Download Kepler`s 3rd Law Applied to our Solar System

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
Let’s set the centripetal force equal
to the gravity force
m1
Fc = Fg
For the centripetal force,
m v2
we will not use
Fc =
r
4 π2 m r
Instead we will use Fc =
T2
Fc = Fg
4 π2 m1 r
T2
=
G m1 m2
r2
=
m1 = mass of planet
MS = mass of Sun
r = mean orbital
distance
G m1 MS
r2
4 π2 m1 r
T2
4 π2 r
=
=
T2
G m1 MS
r2
m1
G MS
r2
4 π2 r3 = G MS T2
4 π2 T2
4 π2 T2
m1 = mass of planet
r3
T2
MS = mass of Sun
r = mean orbital
distance
=
G MS
4 π2
= constant
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
r3
T2
=
=
G MS
4 π2
m1
( 6.67 x 10-11 Nm2/kg2)( 1.99 x 1030 kg)
4 π2
r3
T2
= 3.36 x 1018
MS = 1.99 x 1030 kg
for any planet in the SS
(distance measured in meters, periods measured in seconds)
Planet
Data for our
Solar System:
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Mean distance
from Sun (m)
Orbital Period (s)
5.79 x 1010
1.082 x 1011
1.496 x 1011
2.279 x 1011
7.786 x 1011
1.4335 x 1012
2.8725 x 1012
4.4951 x 1012
7.60 x 106
1.941 x 107
3.156 x 107
5.936 x 107
3.742 x 108
9.285 x 108
2.643 x 109
5.167 x 109
source:
NASA webpage
r3
T2
= 3.36 x
(distance measured in meters,
periods measured in seconds)
1018
Test with orbital data from Earth:
r3
T2
=
( 1.496 x 1011 )3
( 3.156 x 107 )2
=
r = 1.496 x 1011 m
T = 3.156 x 107 s
from Table
3.348 x 1033
9.960 x 1014
= 3.36 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
r3
T2
r3
T2
= 3.36 x 1018
=
( 4.4951 x 1012 )3
( 5.167 x 109 )2
=
9.083 x 1037
2.270 x 1019
r = 4.4951 x 1012 m
T = 5.167 x 109 s
= 3.40 x 1018
Check it one more time, with your favorite planet
(other than Earth)
from Table
r3
T2
=
G MS
4 π2
= constant
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)