The time period of revolution of a satellite | vedclass

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(C) Let the time period $T$ be proportional to $R^a M^b G^c$,so $T = K R^a M^b G^c$.Writing the dimensional formulas for each quantity:$[T] = [T]^1$$[

Vedclass · Accepted Answer

$K \sqrt{\frac{R^3}{GM}}$

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(C) Let the time period $T$ be proportional to $R^a M^b G^c$,so $T = K R^a M^b G^c$.Writing the dimensional formulas for each quantity:$[T] = [T]^1$$[R] = [L]^1$$[M] = [M]^1$$[G] = [M^{-1} L^3 T^{-2}]$Substituting these into the equation: $[T]^1 = [L]^a [M]^b [M^{-1} L^3 T^{-2}]^c = [M]^{b-c} [L]^{a+3c} [T]^{-2c}$.Comparing the powers of $M$,$L$,and $T$ on both sides:For $T$: $-2c = 1 \Rightarrow c = -1/2$.For $M$: $b - c = 0 \Rightarrow b = c = -1/2$.For $L$: $a + 3c = 0 \Rightarrow a = -3c = -3(-1/2) = 3/2$.Substituting these values back into the expression: $T = K R^{3/2} M^{-1/2} G^{-1/2} = K \sqrt{\frac{R^3}{GM}}$.

The time period of revolution of a satellite $(T)$ around the earth depends on the radius of the circular orbit $(R)$,mass of the earth $(M)$,and universal gravitational constant $(G)$. The expression for $T$,using dimensional analysis is ($K$ is a constant of proportionality):

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