Matt Springer is a graduate student of physics at Texas A&M university. He is also an occasional writer and tinkerer, and he is probably too curious for his own good.
when you order from Amazon, and a fraction of the purchase price will be sent to me at zero cost to you. Much obliged, and thanks for your patronage.Happy new year! While we're thinking about years, why don't we think about one of the first guys to explore the physical reason behind the year?
This nattily dressed gentleman is Johannes Kepler, who worked out three basic mathematical laws of planetary motion around the year 1605. He did so from scratch, purely based on phenomenological examination of astronomical data. Newton wouldn't figure out the theory behind these laws for about another century, but with Newton's classical mechanics we can show that his basic laws of motion imply that Kepler's laws must be true.
The third of Kepler's laws tells us how the length of a year varies among the different planets. Since a year on a particular planet is just the length of time required to make an orbit, this law relates the time per orbit - the period - to the distance between the planet and the sun. In mathematical language:
The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
Planets move in ellipses, and the semi-major axis is just the distance to the center of the ellipse at the farthest point. For a period T and a semi-major axis r, we can write this law more compactly as:
Double the distance to the sun r and you'll have increased the year T by a factor of 2^1.5 = 2.83 or so. For instance, the semi-major axis of Mars is about 1.524 times larger than that of earth, and so the Martian year ought to be 1.524^1.5 = 1.881 times larger than the earth's 1 year. As indeed is exactly is - the Martian year is just under 687 earth days.
So can we prove this relationship from Newton's laws? Of course! We can save ourselves a lot of trouble with the simplifying assumptions that the orbits are exactly circular. This is not far from true for the planets, and the intrepid are welcome to try the more challenging but perfectly doable proof for the general elliptical case. We'll just do the circular case. We know by Newton's law of gravitation, the gravitational force on a planet of mass m by the sun (mass M) is given by:
Where G is the universal gravitational constant and r is the radius of the orbit. Newton's laws of motion also required that the force needed to keep an object moving in a uniform circle is equal to:
Both the above expressions equal F, and so they equal each other. Before writing that down, let's figure out how to get rid of the v (which isn't in Kepler's laws explicitly) and replace it with T the orbital period. We know that by definition the velocity is distance/time. The distance is the circumference of the orbit which is just 2*pi*r, and the time is the period T. Thus v = 2*pi*r/T. Swap that into the above:
Fiddle with the algebra a bit and simplify, and we can write that as:
The m's cancel, and everything remaining but r and T are constants. And they're related in exactly the way Kepler wrote. He was a pretty smart guy.
Physical Science
Comments
And, if you write the law as R^3/T^2 is a constant, you can show that the ratio of the constants (for different bodies that have orbiting satellites) is the ratio of the masses of the respective bodies. It even works for the Solar system revolving around the galactic center!
This is so simple and neat that even my high school students get a big kick out of it.
Posted by: joemac53 | January 2, 2010 5:27 PM
Reading this prompted me to do a little research on the subject of Earth's orbital properties.
Interesting to report that tomorrow (3rd Jan) is perihelion.
Posted by: dominich | January 2, 2010 5:54 PM
You suck.com
Posted by: Anonymous | August 30, 2010 11:50 PM
The main motive for Kepler's discoveries was to adjust the recorded observations to take account of Copernicus's discovery that the Earth being the observation point was not stationary but moved round the Sun.
Posted by: Peter L. Griffiths | April 7, 2011 11:29 AM
Further to my comment of 7 April 2011, two of Kepler's discoveries are contained in the Introduction to Astronomia Nova (1609), they are that the velocity of the planets is inversely related to their distances from the Sun and that the shape of the planetary orbit is elliptical. Greater mathematical precision is given to these discoveries in Kepler's later works also the important concept of the foci.
Posted by: Peter L Griffiths | May 12, 2011 10:21 AM
Further to my comment of 12 May 2011, Galileo's law of falling bodies v^2=d can be reconciled with Kepler's distance law v^2=1/r where L indicates a small change as follows v^2+Lv^2=d+Ld=1/(r-Ld). This is the usual method of measuring velocity distance per unit time. For the reciprocal method of measuring the same velocity, time per unit distance we have v^2+Lv^2=r+Lr=1/(d-Lr). d+r equals the length of the major axis of the elliptical orbit, so that any change of d equals an opposite change of r.
Posted by: Peter L. Griffiths | May 24, 2011 11:09 AM
Kepler's area law for time taken can be initially expressed as the area of an isosceles right angle triangle
t=rXrX(1/2). If the 1/2 becomes a power we have t=rXr^(1/2) which is Kepler's distance law v=r/t=1/r^(1/2) which applies throughout the whole universe.
Posted by: Peter L.Griffiths | February 8, 2012 11:50 AM