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 radi

Question

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 $T = \dfrac{k}{R}\sqrt{\dfrac{r^3}{g}}$, where $k$ is a dimensionless constant and $g$ is acceleration due to gravity.

Accepted Answer

> [!TLDR]
> Using dimensional analysis, we relate the period $T$ to the relevant physical quantities $r$, $R$, $g$, and a dimensionless constant $k$, and derive $T = \frac{k}{R}\sqrt{\frac{r^3}{g}}$.

The problem asks us to show that the period $T$ of a satellite in a circular orbit can be expressed in the given form using dimensional analysis. This is a classic exercise in checking how physical quantities combine to give a correct relationship — without solving any differential equations.

The key idea is that the period $T$ must depend on the orbit radius $r$, the planet’s radius $R$, and the acceleration due to gravity $g$ at the planet’s surface. Why $g$? Because gravity is the force keeping the satellite in orbit, and $g$ is a measure of the planet’s gravitational pull at its surface. The planet’s mass $M$ is already hidden inside $g$ (since $g = GM/R^2$), so we don’t need $M$ separately.

We also know from Kepler’s Third Law that $T^2 \propto r^3$, so the dimensional analysis must respect that.

Let’s go step by step.

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1. **List the quantities and their dimensions**

- Period $T$: dimension $[T]$
   - Orbit radius $r$: dimension $[L]$
   - Planet radius $R$: dimension $[L]$
   - Acceleration due to gravity $g$: dimension $[L T^{-2}]$
   - Dimensionless constant $k$: no dimension

We assume a product form:
   $$ T = k \, r^a \, R^b \, g^c $$
   where $a$, $b$, $c$ are exponents to be found.

2. **Write the dimensional equation**

$$ [T] = [L]^a \, [L]^b \, ([L T^{-2}])^c = [L]^{a+b+c} \, [T]^{-2c} $$

For this to be dimensionally consistent, the exponents of $L$ and $T$ on both sides must match.

3. **Equate exponents**

- For time $T$: $1 = -2c \quad \Rightarrow \quad c = -\frac{1}{2}$
   - For length $L$: $0 = a + b + c \quad \Rightarrow \quad a + b - \frac{1}{2} = 0 \quad \Rightarrow \quad a + b = \frac{1}{2}$

We have one equation for two unknowns — this is expected because dimensional analysis alone cannot determine both $a$ and $b$

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