Question

For elliptical orbits, the ratio $\frac{T^2}{a^3}$ is same for every planetary orbit around a massive body, and its value is : [ where $a$ is semi-major axis of the ellipse, $T$ is time period of motion and $y_r$ is year]

• A. \(1.99 \times 10^{-34} \text{ } y_r^2 \text{ m}^{-3} \)
• B. \(2.99 \times 10^{-34} \text{ } y_r^2 \text{ m}^{-3} \)
• C. \(1.99 \times 10^{-28} \text{ } y_r^2 \text{ m}^{-3} \)
• D. \(2.99 \times 10^{-28} \text{ } y_r^2 \text{ m}^{-3} \)

Hint:

The constant $\frac{T^2}{a^3}$ only depends on the mass of the central body. For all planets orbiting the same Sun, this ratio remains identical.

Answer 1

The Correct Option is B

Concept: Kepler's Third Law states that the square of the period of revolution ($T$) of a planet is proportional to the cube of the semi-major axis ($a$) of its orbit: \[ T^2 = \left( \frac{4\pi^2}{GM} \right) a^3 \implies \frac{T^2}{a^3} = \frac{4\pi^2}{GM} \] For the Solar System, $M$ is the mass of the Sun ($M_s \approx 1.99 \times 10^{30} \text{ kg}$).

Step 1: Calculate the value in SI units ($s^2/m^3$).
Using $G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$ and $M_s = 1.99 \times 10^{30} \text{ kg}$: \[ \frac{T^2}{a^3} = \frac{4 \cdot (3.14)^2}{(6.67 \times 10^{-11}) \cdot (1.99 \times 10^{30})} \approx 2.97 \times 10^{-19} \text{ s}^2/\text{m}^3 \]

Step 2: Convert $s^2$ to $y_r^2$.
$1 \text{ year} \approx 3.15 \times 10^7 \text{ s}$, so $1 \text{ s} \approx \frac{1}{3.15 \times 10^7} \text{ year}$. \[ \frac{T^2}{a^3} \approx 2.97 \times 10^{-19} \cdot \left( \frac{1}{3.15 \times 10^7} \right)^2 \approx \frac{2.97 \times 10^{-19}}{9.92 \times 10^{14}} \] \[ \frac{T^2}{a^3} \approx 2.99 \times 10^{-34} \text{ } y_r^2 \text{ m}^{-3} \]