Question

1. Show that the basis of dimensional analysis that the following relations are correct.
a). v2-u2 = 2aS, Where (u) is the initial velocity, (v) is final velocity, (a) is acceleration of the body and (S) is the distance moved.
b). p = 3g/4rG, where (p) is the density of earth, (G) is the gravitational constant, (r) is the radius of the earth and (g) is acceleration due to gravity.

Answer 1

Answer on Question #64436, Physics / Mechanics | Relativity

1. Show that the basis of dimensional analysis that the following relations are correct.

a). v2-u2 = 2aS, Where (u) is the initial velocity, (v) is final velocity, (a) is acceleration of the body and (S) is the distance moved.

b). $\rho = 3g/4rG$, where (p) is the density of earth, (G) is the gravitational constant, (r) is the radius of the earth and (g) is acceleration due to gravity.

Solution:

a) $v^2 - u^2 = 2aS$

dim $v = L \times T^{-1}$ (1)

Of (1) $\Rightarrow$ dim $v^2 = L^2 \times T^{-2}$ (2)

dim $u = L \times T^{-1}$ (3)

Of (3) $\Rightarrow$ dim $u^2 = L^2 \times T^{-2}$ (4)

Of (2) and (4) $\Rightarrow$ dim $v^2 - u^2 = L^2 \times T^{-2}$ (5)

dim $a = L \times T^{-2}$ (6)

dim $S = L$ (7)

Of (6) and (7) $\Rightarrow$ dim $2aS = L \times T^{-2} \times L = L^2 \times T^{-2}$ (8)

Of (5) and (8) $\Rightarrow$ dim $v^2 - u^2 = \dim 2aS$

b) $\rho = 3g/4rG$

dim $\rho = M \times L^{-3}$ (1)

dim $g = L \times T^{-2}$ (2)

dim $r = L$ (3)

dim $G = M \times L \times T^{-2} \times L^2 \times M^{-2} = L^3 \times T^{-2} \times M^{-1}$ (4)

Of (2), (3), (4) $\Rightarrow$ dim $3g/4rG = L \times T^{-2} / L \times L^3 \times T^{-2} \times M^{-1} = M \times L^{-3}$ (5)

Of (1) and (5) $\Rightarrow$ dim $\rho = \dim 3g/4rG$

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