keplers second law

[abvegto]

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...

[jimmyhelu]

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.