Essential idea: The Newtonian idea of gravitational

force acting between two spherical bodies and the

laws of mechanics create a model that can be used

to calculate the motion of planets.

Nature of science: Laws: Newton’s law of gravitation

and the laws of mechanics are the foundation for

deterministic classical physics. These can be used

to make predictions but do not explain why the

observed phenomena exist.

Topic 6: Circular motion and gravitation

6.2 – Newton’s law of gravitation

Understandings:

• Newton’s law of gravitation

• Gravitational field strength

Applications and skills:

• Describing the relationship between gravitational force

and centripetal force

• Applying Newton’s law of gravitation to the motion of

an object in circular orbit around a point mass

• Solving problems involving gravitational force,

gravitational field strength, orbital speed and orbital

period

• Determining the resultant gravitational field strength

due to two bodies

Topic 6: Circular motion and gravitation

6.2 – Newton’s law of gravitation

Guidance:

• Newton’s law of gravitation should be extended to

spherical masses of uniform density by assuming

that their mass is concentrated at their centre

• Gravitational field strength at a point is the force per

unit mass experienced by a small point mass at that

point

• Calculations of the resultant gravitational field strength

due to two bodies will be restricted to points along

the straight line joining the bodies

Topic 6: Circular motion and gravitation

6.2 – Newton’s law of gravitation

Data booklet reference:

𝐹 = 𝐺𝑀𝑚

𝑟2

𝑔 =𝐹

𝑚

𝐹 = 𝐺𝑀

𝑟2

Theory of knowledge:

• The laws of mechanics along with the law of gravitation create the

deterministic nature of classical physics. Are classical physics

and modern physics compatible? Do other areas of knowledge

also have a similar division between classical and modern in

their historical development?

Topic 6: Circular motion and gravitation

6.2 – Newton’s law of gravitation

Utilization:

• The law of gravitation is essential in describing the

motion of satellites, planets, moons and entire

galaxies

• Comparison to Coulomb’s law (see Physics sub-topic

5.1)

Aims:

• Aim 4: the theory of gravitation when combined and

synthesized with the rest of the laws of mechanics

allows detailed predictions about the future position

and motion of planets

Topic 6: Circular motion and gravitation

6.2 – Newton’s law of gravitation

Field (Fundamental) Forces

Gravitational Force

Gravitational force is cumulative and extended to infinity. The gravitational force between two object

is due to the cumulative effect of billions of billions of the atoms made up both bodies. This means

that the larger the body (contain more matter), the stronger the force. But on the scale of individual

particles, the force is extremely small, only in the order of 10 -38 times that of the strong force.

Electromagnetic Force

The force is long range, in principle extending over infinite distance. However, the strength can

quickly diminishes due to shielding effect. Many everyday experiences such as friction and air

resistance are due to this force. This is also the resistant force that we feel, for example, when

pressing our palm against a wall. This is originated from the fact that no two atoms can occupy the

same space. However, its strength is about 100 times weaker within the range of 1 fm, where the

strong force dominates. But because there is no shielding within the nucleus, the force can be

cumulative and can compete with the strong force. This competition determines the stability

structure of nuclei.

Weak Nuclear Force

Responsible for nuclear beta decay and other similar decay processes involving

fundamental particles. The range of this force is smaller than 1 fm and is 10-7 weaker than

the strong force

Strong Nuclear Force

This force is responsible for binding of nuclei. It is the dominant one in reactions and decays of most

of the fundamental particles. This force is so strong that it binds and stabilize the protons of similar

charges within a nucleus. However, it is very short range. No such force will be felt beyond the order

of 1 fm (femtometer or 10-15 m).

One of the most significant intellectual achievements in the

history of thought. It is universal – it applies to all objects

regardless of their location anywhere in the Universe. Every

object in the universe attracts every other object.

Gravitational force between two point masses m1 and m2 is

proportional to their product, and inversely proportional to the

square of their separation r.

1 2

2

m mF = G

r

G = 6.67x10-11 Nm2/ kg2 – “Universal gravitational constant”

the same value anywhere in the universe - very small value

– no significant forces of attraction between ordinary sized objects.

r

The actual value of G, the universal gravitational constant, was not known

until Henry Cavendish conducted a tricky experiment in 1798 to find it.

Newton’s law of gravitation

The earth, planets,

moons, and even

the sun, have many

layers – kind of

like an onion:

In other words,

NONE of the

celestial bodies we

observe are point

masses.

Given that the law is called the universal law of

gravitation, how do we use it for planets and such?

Newton’s law of gravitation

Newton spent much time developing integral calculus to prove that

“A spherically symmetric shell of mass M acts as if all of its

mass is located at its center.”

Thus F = Gm1m2 / r 2 works not only for point masses, which have

no radii, but for any spherically symmetric distribution of mass at

any radius like planets and stars.

m

M

r

-Newton’s shell theorem.

EXAMPLE: The earth has a mass of M = 5.981024kg and the

moon has a mass of m = 7.361022kg. The mean distance

between the earth and the moon is 3.82108m. What is the

gravitational force between them?

SOLUTION: F = GMm / r 2.

F = (6.67×10−11)(5.981024 )(7.361022 ) / (3.82108)2

F = 2.011020 N.

r is the distance between the centers of the masses.

m1 m2

rF12 F21

FYI The radius of each

mass is immaterial.

What is the speed of the moon in its orbit about earth?

SOLUTION: FC = FG = mv 2 / r .

2.011020 = ( 7.361022 ) v 2 / 3.82108 v = 1.02103 ms-1.

What is the period of the moon (in days) in its orbit about earth?

SOLUTION: v = 2r / T.

T = 2r / v = 2( 3.82108 ) / 1.02103

= (2.35 106 s)(1 h / 3600 s)(1 d / 24 h) = 27.2 d.

Gravitational field strength

Suppose a mass m is located a distance r from a another mass M.

The gravitational field strength g is the force per unit mass acting

on m due to the presence of M.

𝑔 = 𝐹

𝑚The units are newtons per kilogram (N kg -1 = m s -2).

Suppose a mass m is located on the surface of a planet of radius R.

We know that it’s weight is F = mg.

But from the law of universal gravitation, the weight of m is equal to its

attraction to the planet’s mass M and equals F = GMm / R 2.

mg = GMm / R 2.

𝑔 =𝐺𝑀

𝑅2

gravitational field strength at surface of

a planet of mass M and radius R

Double the distance from the centre, g is 4 times less,

and so is weight

PRACTICE: A 525-kg satellite is launched from the

earth’s surface to a height of one earth radius above the

surface. What is its weight (a) at the surface, and (b) at

altitude?

SOLUTION:

(a) AT SURFACE: gsurface = 9.8 m s-2.

F = mg = (525)(9.8) = 5145 N. W=5100 N

(b) AT ALTITUDE: gsurface+R = 2.45 m s-2.

F = mg = (525)(2.46) = 1286.25 N W=1300N

Gravitational field strength

Compare the gravitational force formula

F = GMm / r 2 (Force)

with the gravitational field formula

g = GM / r 2 (Field)

Note that the force formula has two masses, and the force is the

result of their interaction at a distance r.

Note that the field formula has just one mass – namely the mass

that “sets up” the local field in the space surrounding it. It “curves”

it.

The field view of the universe (spatial disruption by a

single mass) is currently preferred over the force view

(action at a distance) as the next slides will show.

Consider the force view.

In the force view, the masses know the locations of each

other at all times, and the force is instantaneously felt by both

masses at all times.

This requires the “force signal” to

be transferred between the masses

instantaneously.

Einstein’s special theory of relativity

states unequivocally that

the fastest any signal can travel

is at the (finite) speed of light c.

SUN

Thus the action at a distance “force signal” will be slightly

mdelayed in telling the orbital mass when to turn.

The end result would have to be an expanding spiral

mmotion, as illustrated in the following animation:

We do not observe planets leaving

mtheir orbits as they travel around the

msun.

Thus force as action at a distance

doesn’t work if we are to believe

special relativity.

And all current evidence points to

mthe correctness of special relativity.

SUN

So how does the field view take care of this “signal lag”

problem?

Simply put – the gravitational field distorts the space around

the mass that is causing it so that any other mass placed at

any position in the field will “know” how to respond

immediately.

Think of space as a stretched

rubber sheet – like a drum head.

Bigger masses “curve” the rubber

sheet more than smaller masses.

The next slide illustrates this gravitational “curvature” of the

space surrounding, for example, the sun.

each mass “feels” a different “slope” and must travel

mat a particular speed to stay in orbit.

The field view eliminates the need for long distance

msignaling between two masses. Rather, it distorts the

mspace about one mass.

(a) The field arrow is bigger for m2 than m1. Why?

(b) The field arrow always points to M. Why?

M

In the space surrounding the mass M which sets up

mthe field we can release “test masses” m1 and m2

mas shown to determine the strength of the field.

m1

m2 g1g2

(a) Because g = GM / r2.

It varies as 1 / r2.

