Question

A planet moves around the sun in nearly circular orbit. Its period revolution 'T' depends upon:(i)radius 'r' of orbit (ii)mass 'M' of the sun and (iii)the gravitational constant G. Show dimensionally that T^2 is directly proportional to r^3

Answer 1

Answer:

The square of time period is directly proportional to the cube of the radius of the orbit.

Explanation:

A planet moves around the sun in nearly circular orbit. Its period revolution 'T' depends upon radius of orbit, mass of the sun and the gravitational constant.

Let us consider

$T\propto r^a m^b G^c$

$T=kr^a m^b G^c$...(A)

Where, T = time period

r = radius of orbit

m =mass of the sun

G = gravitational constant

Using dimension formula of all terms

$[T]=[L]^{a}[M]^{b}[M^{-1}L^3T^{-2}]^{c}$

$[T]=[L^{a+3c}M^{b-c}T^{-2c}]$

On comparing

b-c=0...(I)

a+3c=0...(II)

-2c=1...(III)

$c =-\dfrac{1}{2}$

Put the value of c in equation (I)

$b+\dfrac{1}{2}=0$

$b =- \dfrac{1}{2}$

Now, put the value of c in equation (II)

$a+3\times\dfrac{-1}{2}=0$

$a=\dfrac{3}{2}$

Now, put the value of a,b and c in equation A

$T=k r^{\frac{3}{2}} m^{\frac{-1}{2}} G^{\frac{-1}{2}}$

$T^{2}=k r^{3} m^{-1}G^{-1}$

$T^{2}\propto r^{3}$

Hence, The square of time period is directly proportional to the cube of the radius of the orbit.

Answer 2

Answer:

hope it will help u thank u