Kepler's Laws

Brahe and Kepler - A feud worthy of Shakespeare...

In Science, students at times can feel that scientists lead dull, isolated lives with very little real excitement. The reality is, scientists lives at times can be in fact full of sworn enemies, royal intrigue, running swordfights, disfiguring injuries and even bizarre death scenes more worthy of fiction...

The image above shows a painting of Tycho Brahe, a astronomer and alchemist born in 1546. Brahe held many positions throughout a very colourful life ( see the video sequence below ) but his most important role from a scientific point of view was as the Imperial Court Astronomer and Astrologer to the Holy Roman Emperor Rudolf II. In this role, Brahe kept highly detailed records of the motion of the ( at that time ) 5 planets of the solar system. These records, collected every night over decades, gave Brahe a massive amount of raw data with which to predict astrological information for the Emperor.

In 1600, Brahe received contact from another astronomer, Johannes Kepler. Kepler wanted access to Brahe's data in order to complete his own research on planetary motion, but Brahe refused access unless Kepler became his assistant. Even after this, Brahe gave Kepler only very limited access and prevented Kepler from completing his work.

It was only after Brahe's death in 1601 ( in unusual circumstances ) that Kepler gained access to Brahe's data, allowing him craft his Three Laws of Planetary Motion.

The video below shows a brief overview of the life of the astrologer and scientist Tycho Brahe :-

Kepler's Laws

Kepler devised Three Laws based upon the data received from Brahe after his death :-

The Planets Move in elliptical orbits with the Sun at one of the two foci.

The radius vector drawn from the Sun to a planet sweeps out equal area in equal time.

The square of the orbital period is proportional to the cube of the semi-major axis ( T2 ∝ r3 ) .

Each of the above laws is based upon empirical observations, and give no explanation as to why the planets follow these paths. The explanation would have to wait until Newton devised his Law of Universal Gravitation ( in fact, Newton used Kepler's Third Law to derive the value of G ).

The diagram below shows a mathematical representation of an Ellipse :-

In case of orbital dynamics, The Sun is present at one focus, with the planet following the path of the ellipse.

Note - At Advanced Higher level calculations will only be completed for circular orbits. A circular orbit is a special case of an an ellipse, where both foci overlap at the centre of the orbit, making the semi-major axis equivalent to the radius of the circle. This allows limited orbital calculations of the planets around the Sun as they follow an approximately circular orbit ( see below ) .

Newton's Law of Universal Gravitation

As has been shown previously, the Gravitational Field Strength varies greatly between locations. In 1687, Isaac Newton published 'The Principia', containing his work on Gravity. Within this publication, Newton had devised a general rule which would allow the calculation of the Force of attraction due to Gravity between any two objects at any separation. This formula is called Newton's Law of Universal Gravitation :-

Where :-

F = Force between the two masses ( N )

G = Gravitational Constant = 6.67x10-11 m3kg-1s-2

m1 = Mass 1 ( kg )

m2 = Mass 2 ( kg )

r = The separation between Masses ( center to center ) ( m )

Note - The above formula works well for all classical mechanics problems, everything that Newton could measure in the 17th century. It is now known, however, that the above formula does not work when dealing with very strong Gravitational fields. In these situations, Einstein's theory of General Relativity must be used.

Example 1 - Gravitational Attraction to the Earth

A pupil of mass 50 kg is sitting on the Earth's surface and is attracted to the Earth which has a mass of 5.97x1024 kg. What is the Gravitational Force that the pupil experiences if the radius of the Earth is 6.38x106 m ?

G = 6.67x10-11 m3kg-1s-2

m1 = 5.97x1024 kg

m2 = 50 kg

r = 6.38x106 m

F = ( 6.67x10-11 x 5.97x1024 x 50 ) / ( 6.38x106 x 6.38x106 )

F = 490 N


W = mg

W = 50 x 9.8

W = 490 N

Note - If the Weight of the pupil is calculated, the same value is found for the Force. This allows the calculation of 'g' at any given point as a rearrangement of w = mg.

