let us period of revolution of a planet at a distance R from a star be T. prove that if it was at a distance of 2R from the star , it's period of revolution will be √8 T.
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1. consider a stone thrown vertically upwards with initial velocity 'u'. It reaches a height 'h' before coming down.
2. The kinematical equation of motion are given as,
v=u+at......(1)
s =ut+1/2at^2.....(2)
v^2= u^2+ 2as........(3)
3 . For upward motion a= -g, v=0
according to equation (1) , time t1
0= u-gt1
t1= u/g .........(4)
according to equation (3) s=h
0^2 -u^2= -2gh
h=u^2/ 2g........(5)
3 For downward motion of the stone
a=g
u=0
according to equation (2) time=t2
h= 0+1/2gt2^2
t2^2= 2h/g
t2√2h/g.......(6)
4. substituting equation (5)in(6)
t2= √2/g×u^2/2g
t2 = u/g.......(7)
Thus, from equation 4 and 7 ,we can calculate that the time taken by the stone to go up is same as the time taken to come down..
2. The kinematical equation of motion are given as,
v=u+at......(1)
s =ut+1/2at^2.....(2)
v^2= u^2+ 2as........(3)
3 . For upward motion a= -g, v=0
according to equation (1) , time t1
0= u-gt1
t1= u/g .........(4)
according to equation (3) s=h
0^2 -u^2= -2gh
h=u^2/ 2g........(5)
3 For downward motion of the stone
a=g
u=0
according to equation (2) time=t2
h= 0+1/2gt2^2
t2^2= 2h/g
t2√2h/g.......(6)
4. substituting equation (5)in(6)
t2= √2/g×u^2/2g
t2 = u/g.......(7)
Thus, from equation 4 and 7 ,we can calculate that the time taken by the stone to go up is same as the time taken to come down..
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