Question 97: Units , Dimensions and Error Analysis

Question

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.

Accepted Answer

To determine which option is correct based on the principle of homogeneity of dimensions, we need to ensure that both sides of the equation have the same dimensions. The time period ($T$) is measured in units of time ($T$), the gravitational constant ($G$) has units of $	ext{m}^3	ext{kg}^{-1}	ext{s}^{-2}$, mass ($M$) has units of mass ($M$), and the radius of orbit ($r$) has units of length ($L$).
Let's analyze each option:
Option A: $T^2=\frac{4 \pi^2 r}{G M^2}$
The dimensions of the right-hand side of the equation are $\frac{L}{\left(\frac{L^3}{MT^2}ight)M^2}=\frac{L}{L^3M^{-1}T^{-2}M^2}=\frac{L}{L^3T^{-2}}=L^{-2}T^2$, which do not match with $T^2$ (time squared). Thus, Option A is incorrect.
Option B: $T^2=4 \pi^2 r^3$
The dimensions of the right-hand side are $L^3$, which clearly do not match the dimensions $T^2$ of the squared time period. So, Option B is incorrect.
Option C: $T^2=\frac{4 \pi^2 r^3}{G M}$
The dimensions of the right-hand side of the equation are $\frac{L^3}{\left(\frac{L^3}{MT^2}ight)M}=\frac{L^3}{L^3T^{-2}}=T^2$, which match the dimensions of the squared time period $T^2$. Therefore, Option C is correct based on the principle of homogeneity of dimensions.
Option D: $T^2=\frac{4 \pi^2 r^2}{G M}$
The dimensions of the right-hand side are $\frac{L^2}{\left(\frac{L^3}{MT^2}ight)M}=\frac{L^2}{L^3M^{-1}T^{-2}M}=L^{-1}T^2$, which do not match with $T^2$ (time squared). Hence, Option D is incorrect.
Thus, based on the principle of homogeneity of dimensions, Option C is the correct one: $T^2=\frac{4 \pi^2 r^3}{G M}$. This equation also corresponds to Kepler's third law of planetary motion, which relates the orbital period of a planet to its orbital radius, considering the mass of the central body.

Answer

$T^2=\frac{4 \pi^2 r}{G M^2}$

Answer

$T^2=4 \pi^2 r^3$

Answer

$T^2=\frac{4 \pi^2 r^3}{G M}$

Answer

$T^2=\frac{4 \pi^2 r^2}{G M}$

Units , Dimensions and Error Analysis

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.

Choose the correct answer: