### Introduction It is generally useful for link budgets to know the range (distance) between earth and a satellite or between two satellites. The derivation is fun because it uses the cosine law, the quadratic formula, and Euler's Identity. The general case involves two satellites $SV_1$ and $SV_2$ orbiting around a body $\text{Earth}$ with radius $r$. Each satellite has a corresponding altitude $h_1$ and $h_2$. The only known angle is at the $SV_1$ and is $\pi /2$ plus the elevation angle $\phi$. The variable to solve for is the slant range $d$. A ground station simply means that $h_1=0$. These variables are generally set up for ground station elevation calculations or LEO satellites that only have GPS antennas facing zenith. [[slant_angle.drawio]] ![[slant_anle_drawing.png|500]] The diagram above has provocative lettering which matches to the following: - $a=r+h_1$ the distance between the center of earth to $SV_1$ - $b=d$ the slant distance to solve for - $c=r+h_2$ the distance between the center of the earth to $SV_2$ - $C = \pi /2+\phi$ the angle at $SV_1$ It simplifies the math to make $a$ and $c$ dependent on the radius of $SV_1$, $r_1$, so we redefine the values are: - $a=r_1$ - $c=r_1+r_\Delta$ where $r_\Delta=r_2-r_1$ ### Derivation Recall the **cosine law** $ c^2=a^2+b^2-2ab\cos{C} $ Filling in the values and moving everything to one side: $ 0=r_1^2+d^2-2d(r_1)\cos(\frac{\pi}{2}+\phi)-(r_1+r_\Delta)^2 $ Recall the identity $\cos(\frac{\pi}{2}+a)=-\sin(a)$ and simplify $ 0=d^2+2r_1\sin(\phi)d +\big(r_1^2-(r_1+r_\Delta)^2 \big) $ Recall the **quadratic equation** and fill in $ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a} $ $ d=\frac{1}{2}\bigg( -2r_1\sin(\phi) \pm \sqrt{\big(2r_1\sin(\phi)\big)^2-4\big(r_1^2-(r_1+r_\Delta)^2\big) } \bigg) $ $ d= \sqrt{ r_1^2(\sin^2(\phi)-1)+(r_1+r_\Delta)^2 }-r_1\sin(\phi) $ Note the $\pm$ was dropped because $d$ is a physical distance and thus cannot be negative. Insert **Euler's Identity** which results in: > [!NOTE] Slant Range Formula >$ >d= >\sqrt{ >(r_1+r_\Delta)^2-r_1^2\cos^2(\phi) >}-r_1\sin(\phi) >$ ### Example 1: Ground Station and Satellite *Given a ground station on earth has a minimum elevation angle of 5 degrees (10 is also common) and the LEO satellite is 1 000 km away, what is the maximum slant range? What is the minimum slant range?* Starting with the maximum slant range, we know: - $\phi=5\degree$ - $r_1=6731 \text{ km}$ radius of earth [[Radius of Earth Derivation]] - $r_\Delta=1000 \text{ km}$ $ d_\text{max}= \sqrt{ (6731+1000)^2-6371^2\cdot\cos^2(5\degree) }-6371\cdot\sin(5\degree) $ $ d_\text{max}= 3 193 \text{ km} $ The minimum slant range occurs when the satellite is right over head. For a ground station this is intuitively the LEO's orbiting altitude, $1000 \text{ km}$ but lets check with the formula: $ d_\text{min}= \sqrt{ (6731+1000)^2-6371^2\cdot\cos^2(90\degree) }-6371\cdot\sin(90\degree) $ $ d_\text{min}=1000\text{ km}=r_\Delta $ ### Example 2: LEO to GPS Satellite with elevation constraint *A LEO satellite orbiting at $1000\text{ km}$ only has GNSS antennas facing Zenith, that is, away from earth. The LEO has a minimum elevation angle of $5\degree$, what is the maximum slant range?* Starting with the knowns: - $\phi=5\degree$ - $r_1=6371+1000=7371 \text{ km}$ - $r_\text{2}=6371+ 20200\text{ km}$ GPS nominal distance from center of earth - $r_\Delta=r_2-r_1=19200\text{ km}$ $ d_\text{max}= \sqrt{ (7371+19200)^2-7371^2\cdot\cos^2(5\degree) }-7371\cdot\sin(5\degree) $ $ d_\text{max}=24894\text{ km} $ ### Example 3: LEO and GEO with no Elevation Constraint *A LEO satellite orbiting at $1000\ \text{km}$ has omnidirectional GNSS antennas, what is the maximum slant range?* For this with need the horizon angle of each satellite. The horizon angle is the angle at which the line of sight vector coming from the satellite is tangent to the surface of the earth. We will get the equation generically and then determine the maximum $d$ ![[horizon-angle-of-two-sv.png|500]] The knowns: - $r_1$ LEO SV orbit - $r_2$ GEO SV orbit - $R$ radius of earth - although we will keep it generic as it can be increased slightly to clear the atmosphere We define $d_1$ and $d_2$ as the distance from the respective SV to the tangent surface of the earth (or $R$) $ {d_1}^2= {r_1}^2-R^2 $ $ {d_2}^2= {r_2}^2-R^2 $ And simply sum the line of sight distances to get $d$ $ d=\sqrt{{r_1}^2-R^2}+\sqrt{{r_2}^2-R^2} $ $ d=\sqrt{7371^2-6371^2}+\sqrt{26571^2-6371^2} $ $ d=29502\ \text{km} $ Notice that $R$ can be greater than the radius of earth, just not smaller. It is useful to include the atmosphere in the radius $R$ to keep the slant range, and signal, fully in space. Finally, the horzin angles $\theta$ can be simply derived and are generally used along with antenna patterns to formalize worst-case link budgets. ![[Connect#Connect]]