1. **Problem:** Use dimensional analysis to find the exponents $a$, $b$, and $c$ in the formula $$t = K G^a M^b R^c,$$ where $t$ is time, $G$ is the gravitational constant, $M$ is mass, and $R$ is radius. 2. **Step 1: Write down the dimensions of each quantity.** - Time, $t$: $[T]$ - Gravitational constant, $G$: $[M^{-1}L^3T^{-2}]$ - Mass, $M$: $[M]$ - Radius, $R$: $[L]$ 3. **Step 2: Write the dimensional equation.** $$[T] = [M^{-1}L^3T^{-2}]^a [M]^b [L]^c = [M^{-a + b} L^{3a + c} T^{-2a}]$$ 4. **Step 3: Equate the powers of each fundamental dimension on both sides.** - For mass $M$: $0 = -a + b \implies b = a$ - For length $L$: $0 = 3a + c \implies c = -3a$ - For time $T$: $1 = -2a \implies a = -\frac{1}{2}$ 5. **Step 4: Substitute $a$ back to find $b$ and $c$.** - $b = a = -\frac{1}{2}$ - $c = -3a = -3 \times -\frac{1}{2} = \frac{3}{2}$ 6. **Final answer:** $$a = -\frac{1}{2}, \quad b = -\frac{1}{2}, \quad c = \frac{3}{2}$$ This means the time period formula is: $$t = K G^{-\frac{1}{2}} M^{-\frac{1}{2}} R^{\frac{3}{2}}$$ This matches the known formula for orbital period where $t \propto \frac{R^{3/2}}{\sqrt{GM}}$.