### Section 4.6 - Interesting Orbital Calculations

If you've made it this far through my space primer, I commend you! I figured I'd finish with a bang by saving the most math-heavy section for the end. I often wish my math and physics background were better so I could more thoroughly understand the calculations real engineers do, but these problems here were a few I've been able to decipher (and had a ton of fun doing so!) Try you hand at them - if I can do it, so can you!

#### Problem 1: Launch Azimuth and Velocity to the ISS

Before I introduce this problem, there's a few more concepts I need to explain. When launching rockets, the launch azimuth is the angle between the North direction and the direction of initial launch (ie the compass heading you head for when you launch)
 North is 0$^\circ$ azimuth, East is 90$^\circ$, South is 180$^\circ$, West is 270$^\circ$

The latitude of the launch site and the launch azimuth determine the maximum north-south latitude range (ie the inclination) the spacecraft can cover. For example, if you launch from Cape Canaveral (latitude = 28.5$^\circ$) at an azimuth of 90$^\circ$ (due East), the max north-south latitude range you can cover will be 28.5$^\circ$ N/S

But altering launch azimuth can change inclination, according to this formula:

$\cos{i} = \cos{\phi}\sin{\beta}$

$i =$ inclincation
$\phi =$ latitude
$\beta =$ azimuth

There's an interesting consequence of this formula, that if the launch latitude is higher than the inclination, the orbit can't be directly reached and will require either a different launch site or a post-launch orbital maneuver. Here's the mathematical proof behind why:

$\beta = \sin^{-1}{(\frac{\cos{i}}{\cos{\phi}})}$, but $\frac{\cos{i}}{\cos{\phi}}$ cannot be greater than 1 or it will fall out of the domain of $\sin^{-1}$

If you're curious about the derivation of the inclination formula, you can read up on it here

Great, now we're ready to start. The following problem is taken from this website:

The ISS orbits at a 51.6$^\circ$ inclination at a speed of 7,730 m/s. A Space Shuttle is launched from Cape Canaveral (latitude = 28.5$^\circ$). What azimuth and launch speed is needed for the shuttle to reach the ISS?

We know from above that $\beta = \sin^{-1}{(\frac{\cos{i}}{\cos{\phi}})}$, so plug in 51.6$^\circ$ for $i$ and 28.5$^\circ$ for $\phi$ and we'll get $\beta$ = 44.98$^\circ$. Here's the issue - this formula is correct in inertial (static) space, but we need to factor in the Earth's rotation. From here on out, we need to think in terms of vector algebra

Inertial velocity is the velocity we need to catch our target, without the help of Earth's rotation. It is defined as the x and y components of azimuth, scaled up to the proper magnitude of $v_{orbit}$ (speed, the scalar of velocity)

$\vec{v}_{inertial} = v_{orbit} \langle\sin(\beta_{inertial}), \cos(\beta_{inertial}) \rangle$

Remember that a vector is just a physical quantity that has both direction and magnitude. In 2D space, vectors have an x and a y component, using the notation $\vec{v} = \langle v_x, v_y \rangle$. Additionally, if you remember the unit circle from trigonometry, given a certain angle, the x and y components of that angle are given by cosine and sine, respectively. But if that's they case, why is our x component given by sine and y given by cosine (why are they flipped?) This is because we're measuring azimuth, with measures angles clockwise from north, which will algebraically flip the sine and cosine as x and y.

 This diagram illustrates what vector we are trying to calculate - the rotating vector

So now that we understand what vector we're trying to calculate, let's do everything algebraically with just the variables. At the end, it'll be easy to plug in actual numbers and get our answer

$\vec{v}_{earth} = v_{eqrot} \langle \cos(\phi),0 \rangle$

Velocity boost from Earth's rotation depends on the cosine of latitude and velocity of Earth's rotation at the equator ($v_{eqrot}$, which happens to be 465 m/s). Y component is 0 because Earth's rotation doesn't affect the north-south direction

$\vec{v}_{rot} = \vec{v}_{inertial} - \vec{v}_{earth}$

The velocity we need to launch to, accounting for Earth's rotation, is simply the velocity we'd need to launch from a static space minus the help we get from Earth's rotation

$\vec{v}_{rot} = v_{orbit} \langle\sin(\beta_{inertial}), \cos(\beta_{inertial}) \rangle - v_{eqrot} \langle \cos(\phi),0 \rangle$
Substitute formulas for $\vec{v}_{inertial}$ and  $\vec{v}_{earth}$

