Lepton

...Consider a planet which moves from A to B in a small time dT...under polar co- ordinate system let the radius vector be r vector from the origin o.

 

that is

OA = Ob = rvector

angle AOX = y

angle BOX = y+dy

therefore change in angle is dy

 

now let the planet move from A to B in small time dT....let it cover a distance dSvector

 

...here dA ( the area covered by the planet in time dT) = 1/2 rvector x dSvector

 

dividing by dT

.... dA/dT = 1/2 rvector x dS/dT

..... dA/dT =1/2 rvector x vvector

 

multiplying and dividing by 'm'.. the mass of the planet

.... dA/dT = 1/2m (r x mv)

.... dA/dT = 1/2m (r x p)............. 1

differentiating 1 again with T we get

..... d²A/dT² = 1/2m [ (dr/dt x p) x (r x dp/dt)]

......d²A/dT^{2 =} 1/2 m {[ v x (mv)] x [(r x f)]}

....................= 1/2 m (r x f).........2

but since the gravitational force is an example for a central force therefore force acts along the direction of r...that is sin theta = 0 therefore r x f = 0

 

therefore 2 becomes

....d²A/dT² = 0 => dA /dT = CONSTANT hence the second law is proved

 

MY QUESTION IS....IN MY EXAM I WROTE THIS PROOF AND I GOT 0 MARKS!..I CANT UNDERSTAND WHERE DID I GO WRONG... OF COURSE THE BOOKISH PROOF IS DEFFERENT....CAN U GUYS PLEASE PIN POINT MY MISTAKE?... and sorry for my english...

There is more gravitational potential energy in bodies that are separated. You know this. That is why you need to work to climb a hill, as you separate yourself from Earth.

 

 

Since there is a law of conservation of energy, any change in the GPE must also amount to an opposite change in the kinetic energy. No other energy forms are involved in planetary motion.