Kepler’s Laws
Planetary & Satellite Motion
Pg 248-251 HRW
History – Johannes Kepler

Born in Germany in 1571

Assistant to Tycho Brahe (Danish)
• Note: Newton was born in 1642
• Brahe, as an astronomer, meticulously
•
gathered planetary data
Kepler analyzed Brahe’s Mars data and
postulated 3 foundation laws of planetary
motion (1609+)
Kepler’s First Law - Ellipses

Planets move in elliptical orbits with the
Sun as one of the foci
What are the implications for conservation of energy?
Conservation of
Energy – satellite
motion

Circular
• PE, KE do not
change

Elliptical
• PE max @ apogee
• Max orbit radius
• KE max @ perigee
• Max speed
Kepler’s Second Law – Equal
Areas

A line from the Sun to a planet sweeps
out equal areas in equal times.
“Conservation of Angular Momentum” v1r1 = v2r2
Kepler’s Third Law – Periods



For all planets, the square of the orbital
period (T) of a planet is proportional to the
cube of the average orbital radius (do), or
T2  do3, or
T2/ do3 = constant = 42/GM, where
• G = 6.67 x 10-11 N-m2/kg2
• M = mass of the foci body (ex. Sun)
4
3
2
T 
do
GM
2
Practice Kepler’s 3rd Law


What is the orbital period (T) of Neptune, in
seconds and years, if its average orbital
radius (do) is 4.495e12 and the mass of the
sun is 1.99e30 kg?
Solve T2 = 4π2/(GM) x do3
• T2 = 4π2/(6.67e-11*1.99e30)*(4.495e12)3
• T2 = 2.70e19
• T = 5.2e9 seconds
• T = ~165 (Earth) years
Example of Kepler’s 3rd Law


Calculate K’s 3rd law constant for Neptune,
given an orbital period (T) of 5.21e9 sec
and an average orbital radius (do) of
4.495e12 m.
Solve: T2/do3
• (5.21e9)2/(4.495e12)3
• = 2.97e-19 sec2/m3
Kepler Practice

The shuttle orbits the Earth at 400 kms
above the surface. If the radius and
mass of the Earth are 6.37 x 106 m and
5.98 x 1024 kg, respectively:
• What is the period of the shuttle’s orbit (in
seconds)?

Solve using K’s 3rd Law T2= 4π2do3/(GM)
• T = 5541 sec
• T = 92.3 minutes
Kepler Practice

Given that the moon’s period about the
earth is 27.32 days and that the earthmoon distance is 3.84 x 108 m,
• Calculate the mass of the earth.

Solve using K’s 3rd Law: T2= 4π2do3/(GM)
• M = 4π2do3/(GT2)
• M = 6.01 x 1024 kg