Q.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.

Concept: Dimensional Analysis — used here to verify the form of a physical relation.

We are given that $T^2 \propto r^3$, so $T \propto r^{3/2}$. The proposed relation is:

Step 1: Write dimensions of each quantity

• $T$: $[T]$
• $R$: $[L]$
• $r$: $[L]$
• $g$: $[LT^{-2}]$
• $k$: dimensionless

Step 2: Find dimensions of the RHS …

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.

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:

where $a$, $b$, $c$ are exponents to be found.

2. Write the dimensional equation

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 $b$ uniquely. We need an extra condition.

4. Use Kepler’s Third Law

   Kepler’s Third Law states that for a satellite orbiting a central body, $T^2 \propto r^3$. That means $T \propto r^{3/2}$. In our expression $T \propto r^a$, so we must have:

This is the physical input that resolves the ambiguity.

5. Find $b$

   From $a + b = \frac{1}{2}$ and $a = \frac{3}{2}$, we get:

   $\frac{3}{2} + b = \frac{1}{2} \quad \Rightarrow \quad b = -1$ …

Method: Dimensional Analysis (Rayleigh's Method)

Concept: Dimensional homogeneity — both sides of a valid physical equation must have the same dimensions.

Step 1: Write the given proportionality

From Kepler's Third Law:

So we can write:

But $T$ also depends on $M$ (planet mass) and $R$ (planet radius), and we need to introduce $g$.

Step 2: Express $g$ in terms of $M$ and $R$

Acceleration due to gravity at the planet's surface:

Step 3: Assume a general product form

Let:

where $k$ is dimensionless.

Step 4: Write dimensional equations

Dimensions:

• $[T] = T$
• $[r] = L$
• $[M] = M$
• $[R] = L$
• $[g] = LT^{-2}$

So:

Step 5: Equate powers of $M$, $L$, $T$

• For M: $b = 0$
• For T: $1 = -2d \quad \Rightarrow \quad d = -\frac{1}{2}$
• For L: $0 = a + c + d \quad \Rightarrow \quad a + c - \frac{1}{2} = 0$ …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting that $g$ depends on $M$ and $R$

The error: Students treat $g$ as an independent fundamental quantity, like mass or length. They write $T \propto r^{3/2} g^{-1/2} R^{-1}$ and stop — but this is not dimensional analysis; it's just pattern-matching the given formula.

Why it's wrong:

$g$ is not a base dimension. Its dimensional formula is $[g] = [LT^{-2}]$, which already contains $M$ and $R$ through the relation $g = \frac{GM}{R^2}$. If you don't account for this, you miss the physics.

How to avoid:

Always ask: "Is this quantity truly independent, or does it come from other variables?"

In orbital problems, $g$ is derived from $M$ and $R$. So when the problem gives $g$, treat it as a shortcut — but in dimensional analysis, you must verify consistency by substituting $[g] = [LT^{-2}]$.

Mistake 2: Assuming $k$ can absorb any mismatch in dimensions

The error: Students get a wrong combination (e.g., $T \propto \sqrt{r^3/g}$ without the $R$ factor) and say "the $R$ can be absorbed into $k$".

Why it's wrong:

$k$ is dimensionless. If your expression has leftover dimensions of length, $k$ cannot fix it. The $R$ in the denominator is essential for dimensional correctness.

How to avoid:

Check dimensions before declaring success. For the given formula:

• $[T] = T$
• $[r^{3/2}] = L^{3/2}$
• $[g^{-1/2}] = (LT^{-2})^{-1/2} = L^{-1/2} T$
• $[R^{-1}] = L^{-1}$

Multiply: $L^{3/2} \times L^{-1/2} \times L^{-1} \times T = L^0 T = T$ ✓

If you drop $R$, you get $L^{3/2} \times L^{-1/2} T = L^1 T$ — wrong dimensions. So $R$ is non-negotiable.

Mistake 3: Confusing $r$ (orbit radius) with $R$ (planet radius)

The error: Students treat $r$ and $R$ as the same, or swap them in the final expression.

Why it's wrong:

$r$ is the distance from the planet's center to the satellite. $R$ is the planet's surface radius. They are physically different — and dimensionally, both are $L$, so dimensional analysis alone cannot distinguish them. The problem tells you which is which.

How to avoid:

Read the problem statement carefully. Here:

• $r$ = orbit radius (variable)
• $R$ = planet radius (constant)

The formula $T = \frac{k}{R}\sqrt{\frac{r^3}{g}}$ uses both — $R$ in the denominator and $r$ inside the square root. Never interchange them.

Mistake 4: Not using Kepler's Third Law as a starting point

The error: Students jump straight to dimensional analysis without first writing $T^2 \propto r^3$ (Kepler's law).

Why it's wrong:

The problem explicitly says "From Kepler's Third law... show using dimensional analysis". Kepler's law gives the functional form: $T \propto r^{3/2}$. Dimensional analysis then determines the constant of proportionality in terms of $g$ and $R$.

How to avoid:

Always start with the given physical law: …