AMU 2026 — science PYQ

1. Identify the Governing Principle:

   According to Kepler's Third Law of Planetary Motion (The Law of Periods), the square of the orbital period ($T$) of a planet is directly proportional to the cube of the semi-major axis (or radius $R$ for a circular orbit) of its orbit:

   $$T^2\propto R^3$$

2. Set up the Ratio Equation:

   Let $T_1$ and $R_1$ be the period and orbital radius of the first planet, and $T_2$ and $R_2$ be those of the second planet.

   From Kepler's third law, we can write:

   $${\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{R_2}{R_1}\right)}^3$$

3. Substitute the Given Values:

   From the question text in image_e5a824.png, we have:

   • $R_1=R$

   • $T_1=T$

   • $R_2=4R$

   Plugging these values into our ratio relation:

   $${\left(\frac{T_2}{T}\right)}^2={\left(\frac{4R}{R}\right)}^3$$

   $${\left(\frac{T_2}{T}\right)}^2=(4{)}^3$$

   $${\left(\frac{T_2}{T}\right)}^2=64$$

4. Solve for the Ratio:

   Taking the square root on both sides to find the ratio of the periods ($\frac{T_2}{T_1}$):

   $$\frac{T_2}{T}=\sqrt{64}$$

   $$\frac{T_2}{T}=8$$

Conclusion:

The ratio of the orbital period of the second planet to the first planet is 8, corresponding to option (c).