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An artificial satellite is revolving around a planet of mass $M$ and radius $R$, in a circular orbit of radius $r$. From Kepler's third Law about the period of a satellite around a common central body, square of the period of revolution $T$ is proportional to the cube of the radius of the orbit $r$. Show using dimensional analysis, that $T=\frac{k}{R} \sqrt{\frac{r^3}{g}}$, where $k$ is dimensionless constant and $g$ is acceleration due to gravity.
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Verified Answer
By Kepler's third law of Planetary motion,
$$
T^2 \propto r^3 \Rightarrow T \propto r^{3 / 2}
$$
As we know that $T$ is a function of $R$ and $g$ or $T$ depends on radius $R$ and $g$.
Let, $T \propto r^{3 / 2} R^a g^b$
$$
\Rightarrow T=k r^{3 / 2} R^a g^b
$$
where, $k$ is a dimensionless constant of proportionality.
Substituting the dimensions of each term in Eq. (i), we get $\begin{aligned} {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}\right] } &=k[\mathrm{~L}]^{3 / 2}[\mathrm{~L}]^{\mathrm{a}}\left[\mathrm{LT}^{-2}\right]^{\mathrm{b}} \\ &=k\left[\mathrm{~L}^{a+b}+3 / 2 \mathrm{~T}^{-2 b}\right] \end{aligned}$
By principle of homoegeneity comparing the powers of same terms, as $\mathrm{M}, \mathrm{L}$ and $\mathrm{T}$, so we get
$$
\begin{aligned}
&a+b+3 / 2=0 \\
&-2 b=1 \Rightarrow b=-1 / 2
\end{aligned}
$$
From Eq. (ii), we get
$$
a-1 / 2+3 / 2=0 \Rightarrow a=-1
$$
Substituting the value of $a$ and $b$ in Eq. (i), we get
$$
\begin{aligned}
&T=k r^{3 / 2} R^{-1} g^{-1 / 2} \\
&T=\frac{k}{R} \sqrt{\frac{r^3}{g}}
\end{aligned}
$$
$$
T^2 \propto r^3 \Rightarrow T \propto r^{3 / 2}
$$
As we know that $T$ is a function of $R$ and $g$ or $T$ depends on radius $R$ and $g$.
Let, $T \propto r^{3 / 2} R^a g^b$
$$
\Rightarrow T=k r^{3 / 2} R^a g^b
$$
where, $k$ is a dimensionless constant of proportionality.
Substituting the dimensions of each term in Eq. (i), we get $\begin{aligned} {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}\right] } &=k[\mathrm{~L}]^{3 / 2}[\mathrm{~L}]^{\mathrm{a}}\left[\mathrm{LT}^{-2}\right]^{\mathrm{b}} \\ &=k\left[\mathrm{~L}^{a+b}+3 / 2 \mathrm{~T}^{-2 b}\right] \end{aligned}$
By principle of homoegeneity comparing the powers of same terms, as $\mathrm{M}, \mathrm{L}$ and $\mathrm{T}$, so we get
$$
\begin{aligned}
&a+b+3 / 2=0 \\
&-2 b=1 \Rightarrow b=-1 / 2
\end{aligned}
$$
From Eq. (ii), we get
$$
a-1 / 2+3 / 2=0 \Rightarrow a=-1
$$
Substituting the value of $a$ and $b$ in Eq. (i), we get
$$
\begin{aligned}
&T=k r^{3 / 2} R^{-1} g^{-1 / 2} \\
&T=\frac{k}{R} \sqrt{\frac{r^3}{g}}
\end{aligned}
$$
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