- #1

mich

**[SOLVED] Keppler's 2nd & 3rd**

While I have the correct topic, I'm not sure if this belongs toOriginally posted by ObsessiveMathsFreak

Ok the above statement is not STRICTLY true. It should read...

the crackpot theory or not. I appologize if I haven't placed my post in the correct category.

I'm having a bit of a problem concerning Keppler's 2nd.law.

" ...moving planets sweeps out equal areas at equal time."

This clearly states that a planet which approaches or get's farther

from the sun will experience an increase or a decrease of velocity

by a specific amount.

Let a planet revolve around the sun at R=1, if the orbit is circular, then the distance traveled would be 2*pi*R= 2*pi. Let us claim the period to be 1 unit of time.It's velocity would then be 2*pi/1.

According to the 2nd law, it would take a radius of sqrt2 in order

for the period to be equal to 2. the velocity would then be

2*sqrt2*pi/2; or simply sqrt2*pi.

If we continue to increase the radius, we get:

for R=sqrt3 T=3; v=2*sqrt3*pi/3

for R=sqrt4 T=4; v=2*sqrt4*pi/4 = 4*pi/4 = pi (notice that when the radius is doubled, the velocity, being directly affected by the gravitational force,is halved. This "could" seem to claim as you did,that gravitational force is inversally proportional to R...not R^2.

Nevertheless,My question would be: could Newton's 3rd law be centered more on the kenetic energy of a planet relative to it's distance from the gravitational force? Force=Kenetic Energy/R?

F=mv^2/2R

R=1 T=1 V=2 F= m*4/2*1 = 2 (let m=1)

R=2 T=4 v=1 F= m*1/2*2 = 1/4

R=3 T=9 v=.66 F= m*.4356/2*3 = .0726

R=4 T=16 v=.5 F= m*.25/2*4 = .031

R=8 T=64 v=.25 F= m* .0625/2*8 = .0039

This would show that when the radial distance is doubled, the kenetic energy is halved, and the Force is reduced by a factor of 8.