If you've found your way here, you are obviously one of the brave souls who dare to tackle orbital mechanics the old fashioned way — with grit and determination. Kudos!
To help you along your journey, there are examples included with many of the formulas (and more forthcoming).
Enjoy! (And, of course, good luck!)
\[\bar{p} = m \bar{V}\]
Where:
$\bar{p} = linear \ momentum \ vector \ (kg \cdot m/s)$
$m = mass \ (kg)$
$\bar{V} = velocity \ vector \ (m/s)$
How fast would a 5 kg bowling ball have to be rolling to have the same linear momentum as a 25,000 kg truck moving at 1 m/s?
Given:
$m_{ball} = 5 \ kg$
$m_{truck} = 25,000 \ kg$
$V_{truck} = 1 \ m/s$
Find:
$V_{ball} \ to \ equal \ the \ linear \ momentum \ of \ the \ truck$
Solution:
$p_{truck} = m_{truck} V_{truck}$
$p_{truck} = (25,000 \ kg)(1 \ m/s)$
$p_{truck} = 25,000 \ kg \cdot m/s$
$V_{ball} = \frac{p_{truck}}{m_{ball}}$
$V_{ball} = \frac{25,000 \ kg \cdot m/s}{5 \ kg}$
$V_{ball} = 5,000 \ m/s$
Answer:
$V_{ball} = 5,000 \ m/s$
\[\bar{H} = I \bar{\Omega}\]
Where:
$\bar{H} = angular \ momentum \ vector \ (kg \cdot m^{2}/s)$
$I = moment \ of \ inertia \ (kg \cdot m^{2})$
$\Omega = angular \ velocity \ vector \ (rad/s)$
A football thrown in a perfect spiral has a moment of inertia of 0.001 kg · m2. If the football is spinning at 60 rpm, what is its angular momentum?
Given:
$I = 0.001 \ kg \cdot m^{2}$
$\bar{ \Omega} = 60 \ rpm \Rightarrow (60) \left( \frac{2 \pi}{60} \ rad/s \right) = 6.283 \ rad/s$
Find:
$H$
Solution:
$\bar{H} = I \bar{ \Omega}$
$H = (0.001 \ kg \cdot m^{2})(6.283 \ rad/s)$
$H = 0.006283 \ kg \cdot m^{2}/s$
Answer:
$H = 0.006283 \ kg \cdot m^{2}/s$
\[\bar{H} = \bar{R} \times m \bar{V}\]
** Note: this is a cross product **
Where:
$\bar{H} = angular \ momentum \ vector \ (kg \cdot m^{2}/s)$
$\bar{R} = position \ (or, \ moment \ arm) \ (m)$
$m = mass \ (kg)$
$\bar{V} = velocity \ vector \ (m/s)$
Imagine you are spinning a 0.30 kg mass (attached to the end of a 0.5 m string) over your head. If you let go of the string the mass will sail off on a tangent of 2 m/s. What was the angular momentum of the spinning mass before you let go?
Given:
$\ m = 0.30 \ kg$
$R = 0.5 \ m$
$V = 2 \ m/s$
Find:
$H$
Solution:
$ \bar{H} = \bar{R} \times \bar{m} \bar{V}$
$H = RmV$
$H = (0.5 \ m)(0.30 \ kg)(2 m/s)$
$H = 0.30 \ kg \cdot m/s$
Answer:
$H = 0.30 \ kg \cdot m/s$
\[\bar{F} = m \bar{a}\]
Where:
$\bar{F} = force \ vector \ (kg \cdot m/s^{2}, \ or, \ N \ (Newtons))$
$m = mass \ (kg)$
$\bar{a} = acceleration \ (m/s^{2})$
A hockey player is able to apply a 100 N force to 0.170097 kg hockey puck for a total of 0.1 seconds. Ignoring gravity, how fast will the hockey puck be going?
