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Friday, 19 September 2014

Kepler's third law, and how orbiting bodies are similar to freely falling bodies



One of the most interesting things about modern physics is that, sometimes very simple and well-known premises lead us to mind-blowing conclusions. These are usually called Veridical paradoxes, and are among the most important tools we have to make sense of the seemingly limitless world all around us. 

And one such conclusion is the case of freely falling bodies. As we all know, Kepler's third law of planetary motion roughly applies to any system of two bodies (for simplification), where one body revolves around the other. As such, the simply by magnificent relationship: 

t^2 / s^3 = constant 

holds for the system, where: 

t -> time period of revolution of the body 

s -> semi-major axis of the orbit, aka the average distance between the two bodies (roughly). 

Formulation of Kepler's third law for orbitting bodies 

This is basic physics, but still let's brush it up a bit. So, let the mass of the sun be M, and that of the earth be m, the semi-major axis of earth's orbit be R, and average speed of the earth's revolution be v. 

Then, from basic Newtonian mechanics: 

G * M * m / R^2 = m * v^2 / R

orv^2 = G * M / R, (i)

where G is the universal constant of gravitation. The reason behind this statement is basic physics, the earth more or less stays at the same distance from the sun, and naturally we can approximately say that the centrifugal force of the system balances the attraction between the sun and the earth. 

Now, the orbit of the earth is approximately circular (elliptical, more precisely, but for our purpose it's enough). So, the time period of the earth's revolution will be given as: 

T = distance to be covered / earth's speed of revolution 
    = 2 *pi * R / v

orv = 2 * pi * R / T

orv^2 = 4 * pi^2 * R^2 / T^2 (ii

Then, from (i) = (ii), 

G * M / R = 4 * pi^2 * R^2 / T^2

orT^2 = 4 * pi^2 * R^3 / ( G * M )

orT^2 / R^3 = 4 * pi^2 / GM = k

So, T^2 is proportional to R^3, and this verifies Kepler's third law for the earth-sun system. 

The earth as a freely falling body 

Now, a simple conclusion we get from the above equation is that, the mass of the earth doesn't have any effect on the balance between the forces, or the velocity or time period of earth's revolution, to be exact. 

Now, let us assume that the earth is a freely falling body, that falls towards the sun from the distance R. 

Then, first of all the force of gravity on the earth would be: 

F = G * M * m / R^2

and, from Newton's second law of motion, approximate F = m * a, 

So, m * a = G * M * m / R^2

ora = G * M / R^2

So the earth accelerates towards the sun with this value. Now, from the second equation of motion,

S = u * t + ( a * t^2 )/2

For our scenario, u = 0 as the earth is considered to have 0 initial velocity. And S = R, since the total distance the earth would cover (approximately) before falling into the sun would be roughly the same as the radius of the earth's orbit. In addition, the time period of the free fall is taken as T. 

So, R = ( a * T^2 ) / 2
           = G * M * T^2 / 2 * R^2

orT^2 / R^3 = 2/(GM) = constant

So, in this case as well, a relationship qualitative to Kepler's third law holds. 

IMPORTANT: Here, we must understand that the similarity between the two cases is qualitative, and not quantitative. This is because of the following two reasons: 

1. R is NOT equal to S = 2 * pi * R, i.e the distance to be covered by a freely falling body is less than the distance to be covered by an orbitting body. 

2. The value of the two constants is dissimilar. 

Satellites as freely falling bodies 

Since Kepler's third law applies to satellites orbitting the earth, and similar systems, we can say that satellites are thus similar to freely falling bodies. 

However, it is important to understand why the satellites don't fall directly on earth. Since the earth is moving itself, the direction of the acceleration due to gravity on satellites changes. All components balanced out, this usually gives the satellites a motion in the direction perpendicular to the direction of the acceleration. However, thanks to the constant force of gravity pulling the satellites downwards, this translates into the roughly circular orbit of the satellites. 

References 

http://www.phy6.org/stargaze/Sgravity.htm 

http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/orbits.html 

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