$\vec{v}_{rot} = \langle v_{orbit} \sin{(\beta_{inertial})} - v_{eqrot} \cos{(\phi)}, v_{orbit} \cos{(\beta_{inertial})} \rangle$
Combine x and y components

At this point, we have exactly the formula we need to find the speed (magnitude) and azimuth (direction) of the launch we need to reach the ISS. Now we can plug in the numbers and solve

Remember,   $\beta_{inertial} =$ 44.98$^\circ$,   $v_{orbit} =$ 7,730 m/s,   $v_{eqrot} =$ 465 m/s,   $\phi =$ 28.5$^\circ$. We are trying to solve for $\beta_{rot}$ and $v_{rot}$

$\vec{v}_{rot} = \langle 7730 \sin{(44.98^\circ)} - 465 \cos{(28.5^\circ)}, 7730 \cos{(44.98^\circ)} \rangle$

$\vec{v}_{rot} = \langle 5055, 5467 \rangle \thinspace m/s$

$\beta_{rot} = 90^\circ - \tan^{-1}{(\frac{5467}{5055})} = 42.76^\circ$  << Launch to this azimuth

$v_{rot} = \sqrt{5055^2 + 5467^2} = 7446 \thinspace m/s$ << Launch at this speed

$\Delta v = v_{orbit} - v_{rot} = 7730 - 7446 = 284 \thinspace m/s$ << Speed saved by Earth's rotation

#### Problem 2: Escape Velocity at Apogee or Perigee?

We learned from the previous section about Newton's vis-viva equation and how it can be used to calculate escape velocity. The question here is, given a satellite in an elliptical orbit, is it easier to achieve escape velocity at periapsis or apoapsis? This problem is adapted from Orbital Mechanics, Theory and Applications by Tom Logsdon

Assume a satellite is orbiting Earth. Apogee is 4 Earth radii and perigee is 2 Earth radii. Here are some other numbers we'll need:

Gravitational constant: $G = 6.674 \cdot 10^{-11} \thinspace m^3\cdot kg^{-1}\cdot s^{-2}$
Mass of the Earth: $M = 5.972 \cdot 10^{24} \thinspace kg$
Radius of the Earth: $r = 6.371 \cdot 10^6 \thinspace m$

If apogee is $4r$ and perigee is $2r$, we can conclude that the semi-major axis is $3r$

Delta-v needed for escape velocity at apogee can be calculated as such:

$\Delta v_a =$ apogee escape velocity - current velocity at apogee

$\Delta v_a = \sqrt{GM (\frac{2}{4r} - \frac{1}{\infty})} - \sqrt{GM (\frac{2}{4r} - \frac{1}{3r})}$

$\Delta v_a = 5593 - 3229 = 2364 \thinspace m/s$

The same calculation can be applied for determine delta-v needed for escape velocity at perigee

$\Delta v_p =$ perigee escape velocity - current velocity at perigee

$\Delta v_p = \sqrt{GM (\frac{2}{2r} - \frac{1}{\infty})} - \sqrt{GM (\frac{2}{2r} - \frac{1}{3r})}$

$\Delta v_p = 7910 - 6458 = 1451 \thinspace m/s$

Escaping at perigee instead of apogee saves us $912$ m/s in delta-v. In fact, it is always more efficient to escape at periapsis. This concept is known as the Oberth effect, which states that as a spacecraft falls into a gravitational well and accelerates, an engine burn can produce greater mechanical energy than at lower speeds.

#### Problem 3: Hohmann vs Bi-elliptic Transfer Orbits

We learned from a previous section of two methods to increase the altitude of a circular orbit, and that which one is more fuel efficient depends on the ratio of the starting and ending orbit radius. More specifically, the bi-elliptic transfer orbit is generally more fuel efficient than the Hohmann transfer orbit when the ending radius is greater than 11.94x the starting radius. Can we show this in a mathematical example?

