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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.
Options:
where $T$ is time period, $G$ is gravitational constant, $M$ is mass, $r$ is radius of orbit.
Solution:
2654 Upvotes
Verified Answer
The correct answer is:
$T^2=\frac{4 \pi^2 r^3}{G M}$
According to principle of homogeneity dimension of LHS should be equal to dimensions of RHS so option (3) is correct.
$\begin{aligned}
& \mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}} \\
& {\left[\mathrm{T}^2\right]=\frac{\left[\mathrm{L}^3\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right][\mathrm{M}]}}
\end{aligned}$
(Dimension of $\mathrm{G}$ is $\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$ )
$\left[\mathrm{T}^2\right]=\frac{\left[\mathrm{L}^3\right]}{\left[\mathrm{L}^3 \mathrm{~T}^{-2}\right]}=\left[\mathrm{T}^2\right]$
$\begin{aligned}
& \mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}} \\
& {\left[\mathrm{T}^2\right]=\frac{\left[\mathrm{L}^3\right]}{\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right][\mathrm{M}]}}
\end{aligned}$
(Dimension of $\mathrm{G}$ is $\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$ )
$\left[\mathrm{T}^2\right]=\frac{\left[\mathrm{L}^3\right]}{\left[\mathrm{L}^3 \mathrm{~T}^{-2}\right]}=\left[\mathrm{T}^2\right]$
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