Example 2 - Gravitational Attraction between pupils

What is the Gravitational Force between two pupils sitting 2 m apart if both pupils have a mass of 50 kg ?

G = 6.67x10-11 m3kg-1s-2

m1 = 50 kg

m2 = 50 kg

r = 2 m

F = ( 6.67x01-11 x 50 x 50 ) / ( 2 x 2 )

F = 4.17x10-8 N

F = 41.7 nN

Note - As can be seen in the above example, the Gravitational attraction between small objects is negligible but non-zero.

Gravitational Field Diagrams

The Gravitational Field around an isolated Mass can be modelled as shown below :-

The above diagram shows a radial field pattern, with the arrows representing the Force a test mass would experience when placed within the field. The red concentric circles represent points of equipotential, where the Gravitational Force experienced by a point mass is equal.

The field above is non-uniform, and as such the field lines are never parallel and the Gravitational Field Strength changes with distance from the Mass.

However, when taken to a very small section of the field, the field lines are approximately parallel and as such, on a small scale the Field Strength can be approximated as a constant. This allows the use of formulae such as Ep = mgh for small height changes, without requiring a changing value for "g" :-

If the Mass in question is not an isolated Mass, but is interacting with another Gravitational Field, the field lines distort as shown in the diagram below :-

As can be seen above, the closer to the either mass the test mass is placed, the stronger the Gravitational field. Also, there is a point directly between the two masses at which the Gravitational Field Strength is zero. This is due to the two masses applying equal but opposing Gravitational Forces at this point.

The location of this zero point between two masses depends upon the relative field strengths of the masses involved. For example, the diagram below shows the Gravitational Field pattern around the Earth - Moon System :-

As can be seen in the above diagram, the neutral point is not directly between the Earth and the Moon, but at point much closer to the Moon. This is due to the relatively weaker Gravitational Field of the Moon.

Gravitational Field Strength and Altitude

As stated earlier, as the distance from an object increases, the Gravitational Field Strength gets smaller. Due to this, the value of 'g' is reduced at higher altitude. The graph below shows how the value of 'g' varies in relation to height :-

The above relationship shown is an inverse square law relationship. As the distance from the mass is doubled, the value of 'g' is reduced to a quarter.

The value of 'g' below the surface reduces due to the fact that the deeper into a planet the object is, the less mass below to contribute to 'g'. This gives a value of 'g' of zero at the planet's core. The diagram below shows how the value of "g" varies throughout the inside of the Earth :-

Orbital Period of a Satellite

By combining the above formula for the Gravitational Force between objects and the Centripetal Force formula from the dynamics section ( See Radial Acceleration and Centripetal Force ), the orbital Period of any orbiting body can be found :-

Gravitational Force :-

Centripetal Force :-

By combining the above formulae together and by then substituting Velocity, the Period can be found as follows :-

Note - The above formula is a variation on Kepler's third law. It is in fact a reversal of this derivation that Newton used to calculate G in the first place, using Kepler's data for the planets. Kepler's Third Law is generally written in the following form, showing the relationship between T2 and r3 to be a constant :-

Example 3 -

If the orbital Period of a Geostationary satellite is equal to 1 day ( 23.93 hours ), what is the radius of its orbit, if the mass of the Earth is equal to 5.97x1024 kg?

T = 23.93 Hours = 86148 s

G = 6.67x10-11 m3 kg-1 s-2

M = 5.97x1024 kg

r = ?

r3 = ( GMT2 ) / ( 4π2 )

r = 42,144,762 m

r = 42,140 km

Note - In National 5, the altitude of Geostationary satellites was given as 35,786 km. The reason for the difference between values is that the altitude value given in Nat 5 does not take into account the radius of the Earth itself ( 6,378 km ) .