The following problem is taking from Wikipedia's article on the bi-elliptic transfer. The Wikipedia article on the Hohmann may be useful reference too. Assume a satellite is currently in a circular orbit with $r_0 =$ 6,700 km, and we want to increase it to $r_1 =$ 93,800 km. Calculate the cumulative delta-v required for a Hohmann transfer and a bi-elliptic transfer

We'll be needing $G$ and $M$ a lot for this problem too. To keep the formulas concise, I'll leave them as the variables. Refer to the previous problem for the actual values

Hohmann Transfer:

$v_0 = \sqrt{\frac{GM}{6,700,000}} = 7.712 \thinspace km/s$
This is the orbital speed for the original circular orbit (dark blue)

$v_0 = \sqrt{\frac{GM}{93,800,000}} = 2.061 \thinspace km/s$
This is the orbital speed for the new circular orbit (orange)

Now we'll need to find the perigee and apogee speeds for the Hohmann transfer orbit, the elliptical gray orbit. We'll first need to determine its semi-major axis
$a = \frac{6,700 + 93,800}{2} = 50,250 \thinspace km$

$v_{perigee} = \sqrt{GM(\frac{2}{6,700,000} - \frac{1}{50,250,000})} = 10.538 \thinspace km/s$

$v_{apogee} = \sqrt{GM(\frac{2}{93,800,000} - \frac{1}{50,250,000})} = 0.753 \thinspace km/s$

From this, we can determine the delta-v requirements of the two burns:

$\Delta v_1 = 10.538 - 7.712 = 2.826 \thinspace km/s$

$\Delta v_2 = 2.061 - 0.753 = 1.308 \thinspace km/s$

$\Delta v_{total} = 2.826 + 1.308 = 4.134 \thinspace km/s$

Let's see if the bi-elliptic transfer can do better!

Bi-elliptic Transfer:

$\frac{r_1}{r_0} = 14$, so the bi-elliptic transfer should be able to do better since it's greater than 11.94. We do need to choose an apogee for the intermediate orbit, so let's go with $40r_0 = 40 \cdot 6,700 = 268,000 km \thinspace$

Let's start with the semi-major axis and perigee velocity of the green ellipse

$a_{green} = \frac{6,700+268,000}{2} = 50,250 \thinspace km$

$v_{perigee\_green} = \sqrt{GM(\frac{2}{6,700,000} - \frac{1}{137,350,000})} = 10.774 \thinspace km/s$

If the perigee velocity of the green ellipse is 10.774 km/s and the orbital speed of the blue circle is 7.712 km/s, we can determine the delta-v requirement of the first burn

$\Delta v_1 = 10.774 - 7.712 = 3.062 \thinspace km/s$

We need a second burn to widen the orbit again. Let's figure out the speed of the spacecraft at the apogee of the green ellipse.

$v_{apogee\_green} = \sqrt{GM(\frac{2}{268,000,000} - \frac{1}{137,350,000})} = 0.269 \thinspace km/s$

The new orbit (red ellipse) will have a perigee of 93,800 km and an apogee of 268,000 km, so we can calculate the new semi-major axis and apogee velocity

$a_{red} = \frac{93,800+268,000}{2} = 180,900 \thinspace km$

$v_{apogee\_red} = \sqrt{GM(\frac{2}{268,000,000} - \frac{1}{180,900,000})} = 0.878 \thinspace km/s$

Now we can calculate the delta-v requirement of the second burn

$\Delta v_2 = 0.878 - 0.269 = 0.609 \thinspace km/s$

What velocity will the spacecraft have when it reaches the perigee of the red ellipse?

$v_{perigee\_red} = \sqrt{GM(\frac{2}{93,800,000} - \frac{1}{180,900,000})} = 2.509 \thinspace km/s$

Recall that the third burn of a bi-elliptic transfer that increases orbital altitude is a retrograde burn. We need to slow the spacecraft down to re-circularize into the final orbit (orange). We know the orbital velocity of the orange circular orbit is 2.061 km/s, so the delta-v requirement of the third burn is

$\Delta v_3 = 2.509 - 2.061 = 0.448 \thinspace km/s$

Total delta-v of all three burns is

$\Delta v_{total} = 3.062 + 0.609 + 0.448 = 4.119 \thinspace km/s$

Compared to the delta-v requirement of the Hohmann transfer of 4.134 km/s, the bi-elliptic orbit would save 0.015 km/s of delta-v!

Although it doesn't seem like much, for larger orbital transfers (like Earth orbit to Mars orbit) the difference would be more meaningful. If we wanted to, we could further improve the delta-v savings by increasing the apogee of the intermediate orbit. The tradeoff: time. Bi-elliptic transfers take a lot longer, which would be unsuitable for a manned mission