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.

(1) \(\mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{r}}{\mathrm{GM}^{2}}\)

(2) \(\mathrm{T}^{2}=4 \pi^{2} \mathrm{r}^{3}\)

(3) \(\mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{r}^{3}}{\mathrm{GM}}\)

(4) \( \mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{r}^{2}}{\mathrm{GM}} \)

Answer 1

Correct option is :  (3)  \(\mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{r}^{3}}{\mathrm{GM}}\)   

According to principle of homogeneity dimension of LHS should be equal to dimensions of RHS.

\(\mathrm{T}^{2}=\frac{4 \pi^{2} \mathrm{r}^{3}}{\mathrm{GM}}\)

\(\left[\mathrm{T}^{2}\right]=\frac{\left[\mathrm{L}^{3}\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right][\mathrm{M}]}\)

(Dimension of G is\( \left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\))

\(\left[\mathrm{T}^{2}\right]=\frac{\left[\mathrm{L}^{3}\right]}{\left[\mathrm{L}^{3} \mathrm{~T}^{-2}\right]}=\left[\mathrm{T}^{2}\right]\)