A satellite of mass, M revolves around a planet of mass, m in an orbit of radius, R with velocity, V. Show that its period, T is related to the orbital radius, R by the equation T2 = 39.44R3 / Gm [π = 3.14]

The gravitational force between the planet (mass $M$) and the satellite (mass $m$) is given by:

$$F_g = \frac{GMm}{R^2}$$

The centripetal force required to keep the satellite in orbit is:

$$F_c = \frac{mV^2}{R}$$

Setting these two forces equal:

$$\frac{GMm}{R^2} = \frac{mV^2}{R}$$

Cancel $m$ from both sides and multiply by $R$:

$$\frac{GM}{R} = V^2$$ $$V = \sqrt{\frac{GM}{R}}$$

The satellite travels a distance equal to the circumference of its orbit ($2\pi R$) in one period $T$.

$$V = \frac{2\pi R}{T}$$

$$\left(\frac{2\pi R}{T}\right)^2 = \left(\sqrt{\frac{GM}{R}}\right)^2$$ $$\frac{4\pi^2 R^2}{T^2} = \frac{GM}{R}$$

Multiply both sides by $T^2$ and by $R$, and divide by $GM$:

$$T^2 = \frac{4\pi^2 R^2 \cdot R}{GM}$$ $$T^2 = \frac{4\pi^2 R^3}{GM}$$

Calculate $4\pi^2$:

$$4\pi^2 = 4 \times (3.14)^2$$ $$4\pi^2 = 4 \times 9.8596$$ $$4\pi^2 = 39.4384$$

$$T^2 = \frac{39.4384 R^3}{GM}$$

Rounding 39.4384 to two decimal places gives 39.44.
Therefore, we have shown that:

$$T^2 = \frac{39.44 R^3}{GM}$$