(b) Because the

gravitational force is attractive.

The field arrows of the inner ring are longer than the field arrows of

the outer ring and all field arrows point to the centerline.

M

By “placing” a series of test masses about a larger

mmass, we can map out its gravitational field:

SUN

SUN

Gravitational field strength

Simplification: Instead of having arrows at every point

we take the convention of drawing “field lines”

as a single arrow.

Density/concentration of the

field lines convey us comparable

strengths.

The closer together the field

lines, the stronger the field.

In the red region the field lines are closer

together than in the green region.

Thus the red field is stronger than the

green field.

PRACTICE: Sketch the gravitational field about the earth (a) as

viewed from far away, and (b) as viewed “locally” (at the surface).

SOLUTION:

(a)

The closer to the surface we are, the more uniform the field

concentration.

(b)

or

EXAMPLE: Find the gravitational field strength at a

point between the earth and the moon that is right

between their centers.

SOLUTION:

Make a sketch.

Note that r = d / 2 = 3.82108 / 2 = 1.91 108 m.

gm = Gm / r 2

gm = (6.67×10-11)(7.361022)/(1.91108)2 = 1.3510-4 N.

gM = GM / r 2

gM = (6.67×10-11)(5.981024)/(1.91108)2 = 1.0910-2 N.

Finally, g = gM – gm = 1.0810 -2 N,→.

M = 5.981024 kgm = 7.361022 kg

gMgm

d = 3.82108 m

PRACTICE: Two spheres of equal mass

and different radii are held a

distance d apart. The gravitational field

strength is measured on the line joining

the two masses at position x

which varies.

Which graph shows the variation of g with x correctly?

There is a point between M and m where g = 0.

Since g = Gm / R 2 and Rleft < Rright, then gleft > gright at the surfaces

of the masses.

EXAMPLE: Derive Kepler’s law, which states that the period T of an object in a circular orbit about a body of mass M is given by

𝑇2=4𝜋2

𝐺𝑀𝑟3

SOLUTION:

▪ 𝐹𝑐 =𝑚𝑣2

𝑟= 𝐺

𝑀𝑚

𝑟2 𝑣2 = 𝐺𝑀

𝑟

▪2𝜋𝑟

𝑇

2

= 𝐺𝑀

𝑟

▪ 𝑇2 =4𝜋2

𝐺𝑀𝑟3

EXAMPLE: A satellite in geosynchronous orbit takes 24 hours to

orbit the earth. Thus, it can be above the same point of the earth’s

surface at all times, if desired. Find the necessary orbital radius,

and express it in terms of earth radii. RE = 6.37106 m.

SOLUTION:

𝑇 = (24 ℎ)(3600 𝑠 ℎ−1 ) = 86400 𝑠

𝑚𝑣2

𝑟= 𝐺

𝑀𝑚

𝑟2 𝑣2 = 𝐺

𝑀

𝑟

2𝜋𝑟

𝑇

2

= 𝐺𝑀

𝑟

𝑟3 =𝐺𝑀

4𝜋2𝑇2 ⇒ 𝑟3 =

6.6710−115.981024

4𝜋2(86400)2

𝑟 = 42250474 𝑚 = 6.63 𝑅𝐸

FYI

Kepler’s third law originally said that the square of the period was

proportional to the cube of the radius – and nothing at all about

what the constant of proportionality was. Newton’s law of

gravitation was needed for that!

