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A planet moves around the sun in nearly circular orbit. It's the period of revolution ‘ T ’ upon i) radius ‘ r ’ of the orbit ii) Mass ‘ M ’ of the sun and iii) gravitational constant ‘ G ’. Show dimensionally that T2r3 .

Answer
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Hint: Law of Gravitation – Every object in the universe attract other object along the line of center for two objects that is proportional to the product of their masses and inversely proportional to the square of the separation between the two objects.

Complete step by step answer:
According to the law of Gravitation,
 F=GMmr2 where F is the gravitational force, M is the mass of the sun, m is the mass of the planet, r is the distance between the sun and the planet.
It is stated that the planet moves round the sun in a nearly circular orbit.
Let ,
 T=KrxMyGz …(i) where K is a dimensionless constant. The dimensions of various quantities are-
 [T]=T , [r]=L , [M]=M
 [G]=Fr2Mm =MLT2L2MM
 =M1L3T2
By substituting these dimensionally in equation (i), we get
 M0L0T1=MyzLx+3zT2z
Equating the dimensions,
 yz=0 , x+3z=0 , 2z=1
On solving,
 x=32 , y=12 , z=12
 T=Kr32M12G12
Squaring both sides,
 T2=K2R3MG
 T2r3 .

Note:
Planets revolve in elliptical orbits around the sun. It is closest to the sun at perihelion and farthest from the sun at aphelion.
 Applying Kepler’s Third Law – We assume that the planets revolve in circular orbits around the sun.
 FC=FG where FC is the centripetal force and FG is the gravitational force. The centripetal force arises due to the gravitational force.
 mv2r=GMmr2
 v=GMr where v is the velocity with which the planet moves in an orbit around the sun.
 T=2πrv where T is the time period.
 T=2πrGMr
 T=2πr32G12M12
 Tr32
Squaring both sides,
 T2r3
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