A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet?
[Given: Mass of planet = $$8 \times 10^{22}$$ kg, Radius of planet = $$2 \times 10^6$$ m, Gravitational constant G = $$6.67 \times 10^{-11}$$ Nm$$^2$$/kg$$^2$$]

We start by collecting all the numerical data given in the question.

Mass of the planet: $$M = 8 \times 10^{22}\,\text{kg}$$

Radius of the planet: $$R = 2 \times 10^{6}\,\text{m}$$

Height of the circular orbit above the surface: $$h = 20\,\text{km} = 20 \times 10^{3}\,\text{m} = 2.0 \times 10^{4}\,\text{m}$$

Universal gravitational constant: $$G = 6.67 \times 10^{-11}\,\text{N\,m}^2/\text{kg}^2$$

The distance of the spaceship from the centre of the planet is the orbital radius

$$r = R + h = 2 \times 10^{6}\,\text{m} + 2 \times 10^{4}\,\text{m} = 2.02 \times 10^{6}\,\text{m}.$$

For a body in a circular orbit, the gravitational force supplies the required centripetal force. We write this fundamental equality explicitly:

$$\frac{G M m}{r^{2}} = \frac{m v^{2}}{r}.$$

Cancelling the common factor $$m$$ and one power of $$r$$, we obtain the orbital speed $$v$$:

$$v^{2} = \frac{G M}{r} \quad\Longrightarrow\quad v = \sqrt{\frac{G M}{r}}.$$

The time period $$T$$ of one revolution is the distance travelled in one orbit divided by the speed. The path length for one revolution is the circumference $$2\pi r$$, so

$$T = \frac{2\pi r}{v}.$$

Substituting the expression for $$v$$ that we just derived, we have

$$T = \frac{2\pi r}{\sqrt{\dfrac{G M}{r}}} = 2\pi r \sqrt{\frac{r}{G M}} = 2\pi \frac{r^{3/2}}{\sqrt{G M}}.$$

Now we evaluate each factor step by step.

First, compute $$r^{3}$$:

$$r = 2.02 \times 10^{6}\,\text{m},$$ $$r^{3} = (2.02 \times 10^{6})^{3} = 2.02^{3} \times 10^{18} = 8.244 \times 10^{18}\,\text{m}^{3}.$$

Taking the square root gives $$r^{3/2}$$:

$$r^{3/2} = \sqrt{r^{3}} = \sqrt{8.244 \times 10^{18}} = \sqrt{8.244}\,\times\,10^{9} \approx 2.87 \times 10^{9}\,\text{m}^{3/2}.$$

Next, calculate $$G M$$:

$$G M = 6.67 \times 10^{-11}\,\text{N\,m}^{2}\!/\text{kg}^{2}\; \times 8 \times 10^{22}\,\text{kg} = 53.36 \times 10^{11}\,\text{m}^{3}\!/\text{s}^{2} = 5.336 \times 10^{12}\,\text{m}^{3}\!/\text{s}^{2}.$$

We now need $$\sqrt{G M}$$:

$$\sqrt{G M} = \sqrt{5.336 \times 10^{12}} = \sqrt{5.336}\,\times\,10^{6} \approx 2.309 \times 10^{6}\,\text{m}^{3/2}\!/\text{s}.$$

With these two intermediate results, we substitute back into the formula for the period:

$$T = 2\pi\,\frac{r^{3/2}}{\sqrt{G M}} = 2\pi\,\frac{2.87 \times 10^{9}}{2.309 \times 10^{6}} = 2\pi\,\times\,1.243 \times 10^{3} = 6.2832 \times 1.243 \times 10^{3} \approx 7.81 \times 10^{3}\,\text{s}.$$

Thus one revolution takes approximately

$$T \approx 7.81 \times 10^{3}\,\text{s} = 7810\,\text{s}.$$

We convert this period into hours for convenience:

$$1\,\text{hour} = 3600\,\text{s},\quad T = \frac{7810}{3600}\,\text{h} \approx 2.17\,\text{h}.$$

The spaceship completes one revolution in about 2.17 hours. The total time we are interested in is 24 hours, so the number of complete revolutions $$N$$ is

$$N = \frac{24\,\text{h}}{2.17\,\text{h}} \approx 11.06.$$

Since only whole (complete) revolutions count, the spaceship makes 11 complete orbits in 24 hours.

Hence, the correct answer is Option D.