𝑚𝑣2

𝑟= 𝐺

𝑀𝑚

𝑟2 𝑣2 = 𝐺

𝑀

𝑟

2𝜋𝑟

𝑇

2

= 𝐺𝑀

𝑟⇒ 𝑟3 =

𝐺𝑀

4𝜋2 𝑇2

▪ 𝐹𝑐 =𝑚𝑣2

𝑟= 𝐺

𝑀𝑚

𝑟2 𝑣2 = 𝐺𝑀

𝑟

▪2𝜋𝑟

𝑇

2

= 𝐺𝑀

𝑟

▪ 𝑇2 =4𝜋2

𝐺𝑀𝑟3 ⇒ 𝑇 =

4𝜋2

𝐺𝑀

3/2

𝑟3/2

𝑚𝑣2

𝑟= 𝐺

𝑀𝑚

𝑟2 𝑟 = 𝐺

𝑀

𝑣2

𝑟𝑋𝑟𝑦

=

𝐺𝑀𝑣𝑥

2

𝐺𝑀𝑣𝑌

2

=𝑣𝑌

2

𝑣𝑥2 =

2𝜋𝑟𝑌𝑇𝑌

2𝜋𝑟𝑋𝑇𝑋

2

=𝑇𝑋𝑟𝑌𝑇𝑌𝑟𝑋

2

=8𝑟𝑌𝑟𝑋

2

𝑟𝑋𝑟𝑦

3

= 64 ⇒𝑟𝑋𝑟𝑦

= 4

PRACTICE: If the elevator is accelerating upward at 2 ms-2, what will Dobson observe the dropped ball’s acceleration to be?SOLUTION: Since the elevator is accelerating upward at 2 ms-2 to meet the ball which is accelerating downward at 10 ms-2, Dobson would observe an acceleration of 12 ms-2.If the elevator were accelerating downward at 2, he would observe an acceleration of 8 ms-2.

Solving problems involving gravitational field

Consider Dobson inside an elevator which is not moving…

If he drops a ball, it will accelerate downward at 10 ms-2 as expected.

PRACTICE: If the elevator were to accelerate

downward at 10 ms-2, what would Dobson

observe the dropped ball’s acceleration to be?

SOLUTION:

He would observe the acceleration of the ball

to be zero!

He would think that the ball was “weightless!”

FYI

The ball is NOT weightless, obviously. It is

merely accelerating at the same rate as Dobson!

How could you get Dobson to accelerate

downward at 10 ms-2?

Cut the cable!

The “Vomit Comet”

PRACTICE: We have all seen astronauts experiencing

“weightlessness.” Explain why it only appears that they are

weightless.

SOLUTION: The astronaut, the spacecraft, and the tomatoes, are

all accelerating at ac = g.

They all fall together and appear to be weightless.

International Space

Station

In deep space, the r in

F = GMm / r 2

is so large for every m

that F, the force of

gravity, is for all

intents and

purposes, zero.

PRACTICE: Discuss the concept of weightlessness in deep space.

SOLUTION: Only in deep space – which is defined to be far, far

away from all masses – will a mass be truly weightless.

Since the satellite is in circular orbit FC = mv 2/ r.

Since the satellite’s weight is holding it in orbit, FC = mg.

Thus mv 2/ r = mg.

Finally g = v 2/ r.

R

x

(ac = g in circular orbits).

A satellite of mass m orbits a planet of mass M and radius R as shown.The radius of the circular orbit of the satellite is x. The planet may be assumed to behave as a point mass with its mass concentrated at its centre.

(a) Deduce that the linear speed v of the satellite in its

orbit is given by expression 𝑣 =𝐺𝑀

𝑥

𝑣2 =𝐺𝑀

𝑥 𝑣 =

𝐺𝑀

𝑥

𝑔 =𝐺𝑀

𝑥2 𝑎𝑐 =

𝑣2

𝑥=

𝐺𝑀

𝑥2

(b) Derive an expression, in terms of m, G, M and x, for the kinetic energy of the satellite.

𝐹𝑟𝑜𝑚 𝑎 𝑣2 =𝐺𝑀

𝑥

𝐸𝐾 =1

2𝑚𝑣2 =

𝐺𝑀𝑚

2𝑥

It is the gravitational force.

This question is about gravitation. A binary star consists of two stars that each follow circular orbits about a fixed point P as shown. The stars have the same orbital period T. Each star may be considered to act as a point mass with its mass concentrated at its centre. The stars, of masses M1 and M2, orbit at distances R1 and R2 respectively from point P.

(b) By considering the force acting on one of the stars, deduce that

the orbital period T is given by the expression

(a) State the name of the force that provides the centripetal force

for the motion of the stars.

M1 experiences FC = M1v12/ R1.

Since v1 = 2R1/ T, then v12 = 42R1

2/ T 2.

Thus FC = FG M1v12/ R1 = GM1M2 / (R1+R2)

2.

42R1(R1+R2)2 = GM2T

2

T 2 = R1(R1+R2)2

42

GM2

M1(42R1

2/ T 2) / R1 = GM1M2 / (R1+R2)2

Note that FG = GM1M2 / (R1+R2)2.

M2M1

R1

R2

P

M2M1

R1

R2

P From (b)

T 2 = (42 / GM2)R1(R1+R2)2.

From symmetry

T 2 = (42/ GM1)R2(R1+R2)2.

(1 / M2)R1 = (1 / M1)R2

M1 / M2 = R2 / R1

Since R2 > R1, we see that M1 > M2.

(c) The star of mass M1 is closer to the point P than the star of mass M2.

Using the answer in (b), state and explain which star has the larger mass.

(42/ GM2)R1(R1+R2)2 = (42/ GM1)R2(R1+R2)

2