The second Kepler law gives: $$\frac{dA(t)}{dt} = \frac{1}{2}r^2\dot{\theta} = \frac{1}{2}C$$

With A(t) the area swept by the line between the satellite and the Earth.

When the period T is reached, the whole ellipse has been swept. $$A(T) = \pi ab$$ $$CT = 2\pi a b$$

Hence $$\frac{T^2}{a^3} = \frac{4\pi^2b^2}{C^2a}$$

We can then use that the radius of the formula given by Icchyamoy, which is equivalent to assume an elliptical orbit: $$p(\theta) = \frac{B}{1+e\cos(\theta)}$$ Where $B = \frac{C^2}{GM}$ and e is constant.

Geometry gives that $$a = \frac{B}{1-e^2}$$

$$b = \frac{B}{\sqrt{1-e^2}}$$

So $$\frac{b^2}{a} = B = \frac{C^2}{GM}$$

And $$\frac{T^2}{a^3} = \frac{4\pi^2}{GM}$$

The second Kepler law gives: $$\frac{dA(t)}{dt} = \frac{1}{2}r^2\dot{\theta} = \frac{1}{2}C$$

with $A(t)$ the area swept by the line between the satellite and the Earth.

When the period $T$ is reached, the whole ellipse has been swept. $$A(T) = \pi ab$$ $$CT = 2\pi a b$$

Hence $$\frac{T^2}{a^3} = \frac{4\pi^2b^2}{C^2a}$$

We can then use that the radius of the formula given by Icchyamoy, which is equivalent to assume an elliptical orbit: $$p(\theta) = \frac{B}{1+e\cos(\theta)}$$ Where $B = \frac{C^2}{GM}$ and $e$ is constant.

Geometry gives that $$a = \frac{B}{1-e^2}$$

$$b = \frac{B}{\sqrt{1-e^2}}$$

So $$\frac{b^2}{a} = B = \frac{C^2}{GM}$$

And $$\frac{T^2}{a^3} = \frac{4\pi^2}{GM}$$

The second Kepler law gives: $$\frac{dA(t)}{dt} = \frac{1}{2}r^2\dot{\theta} = \frac{1}{2}C$$

With A(t) the area swept by the line between the satellite and the Earth.

When the period T is reached, the whole ellipse has been swept. $$A(T) = \pi ab$$ $$CT = 2\pi a b$$

Hence $$\frac{T^2}{a^3} = \frac{4\pi^2b^2}{C^2a}$$

We can then use that the radius of the formula given by Icchyamoy: $$p(\theta) = \frac{B}{1+e\cos(\theta)}$$ Where $B = \frac{C^2}{GM}$ and e is constant.

Geometry gives that $$a = \frac{B}{1-e^2}$$

$$b = \frac{B}{\sqrt{1-e^2}}$$

So $$\frac{b^2}{a} = B = \frac{C^2}{GM}$$

And $$\frac{T^2}{a^3} = \frac{4\pi^2}{GM}$$

The second Kepler law gives: $$\frac{dA(t)}{dt} = \frac{1}{2}r^2\dot{\theta} = \frac{1}{2}C$$

With A(t) the area swept by the line between the satellite and the Earth.

When the period T is reached, the whole ellipse has been swept. $$A(T) = \pi ab$$ $$CT = 2\pi a b$$

Hence $$\frac{T^2}{a^3} = \frac{4\pi^2b^2}{C^2a}$$

We can then use that the radius of the formula given by Icchyamoy, which is equivalent to assume an elliptical orbit: $$p(\theta) = \frac{B}{1+e\cos(\theta)}$$ Where $B = \frac{C^2}{GM}$ and e is constant.

Geometry gives that $$a = \frac{B}{1-e^2}$$

$$b = \frac{B}{\sqrt{1-e^2}}$$

So $$\frac{b^2}{a} = B = \frac{C^2}{GM}$$

And $$\frac{T^2}{a^3} = \frac{4\pi^2}{GM}$$