Given:
$m_{puck} = 0.170097 \ kg$
$F_{player} = 100 \ N$
$\Delta t = 0.1 \ s$
Find:
$\Delta V_{puck}$
Solution:
$F = ma \Rightarrow F = m \frac{\Delta V}{\Delta t}$
$\Delta V = \frac{F \Delta t}{m}$
$\Delta V = \frac{(100 \ N)(0.1 \ s)}{0.170097 \ kg}$
$\Delta V = 58.79 \ m/s$
Answer:
$\Delta V = 58.79 \ m/s$
Or, 131.5 mph — What a shot!! (A hard slapshot typically clocks in at about 100 mph)
\[F_{g} = \frac{Gm_{1}m_{2}}{R^{2}}\]
Where:
$F_{g} = force \ due \ to \ gravity \ (N)$
$G = universal \ gravitational \ constant \approx 6.67 \times 10^{-11} (N \cdot m^{2}/kg^{2})$
$m_{1}, \ m_{2} = masses \ of \ two \ bodies \ (m)$
$R = distance \ between \ the \ two \ bodies \ (m)$
In the void of interstellar space, two asteroids pass each other at a distance of 200 m. Asteroid 1 has a mass of 2 X 106 kg, while Asteroid 2 has a mass of 8 X 106 kg. What is the force of gravitational attraction between them?
Given:
$m_{Asteroid 1} = 2 \times 10^{6} \ kg$
$m_{Asteroid 2} = 8 \times 10^{6} \ kg$
$R = 200 \ m$
Find:
$F_{g}$
Solution:
$F_{g} = \frac{Gm_{Asteroid 1}m_{Asteroid 2}}{R^{2}}$
$F_{g} = \frac{(6.67 \times 10^{-11} \ N \cdot m^{2}/kg^{2})(2 \times 10^{6} \ kg)(8 \times 10^{6} \ kg)}{(200 \ m)^{2}}$
$F_{g} = 0.02668 \ N$
Answer:
$F_{g} = 0.02668 \ N$
\[a_{g} = \frac{\mu_{earth}}{R^{2}}\]
Where:
$a_{g} = acceleration \ due \ to \ gravity \ (m/s^{2})$
$\mu_{earth} = Gm_{earth} \approx 3.986 \times 10^{14} \ (m^{3}/s^{2}) \ or \ 3.986 \times 10^{5} \ (km^{3}/s^{2})$
$R = distance \ to \ Earth's \ center \ (m \ or \ km)$
\[E = KE + PE\]
Where:
$E = total \ mechanical \ energy \ (kg \cdot m^{2}/s^{2})$
$KE = kinetic \ energy \ (kg \cdot m^{2}/s^{2})$
$PE = potential \ energy \ (kg \cdot m^{2}/s^{2})$
\[E = \frac{1}{2}mV^{2} - \frac{m \mu}{R}\]
Where:
$E = total \ mechanical \ energy \ (kg \cdot km^{2}/s^{2})$
$m = mass \ (kg)$
$V = velocity \ (km/s)$
$\mu = gravitational \ parameter \approx 3.986 \times 10^{5} \ (km^{3}/s^{2})$
$R = position \ (km)$
\[KE = \frac{1}{2}mV^{2}\]
Where:
$KE = kinetic \ energy \ (kg \cdot km^{2}/s^{2})$
$m = mass \ (kg)$
$V = velocity \ (km/s)$
\[PE = - \frac{m \mu}{R}\]
Where:
$PE = potential \ energy \ (kg \cdot km^{2}/s^{2})$
$m = mass \ (kg)$
$\mu = gravitational \ parameter \approx 3.986 \times 10^{5} \ (km^{3}/s^{2})$
$R = distance \ from \ Earth's \ center \ (km)$
\[\ddot{\bar{R}} + \frac{\mu}{R^{2}} \frac{\bar{R}}{R} = 0\]
** Note: $\ddot{\bar{R}}$ is the engineering convention for the second derivative of $\bar{R}$ with respect to time, better known as acceleration, or, $\bar{a}$ **
Where:
$\ddot{\bar{R}} = spacecraft's \ acceleration \ (km/s^{2})$
$\mu = gravitational \ parameter = 3.986 \times 10^{5} (km^{3}/s^{2}) \ for \ Earth$
$\bar{R} = spacecraft's \ position \ vector \ (km)$
$R = magnitude \ of \ spacecraft's \ position \ vector \ (km)$
$\bar{R} = spacecraft's \ position \ vector, \ measured \ from \ Earth's \ center$
$\bar{V} = spacecraft's \ velocity \ vector$
$F \ and \ F^{1} = primary \ and \ vacant \ foci \ of \ the \ ellipse$
$R_{p} = radius \ of \ periapsis$
$R_{a} = radius \ of \ apoapsis$
$2a = major \ axis$
$2b = minor \ axis$
$2c = distance \ between \ the \ foci$
$a = semimajor \ axis$
$b = semiminor \ axis$
$\nu = true \ anomaly$
$\phi = flight \ path \ angle$
$e = eccentricity$
\[R = \frac{a(1 - e^{2})}{1 + e \cos \nu}\]
Where:
$R = magnitude \ of \ the \ spacecraft's \ position \ vector \ (km)$
$a = semimajor \ axis \ (km)$
$e = eccentricity \ (unitless)$
$\nu = true \ anomaly \ (deg \ or \ rad)$
\[\varepsilon = \frac{V^{2}}{2} - \frac{\mu}{R}\]
Where:
$\varepsilon = spacecraft's \ specific \ mechanical \ energy \ (km^{2}/s^{2})$
$V = spacecraft's \ velocity \ (km/s)$
$\mu = gravitational \ parameter \approx 3.986 \times 10^{5} \ (km^{3}/s^{2}) \ on \ Earth$
$R = spacecraft's \ distance \ from \ Earth's \ center \ (km)$
\[\varepsilon = - \frac{\mu}{2a}\]
$\varepsilon = spacecraft's \ specific \ mechanical \ energy \ (km^{2}/s^{2})$
$\mu = gravitational \ parameter = 3.986 \times 10^{5} \ (km^{3}/s^{2})$
$a = semimajor \ axis \ (km)$
What is the specific mechanical energy of an orbit with a semimajor axis of 46,320 km?
