Kepler’s Laws

Newton’s Universal Law of gravitation

Gravitational and inertia mass

Gravitational potential energy

Chapter 11 – Gravity Lecture 1 April 6, 2010

Kepler’s (empirical) Laws• Kepler’s First Law:

– All planets move in elliptical orbits with the sun at one focus

• Kepler’s Second Law:– The radius vector drawn from the Sun to a planet sweeps out equal areas in equal times, dA/dt = const. (the law of equal areas)

• Kepler’s Third Law: – The square of the orbital period of any planet is proportional to the

cube of the semimajor axis of the elliptical orbit.

Today we can understand the physical reasons for these laws …

Let’s remind us first of the geometry of the ellipse and then discuss the three laws.

T 2 =Cr3

Ellipses

e = c/a

Planetary Orbits

Kepler found that the orbit of Mars was an ellipse, not a circle.

Kepler had access to very good data from the astronomer Tycho Brahe in Prague. See table for today’s data.

Kepler’s Third Law

After many years of work Kepler found an intriguing correlation between the orbital periods and the length of the semimajor axis of orbits.

T 2 =Cr3

Of the satellites shown revolving around Earth, the one with the greatest speed is

• 1 • 2 • 3 • 4 • 5

Of the satellites shown revolving around Earth, the one with the greatest speed is

• 1 • 2 • 3 • 4 • 5

The constant-area law.

The orbits of two planets orbiting a star are shown. The semimajor axis of planet A is twice that of planet B. If the

period of planet B is TB, the period of planet A is

B

B

B

B

B

TT

TT

T

4.E3.D

3.C22.B

2.A

rA A

B

rB

A’

The orbits of two planets orbiting a star are shown. The semimajor axis of planet A is twice that of planet B. If the

period of planet B is TB, the period of planet A is

B

B

B

B

B

TT

TT

T

4.E3.D

3.C22.B

2.A

rA A

B

rB

A’

Newton’s Universal Law of Gravity

m M

F12 F21

r

Newton’s Law of Gravity• Newton’s law of gravity will provide a physical theory of Kepler’s laws.

Fur

12 =GmMR 2 r̂12

m M

F12 F21

r

G =6.67 ⋅10−11 Nm2

kgFg =G

mMr2

Magnitude of force

The Cavendish experiment

04/22/23 Physics 201, UW-Madison 16

Measuring G• G was first measured by

Henry Cavendish in 1798• The apparatus shown here

allowed the attractive force between two spheres to cause the rod to rotate

• The mirror amplifies the motion

• It was repeated for various masses

Magnitude = Gm Mr2

G =6.67 ×10−11 Nm 2 / kg2

Gravitational and inertial mass

• Gravitation is a force that acts on the gravitational mass (the masses are the source)

• Newton’s Law of motion acts on the inertial mass

• In principle, they are not necessarily related, that the gravitational mass mg is not the same as mi

(but they are, up to the current experimental accuracy)

Fg =Gm gMg

r2

F =m ia

Kepler’s Third Law derived from Newton’s Law

Centripetal force = gravitational force

Easily derived for a circular orbit

Fcent =mv2

R=mRω 2 Fg =G

mMR 2

mRω 2 =GmMR 2

R4π 2

T 2 =GMR 2

ω =2πT

R3

T 2 =GM4π 2 =const.

T 2 =4π 2

GMS

R 3Kepler’s Third Law

Kepler’s Third Law derived from Newton’s Law

Extension for elliptical orbits, (Without proof R a)

T 2 =4π 2

GMS

a3Kepler’s Third Law

T 2 =4π 2

GMS

R 3

Where a is the semimajor axis

A woman whose weight on Earth is 500 N is lifted to a height of two Earth radii above the surface of Earth. Her weight

• decreases to one-half of the original amount. • decreases to one-quarter of the original amount. • does not change. • decreases to one-third of the original amount. • decreases to one-ninth of the original amount.

A woman whose weight on Earth is 500 N is lifted to a height of two Earth radii above the surface of Earth. Her

weight

A. decreases to one-half of the original amount. B. decreases to one-quarter of the original amount. C. does not change. D. decreases to one-third of the original amount. E. decreases to one-ninth of the original amount.

From work to gravitational potential energy.

In the example before, it does not matter on what path the person is elevated to 2 Earth radii above. Only the final height (or distance) matters for the total amount of work performed.

Potential Energym

M

r

Force (1,2) = −Gm Mr1,22

r̂1,2 G =6.67 ×10−11 Nm 2 / kg2

Work done to bring mass m from initial to final position.

PE =−W =−rFgdrr

i

f

∫ =− −GmMr2

⎛⎝⎜

⎞⎠⎟

i

f

∫ dr⇓

=GmM r−2i

f

∫ dr=−GmM1rf

−1ri

⎛⎝⎜

⎞⎠⎟

PE =−GMmr

PE(r=∞)=0

U(r)=−GMmrZero point is arbitrary. Choose zero at infinity.

rr

12

1

2

Demo