Question

Applying the principle of homogeneity of dimensions, determine which one is correct. Where \( T \) is the time period, \( G \) is the gravitational constant, \( M \) is the mass, and \( r \) is the radius of the orbit.

• A. \( T^2 = \frac{4\pi^2 r}{GM^2} \)
• B. \( T^2 = 4\pi^2 r^3 \)
• C. \( T^2 = \frac{4\pi^2 r^3}{GM} \)
• D. \( T^2 = \frac{4\pi^2 r^2}{GM} \)

Answer 1

The Correct Option is C

According to the principle of homogeneity of dimensions, the dimensions on the left-hand side (LHS) must match those on the right-hand side (RHS).
1. Check Dimensions of Each Term in Option (3):
Consider:
\[ T^2 = \frac{4\pi^2 r^3}{GM}. \] - The dimensions of \( T^2 \) are \([T^2]\).
- The dimensions of \( G \) (gravitational constant) are \([M^{-1}L^3T^{-2}]\).
- The dimensions of \( M \) are \([M]\).
- The dimensions of \( r \) (radius) are \([L]\).
2. Dimensional Analysis:
Substitute the dimensions into RHS:
\[ \left[\frac{L^3}{M \times M^{-1}L^3T^{-2}}\right] = [T^2]. \] Since both sides have the dimension of \([T^2]\), option (3) is dimensionally correct.
Answer: \( \frac{4\pi^2 r^3}{GM} \)

Answer 2

Step 1: Understand the problem
We are asked to determine which equation is dimensionally correct using the principle of homogeneity of dimensions. The given physical quantities are:
\(T\) → time period (dimension of time),
\(G\) → universal gravitational constant,
\(M\) → mass of the central body,
\(r\) → radius of the orbit.

We need an equation that is dimensionally consistent (both sides having the same dimensions).

Step 2: Recall dimensions of the quantities involved
1) \([T] = [T]\)
2) \([G] = [M^{-1} L^3 T^{-2}]\)
3) \([M] = [M]\)
4) \([r] = [L]\)

Step 3: Write the proposed formula
The given possible relation (Kepler’s third law) is:
\[ T^2 = \frac{4\pi^2 r^3}{GM} \] We will test its dimensional correctness.

Step 4: Perform dimensional analysis
On the right-hand side:
\[ \frac{r^3}{GM} \Rightarrow \frac{[L^3]}{[M^{-1}L^3T^{-2}] \cdot [M]} = [L^3] \times [M][M^{1}L^{-3}T^2] = [T^2] \] Thus, the right-hand side has the dimension of time squared, i.e. \([T^2]\).

On the left-hand side, we have \(T^2\), which also has dimension \([T^2]\).

Since both sides have the same dimension, the equation is dimensionally consistent.

Step 5: Physical interpretation
The equation \(T^2 = \frac{4\pi^2 r^3}{GM}\) represents the correct form of Kepler’s third law for planetary motion, showing that the square of the orbital period is proportional to the cube of the orbital radius. The constant of proportionality depends on the gravitational constant \(G\) and the mass \(M\) of the central object.

Final Answer:

\( T^2 = \frac{4\pi^2 r^3}{GM} \)