PHYS 20 LESSONS Unit 5: Circular Motion Gravitation Lesson 5: Gravitation.

PHYS 20 LESSONS

Unit 5: Circular Motion

Gravitation

Lesson 5: Gravitation

Reading Segment #1:

Early TheoriesKepler’s Laws

To prepare for this section, please read:

Unit 5: pp. 16-17

Introduction to Gravitation

Throughout history, people have wondered about how

and why the stars and planets move.

Early Greeks (6th / 7th century B.C.)

They believed that the stars revolved around the Earth

in perfect circles.

This is called the geocentric view of the heavens.

"geo" = Earth

"centric" = centre So, Earth-centred

Copernicus (1543)

Copernicus suggested that the planets revolve about the

sun in perfect circles.

This is called the heliocentric ("sun-centred") view

of the heavens.

This was a key advance for modern understanding.

D1. Kepler's Laws

In the 16th century, Tyco Brahe carefully observed

planetary motion for 20 years.

He then gave all of this data to his student, Johann Kepler

In 1609, Kepler found patterns in the data.

He summarized these patterns into 3 laws.

Law #1: Elliptical Orbits

- the planets revolve about the sun in ellipses

- the sun is one of the two foci of the ellipse

i.e.

F1 F2

Law #2: Equal Areas

- the planets sweep out equal areas in equal periods

of time

i.e.

A1 A2

A1 = A2

Animation

Kepler's 2nd Law:

http://www.walter-fendt.de/ph11e/keplerlaw2.htm

Law #3: Kepler's Equation

The ratio T2 / R3 is constant for all planets revolving about

the sun (or about any central mass).

i.e. k = T12 = T2

2

R13 R2

3

where

R is the average radius of the oribit (in m)

T is the orbital period, or time for 1 lap (in s)

k is Kepler's constant

Ex. 1 An asteroid has a period of 8.1 107 seconds.

What is its mean (average) radius around the sun?

Note: Kepler's constant for objects around the sun

is 2.985 10-19 s2 / m3

k = T2

R3

k R3 = T2

R3 = T2

k

R = T2 = (8.1 107 s)2

3 k 3 (2.985 10-19 s2 / m3)

= 2.8 1011 m

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 5 p. 19 #1 - 4

Reading Segment #2:

Inverse Square Law

To prepare for this section, please read:

Unit 5: p.18

Inverse Square Law (Newton, 1687)

Kepler could not explain why the planets moved this way,

but Isaac Newton could.

When an apple fell on Newton's head due to Earth's gravity,

he wondered:

Could the Moon be held in orbit due to Earth's gravity

as well?

Is it not the same kind of force (Fg) ?

Consider the Moon going in uniform circular motion

around the Earth: Fg

Moon (m)

ac

Earth

The force of gravity is the centripetal force,

holding the Moon in orbit.

It follows that

Fg = Fc

Fg = m 4 2 r

T2

It follows that

Fg = Fc

Fg = m 4 2 r

T2

Now, based on Kepler's 3rd law,

k = T2

R3

or T2 = k R3

Combining Fg = m 4 2 r and T2 = k R3

T2

we find

Fg = m 42 R

k R3

or

Fg = m 42

k R2

where R is the distance from the centre of the planet

Newton's Conclusion: Based on Fg = m 42

k R2

Fg has an inverse square relationship with R.

i.e. Fg 1

R2

Newton was able to use this formula to accurately predict

the centripetal acceleration of the Moon (ac = 2.7 10-3 m/s2)

Ex. 2 An object experiences a Fg of 18 N at a distance D

from the centre of the Earth. What would this Fg be

if the distance was 6D ?

Based on Newton's inverse square law,

Fg 1

R2

So, if R 6, then Fg 62

Thus, the new Fg would be 18 N = 0.50 N

62