The planet Mars has two moons, if one of them has a period 7 hours, 30 minutes and an orbital radius of $$9.0 \times 10^3$$ km. Find the mass of Mars.
$$\left\{\text{Given } \frac{4\pi^2}{G} = 6 \times 10^{11} \text{ N}^{-1} \text{ m}^{-2} \text{ kg}^2\right\}$$

We are told that one of Mars’ moons has an orbital period of 7 hours 30 minutes and an orbital radius of $$9.0 \times 10^3$$ km. First we convert all the given data into SI units.

The period is 7 hours 30 minutes. Since 30 minutes is half an hour, this is a total of 7.5 hours. Converting hours to seconds we use the relation $$1 \text{ hour} = 3600 \text{ s}.$$ So $$T = 7.5 \times 3600 \text{ s} = 27000 \text{ s} = 2.7 \times 10^4 \text{ s}.$$

The radius is $$9.0 \times 10^3$$ km. Knowing that $$1 \text{ km} = 1000 \text{ m},$$ we get $$r = 9.0 \times 10^3 \times 10^3 \text{ m} = 9.0 \times 10^6 \text{ m}.$$

For a satellite orbiting a planet, Kepler’s third law (derived from Newton’s law of gravitation) states $$T^2 = \frac{4\pi^2}{G M}\,r^3,$$ where $$M$$ is the mass of the planet. We are asked to find $$M$$, so we solve this equation for $$M$$. Multiplying both sides by $$G M$$ and then dividing by $$T^2$$ gives

$$M = \frac{4\pi^2 r^3}{G T^2}.$$

The problem supplies the convenient constant $$\frac{4\pi^2}{G} = 6 \times 10^{11}\;\text{N}^{-1}\text{ m}^{-2}\text{ kg}^2.$$ Using this, we can rewrite the expression for the mass as

$$M = \left(\frac{4\pi^2}{G}\right)\frac{r^3}{T^2}.$$

Now we compute each part step by step. First, the cube of the radius:

$$r^3 = \left(9.0 \times 10^6\right)^3 = 9^3 \times 10^{18} = 729 \times 10^{18} = 7.29 \times 10^{20} \text{ m}^3.$$

Next, the square of the period:

$$T^2 = \left(2.7 \times 10^4\right)^2 = 2.7^2 \times 10^{8} = 7.29 \times 10^{8} \text{ s}^2.$$

We now form the ratio $$\dfrac{r^3}{T^2}$$:

$$\frac{r^3}{T^2} = \frac{7.29 \times 10^{20}}{7.29 \times 10^{8}} = 10^{12} \text{ m}^3\text{ s}^{-2}.$$

Finally, substituting this result into the mass formula, we get

$$M = \left(6 \times 10^{11}\right)\left(10^{12}\right) \text{ kg} = 6 \times 10^{23} \text{ kg}.$$

Hence, the correct answer is Option D.