## 6 Propagation Time for a Signal, Emitted and Intercepted by Satellites on a Space-Oriented Orbit
### Second Formulation of the Theorem for Four-Dimensional Uniformization
In the first approach (see Section 10.1), the initial integral \(\widetilde{J}_{0}^{(q)}(y)\) was brought to the Weierstrass form (144) (multiplied by the conformal factor \(\sqrt{a}=\sqrt{\frac{g_{\mathcal{G}_{3}}}{8^{2}}.\frac{(q^{4}-q^{2}+1)}{(2q^{ 4}-5q^{2}+2)}}\)) by means of the transformation (145) \(x=\frac{a}{y^{2}}+b\), as a result of which, the integral acquires the form (144). This will be Equation (228) below. The second approach is based on the method of \"four-dimensional uniformization\", as a result of which, the integral is brought to the second representation (203) \(I_{0}^{(3)}(x)=\int_{s}^{\infty}\frac{d\sigma}{\sqrt{4\sigma^{3}-\overline{ \overline{g}}_{2}\sigma^{\prime}-\overline{\overline{g}}_{3}}}\). This will be Equation (229) below. As a result of the two representations, the following two equalities can be written:  
\[\widetilde{J}_{0}^{(q)}(y)=\int\frac{dy}{\sqrt{(1-y^{2})(1-q^{2}y^{2})}}=- \sqrt{a}\int\frac{dx}{\sqrt{4x^{3}-g_{2}x-g_{3}}} \tag{228}\]  
\[=\int\frac{d\sigma}{\sqrt{4\sigma^{3}-\overline{\overline{g}}_{2}\sigma^{ \prime}-\overline{\overline{g}}_{3}}}. \tag{229}\]  
Now, we shall use the following theorem, proved in the monograph by Hancock [67], in order to transform the integral (228). **Theorem 11**. : _The elliptic integral in the Weierstrass representation (144) (equation (A) in (228) in terms of the \"\(x\)\"-variable) can be written in terms of the variable \(\sigma\) as_  
\[\int\frac{dx}{\sqrt{4x^{3}-g_{2}x-g_{3}}}=-\int\frac{d\sigma}{\sqrt{4\sigma^{ 3}-g_{2}\sigma^{\prime}-g_{3}}} \tag{230}\]  
_after performing the variable transformation_  
\[I_{0}^{(4)}(y)=\int\frac{dy}{\sqrt{a_{0}y^{4}+a_{1}y^{3}+a_{2}y^{2}+a_{3}y+a_{ 4}}}, \tag{222}\]0_where \(x_{0},x_{1}\) and \(x_{2}\) are the roots of the cubic polynomial \(4x^{3}-g_{2}x-g_{3}=0\)._  
In the original formulation of the theorem in [67] (page 29), the transformation \(t=e_{1}+\frac{(e_{2}-e_{1})(e_{3}-e_{1})}{s-e_{1}}\) has been applied to the integral \(\int\frac{dt}{\sqrt{(t-e_{1})(t-e_{2})(t-e_{3})}}\). Note, however, the minus sign in front of the right term in Equation (230), which is important and can easily be confirmed by a direct calculation. Applying, again, the above theorem with respect to Equations (229) and also (203), the equality (230) can be written as  
\[I_{0}^{(4)}(y)=\int\frac{dy}{\sqrt{a_{0}y^{4}+a_{1}y^{3}+a_{2}y^{2}+a_{3}y+a_{ 4}}}, \tag{222}\]1  
where the variable transformation \(\sigma=\frac{1}{\sqrt[3]{a}}\overline{\sigma}\) has been performed and the notations  
\[I_{0}^{(4)}(y)=\int\frac{dy}{\sqrt{a_{0}y^{4}+a_{1}y^{3}+a_{2}y^{2}+a_{3}y+a_{ 4}}}, \tag{222}\]2  
are introduced. From Equations (230) and (232) (combined with (203)), we obtain  
\[I_{0}^{(4)}(y)=\int\frac{dy}{\sqrt{a_{0}y^{4}+a_{1}y^{3}+a_{2}y^{2}+a_{3}y+a_{ 4}}}, \tag{222}\]3  
Note the important peculiarity of Equation (234): The Weierstrass invariants on the left-hand side of (234) are \(g_{2}\), \(g_{3}\), while on the right-hand side, they are \(\widehat{g}_{2}\equiv\overline{g}_{2}a^{\frac{3}{2}}\), \(\widehat{g}_{3}\equiv\overline{g}_{3}a\). Algebraic Roots of the Cubic Equations and Comparison of the Weierstrass Invariants--The Case of Positive Discriminant  
From the Cardano formulae, we have, for the roots of the cubic polynomial \(4\sigma^{3}-g_{2}\sigma-g_{3}=0\) (\(i=1,2,3\)), the following formulae (see the monographs [61; 68]):  
\[I_{0}^{(4)}(y)=\int\frac{dy}{\sqrt{a_{0}y^{4}+a_{1}y^{3}+a_{2}y^{2}+a_{3}y+a_{ 4}}}, \tag{222}\]4  
\[I_{0}^{(4)}(y)=\int\frac{dy}{\sqrt{a_{0}y^{4}+a_{1}y^{3}+a_{2}y^{2}+a_{3}y+a_{ 4}}}, \tag{222}\]5  
This is only the first root of the cubic equation \(4\sigma^{3}-g_{2}\sigma-g_{3}=0\). Further, we shall investigate the case of a positive discriminant \(\Delta=\frac{g^{2}}{4}+\frac{p^{3}}{2\overline{\sigma}}>0\) for the cubic equation \(x^{3}+px+q=0\)[61].