1. Kepler’s Third Law Applied to Our Solar System
First, let’s derive Kepler’s 3rd Law mathematically:
In our Solar System, the planets are in elliptical orbits around the
Sun (according to Kepler’s 1st Law), but the ellipses are very
nearly circles
Let’s assume the orbits are circles
Then a centripetal forced is required
to keep the planet in circular motion
This centripetal force is provided
by the gravity force between the
Sun and the planet

2. Let’s set the centripetal force equal to the gravity forcem1
Fc = Fg
For the centripetal force,
we will not use Fc =
m v2 r
m1 = mass of planet
4 π2 m r
Instead we will use Fc =
MS = mass of Sun T2
r = mean orbital
distance
Fc = Fg
4 π2 m1 r
G m1 m2
G m1 MS = = T2 r2 r2

3. So, for any planet in the Solar System, the ratio of the cube of
4 π2 m1 r
G m1 MS m1 = T2 r2
4 π2 r
G MS = T2 r2
4 π2 r3 = G MS T2
m1 = mass of planet
4 π π2
T T2
MS = mass of Sun r3
G MS
r = mean orbital
distance =
= constant T2
4 π2
So, for any planet in the Solar System, the ratio of the cube of
the mean orbital distance to the square of the orbital period is
always the same number
; let’s determine what the number is, and then apply it to the planets

4. r3 G MS = T2 4 π2 ( 6.67 x 10-11 Nm2/kg2)( 1.99 x 1030 kg) = 4 π2 r3
MS = x 1030 kg
4 π2 r3
= x 1018
for any planet in the SS T2
(distance measured in meters, periods measured in seconds)
Mean distance
Planet from Sun (m) Orbital Period (s)
Mercury x x 106
Venus x x 107
Earth x x 107
Mars x x 107
Jupiter x x 108
Saturn x x 108
Uranus x x 109
Neptune x x 109
Data for our
Solar System:
source:
NASA webpage

5. Test with orbital data from Earth: r = 1.496 x 1011 m
(distance measured in meters,
periods measured in seconds)
= x 1018 T2
Test with orbital data from Earth:
r = x 1011 m
from Table
T = x 107 s r3
( x 1011 )3
3.348 x 1033 = = T2
( x 107 )2
9.960 x 1014
= x 1018
Just for fun, let’s do it again, for another planet
Let’s check Neptune, just because it has such a cool blue color

6. Check it one more time, with your favorite planet (other than Earth)
= x 1018 T2 r3
( x 1012 )3 = T2
( x 109 )2
9.083 x 1037 =
2.270 x 1019
r = x 1012 m
T = x 109 s
= x 1018
from Table
Check it one more time, with your favorite planet
(other than Earth)

7. r3
G MS =
= constant T2
4 π2
Kepler’s 3rd Law holds for any orbital system
; the one we tested was the system of our Sun and the planets, but it also holds for
the system of Jupiter and its moons (the Jovian system)
The constant will be different; to find the constant for the Jovian
system, use the mass of Jupiter for MS in the equation above
You can find the constant for the Earth-Moon system by using the
mass of Earth for MS in the equation above
; then test it with the data from our Moon
; re-test with data from another satellite, such as the International Space Station (you’ll have to look up the data
for the distance and period)