JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Gravitation

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Applying the principle of homogeneity of dimensions, determine which one is correct. where T is time period, G is gravitational constant,

M

is mass,

r

is radius of orbit.

(A) $T^{2} = \frac{4 π ^{2} r}{G M ^{2}}$
(B) $T^{2} = 4 π^{2} r^{3}$
(C) $T^{2} = \frac{4 π ^{2} r ^{3}}{GM}$ ✓ Correct
(D) $T^{2} = \frac{4 π ^{2} r ^{2}}{GM}$

Answer: (C)

Solution

Check dimensional homogeneity of

T^{2} = \frac{4 π ^{2} r ^{3}}{GM}

[G] = M^{- 1} L^{3} T^{- 2}

, so

[\frac{r ^{3}}{GM}] = \frac{L ^{3}}{M ^{- 1} L ^{3} T ^{- 2} \cdot M} = \frac{L ^{3}}{L ^{3} T ^{- 2}} = T^{2}

This matches

[T^{2}]

on the left side, so this is the dimensionally consistent (and physically correct, Kepler's third law) relation; the other options fail this check.

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option

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