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Kepler
and his
3 planetary laws
Solar System Kepler wanted to be able to explain how the solar
according to system behaved, and he wanted reasons that
Kepler
explained this
How did the planets move (so as match the way we
saw them moving)?
Perhaps their orbits were elliptical not circular
Why were there only 6 planets? Why not 4? Or 9?
There are only 5 perfect 3D figures; if they are
different sizes placed was a perfect shape
between each, this meant that there had to be 6,
and only 6, planets
What was their spacing? Why this spacing?
Maybe the 5 perfect objects, in the correct order,
would give the correct distances between planets
Solar System
according to
Kepler
Kepler’s Laws summary
1. Planetary orbits
are elliptical
2. Each orbit
sweeps equal
areas in equal
times
3. The square of the
orbital period is
proportional to the
cube of the mean
distance from the Sun
Sun
Law 1. Planetary
orbits are elliptical
All planets have
elliptical orbits
around the Sun,
with Sun at one
focus, (not in
the centre).
Sun
Law 2. Each orbit sweeps equal areas in equal times
Here’s a graphic showing Kepler’s 2nd Law
Law 3. The square of the orbital period is proportional
to the cube of the mean distance from the Sun
Examples and questions to go
with using Kepler’s 3rd Law
Use Kepler’s third law for one body in the form: (T)2 = (r)3
where T is given in years and r is given in Astronomical Units.
How do we use this?
(a) You are given the distance from the Sun and asked to find the orbit time
(i) Make sure the distance is in AU (N.B. 1AU = 150,000,000 km)
(ii) Cube it
(iii) Take the square root
(iv) This is the Period in years
Asteroid Fred is 600,000,000 km from the Sun. What is its orbit (in years)?
(i) Convert distance to AU.
i.e. 600,000,000/150,000,000 = 4AU
(ii) Cube it
i.e.4 cubed = 64
(iii) Take the square root
i.e. square root of 64 is 8
(iv) Fred takes 8 years to orbit the Sun
3.22 Use Kepler’s third law for one body in the form: (T)2 = (r)3
where T is given in years and r is given in Astronomical Units.
How do we use this?
(b) You are given the orbit time and asked to find the distance from the Sun
(i) Make sure the orbit time is in years
(ii) Square it
(iii) Take the cube root
(iv) This is the mean distance to the Sun (in AU)
(you could x150,000,000 to convert to km)
Asteroid Wilma takes 27 years to orbit the Sun. What is its distance from it?
(i) Make sure its in years
i.e. it is 27 years
(ii) Square it
i.e. 27 squared = 729
(iii) Take the cube root
i.e. cube root of 729 is 9
(iv) Wilma is an average distance of 9 AU from the Sun
(or 9 x 150,000,000 or 1,350,000,000 km from the Sun)
More Qs
Asteroid Ben is 25AU from the Sun. What is its orbit (in years)?
Dist3 = Per2
Dist = 25AU.
Cube it = 15625.
Find the square root = 125
The orbit is 125 years
Asteroid Jerry takes 64 years to orbit the Sun. What is its distance from it?
Per2 = Dist3 Period = 64. Square it = 4096.
Find the cube root = 16
The asteroid is 16AU from the Sun
Harder Qs
A new planet is discovered and photographed. One year later, it is once again
photographed, and it seems to have moved across the sky by 1 degree.
Roughly, how far away is it from the Sun (to the nearest AU) ?
1 degree a year means about 360 years to complete one orbit.
Square it = 129,600.
Find the cube root = 50.606
This planet is about 51AU from the Sun
Another version of Kepler’s Law is (T1/T2)2 = (r1/r2)3
The ratio of the Periods squared = the ratio of the distances cubed
Example of how to use this
Q1. The planets Laurel and Hardy are in orbit around a star. Hardy is 9 times
further away from this star than Laurel is. Laurel takes 3 years to orbit the star.
How long does Hardy take to orbit ?
A1
Ratio of distances = 9
Ratio of distances cubed = 9 cubed = 729
Ratio of the Periods squared = the ratio of the distances cubed = 729
Ratio of the Periods = square root of 729 = 27
Therefore Hardy takes 27 times longer to orbit than Laurel does
So as Laurel takes 3 years, Hardy must take 81 years
Another version of Kepler’s Law is (T1/T2)2 = (r1/r2)3
The ratio of the Periods squared = the ratio of the distances cubed
Example of how to use this
Q2. The other planets called Abbott and Costello orbit another star. Abbott is 4
times further away from this star than Costello is. Costello takes 5 years to orbit
the star. How long does Abbott take to orbit ?
A2
Ratio of distances = 4
Ratio of distances cubed = 4 cubed = 64
Ratio of the Periods squared = the ratio of the distances cubed = 64
Ratio of the Periods = square root of 64 = 8
Therefore Abbott takes 8 times longer to orbit than Costello does
So as Costello takes 5 years, Abbott must take 40 years