in a circular orbit of radius

(112)

around a much larger mass

(113)

The required centripetal force is

(114)

Gravity supplies this force:

(115)

Cancel

(116)

and one factor of

(117)

(118)

Therefore

(119)

23 Circular orbital period

For a circular orbit,

(120)

Substitute

(121)

(122)

Therefore

(123)

This is the circular-orbit form of Kepler’s third law.

A later article will derive the more general orbital result.

24 Example 5: low circular orbit speed near Earth

Using

(124)

and taking

(125)

the idealized circular speed would be

vcirc	=	(126)
	≈ 7.91 × 103 m∕s.	(127)

This is about

(128)

A real orbit cannot exist at Earth’s solid surface because of terrain and atmosphere, but the calculation establishes the orbital speed scale.

25 Gravitational parameter

The product

(129)

appears repeatedly.

It is often denoted

(130)

The quantity

(131)

is the gravitational parameter of the source body.

Then

(132)

and

(133)

For orbital calculations,

(134)

is often known more accurately than

(135)

and

(136)

separately.

26 Why the force is a central force

A central force has the form

(137)

with direction along the line from a chosen center.

Newtonian point-mass gravity is a central force.

For a fixed source mass at the origin,

(138)

Because

(139)

and

(140)

are parallel or antiparallel,

(141)

Thus gravitational torque about the force center vanishes:

(142)

Therefore angular momentum about that center is conserved.

This connects Newtonian gravitation to the central force mechanics developed earlier.

27 Limits of Newtonian gravitation

Newton’s law is extraordinarily successful when:

• gravitational fields are weak,
• speeds are small compared with the speed of Light,
• relativistic corrections are negligible.

General relativity replaces Newtonian gravitation when stronger gravitational fields, high precision, or relativistic conditions require it.

Within ordinary classical mechanics and most elementary orbital problems, Newtonian gravitation is the correct starting model.

28 Common mistakes

1. Using 1∕r instead of 1∕r² for the gravitational-force magnitude.
2. Forgetting that gravitational force is attractive.
3. Writing a vector equation without first defining the direction of the separation vector.
4. Treating the equal-and-opposite gravitational forces as equal accelerations.
5. Confusing mass with weight.
6. Using altitude h in place of center-to-center radius R + h.
7. Forgetting that the test mass cancels from gravitational acceleration when gravitational and inertial mass are equal.
8. Adding gravitational-force magnitudes instead of vectors.
9. Applying the point-mass formula inside an extended body without checking the mass distribution.
10. Assuming zero net gravitational force means no gravitational forces are present.
11. Treating the massive body as exactly fixed in every two-body problem.
12. Forgetting the sum m₁ + m₂ in the exact relative two-body acceleration.
13. Confusing G with the gravitational parameter μ = GM.
14. Treating g as globally constant rather than an approximation near Earth’s surface.
15. Forgetting that circular-orbit speed follows from gravitational acceleration equaling centripetal acceleration.

29 Practice exercises

1. Derive the dimensions of G from Newton’s gravitational law.
2. If one mass is doubled and the separation is tripled, determine the new force relative to the original force.
3. Starting from r₁₂ = r₂ − r₁, derive the vector force on mass 1 due to mass 2.
4. Show explicitly that the two gravitational forces satisfy Newton’s third law.
5. For m₂ = 5m₁, determine the ratio of the two acceleration magnitudes.
6. Derive the cancellation of test mass from gravitational acceleration.
7. Reproduce the force between two 1 kg masses separated by 1 m.
8. Use superposition to show that two equal masses placed symmetrically about a test mass produce zero net force at the midpoint.
9. Write the gravitational-force integral for a continuous density ρ(r′).
10. Reproduce the simple spherical-Earth value of surface gravitational acceleration.
11. Find the value of g at altitude h = R_E∕2 relative to surface gravity.
12. Derive the exact relative two-body acceleration equation.
13. Derive the circular-orbit speed v_circ = .
14. Derive the circular-orbit form of Kepler’s third law.
15. Explain why gravitational angular momentum about a fixed point source is conserved.

30 Summary

Newton’s law of universal gravitation states

(143)

The force is attractive and acts along the line joining the masses.

A useful vector form is

(144)

The interaction obeys

(145)

When gravitational and inertial mass are equal, the acceleration of a test body near source mass

(146)

is

(147)

For many source masses,

(148)

For a circular orbit around a much larger source mass,

(149)

and

(150)

Newton’s inverse-square law is the foundation for the gravitational fields, extended-body gravity, potential energy, and orbital mechanics developed in the remainder of M07.

References

References