Kepler's Laws Of Planetary Motion

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A newly discovered asteroid orbits the Sun at a mean distance four times that of the Earth from the Sun. How long does this asteroid take to complete one revolution around the Sun?

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NCERT Class 11, Chapter 8 (Gravitation) states Kepler's third law: the square of a planet's orbital period is proportional to the cube of the semi-major axis of its orbit, written $T^2 \propto r^3$. This empirical law, later explained by Newton's gravitation, lets us compare orbits using only distance ratios. Taking Earth as the reference with $T = 1$ year and $r = 1$ astronomical unit, the relation becomes $\left(\dfrac{T_a}{T_E}\right)^2 = \left(\dfrac{r_a}{r_E}\right)^3$. Substituting $\dfrac{r_a}{r_E} = 4$ gives $T_a^2 = 4^3 = 64$, so $T_a = \sqrt{64} = 8$ years. The option $4$ years assumes the period scales linearly with distance, ignoring the law altogether. The option $16$ years uses $4^2$, confusing the cube and the square. The option $64$ years forgets to take the square root of the period-squared value. As a plausibility check, a more distant orbit must have a longer period, and Kepler's law correctly predicts that the asteroid takes several times longer than Earth to circle the Sun, consistent with the eight-year result.

Kepler's Laws Of Planetary Motion

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