Given:
$a = 46,320 \ km$
Find:
$ \varepsilon$
Solution:
$ \varepsilon = - \frac{ \mu}{2a}$
$ \varepsilon = - \frac{3.986 \times 10^{5} \ km^{3}/s^{2}}{2(46,320 \ km)}$
$ \varepsilon = - 4.30 \ km^{2}/s^{2}$
Answer:
$ \varepsilon = - 4.30 \ km^{2}/s^{2}$
\[P = 2 \pi \sqrt{\frac{a^{3}}{\mu}}\]
Where:
$P = orbital \ period \ (s)$
$\pi = 3.14159... \ (unitless)$
$a = semimajor \ axis \ (km)$
$\mu = gravitational \ parameter = 3.986 \times 10^{5} \ (km^{3}/s^{2})$
\[\bar{h} = \bar{R} \times \bar{V}\]
Where:
$\bar{h} = spacecraft's \ specific \ angular \ momentum \ vector \ (km^{2}/s)$
$\bar{R} = spacecraft's \ position \ vector \ (km)$
$\bar{V} = spacecraft's \ velocity \ vector \ (km/s)$
A spacecraft sends the following information to a ground-based tracking station:
$ \bar{R} = 7220 \ \hat{I} + 5477 \ \hat{J} + 223 \ \hat{K} \ km$
$ \bar{V} = 0.34 \ \hat{I} - 0.75 \ \hat{J} - 8.00 \ \hat{K} \ km/s$
What is the specific angular momentum of the spacecraft?
Given:
$ \bar{R} = 7220 \ \hat{I} + 5477 \ \hat{J} + 223 \ \hat{K} \ km$
$ \bar{V} = 0.34 \ \hat{I} - 0.75 \ \hat{J} - 8.00 \ \hat{K} \ km/s$
Find:
$ \bar{h}$
Solution:
$ \bar{h} = \bar{R} \times \bar{V}$
$ \bar{h} = [(5477)(-8.00) - (-0.75)(223)] \ \hat{I} - [(7220)(-8.00) - (0.34)(223)] \ \hat{J} +$
$[(7220)(-0.75) - (0.34)(5477)] \ \hat{K} \ km^{2}/s$
$ \bar{h} = [(-43816) - (-167.25)] \ \hat{I} - [(-57760) - (75.82)] \ \hat{J} +$
$[(-5415) - (1882.18)] \ \hat{K} \ km^{2}/s$
$ \bar{h} = - 43,648.75 \ \hat{I} + 57,835.82 \ \hat{J} - 7,277.18 \ \hat{K} \ km^{2}/s$
Answer:
$ \bar{h} = - 43,648.75 \ \hat{I} + 57,835.82 \ \hat{J} - 7,277.18 \ \hat{K} \ km^{2}/s$
This example illustrates the use of most (if not all) formulas listed in this section. It's very beneficial to see how they all fit together.
A new Earth observation satellite is known to have the following position and velocity vectors:
$\bar{R} = 8228 \hat{I} + 389.0 \hat{J} + 6888 \hat{K} \ km$
$\bar{V} = -0.7000 \hat{I} + 6.600 \hat{J} -0.6000 \hat{K} \ km/s$
What are this satellite's COE's (a, e, i, Ω, ω, and ν)?
Given:
$\bar{R} = 8228 \hat{I} + 389.0 \hat{J} + 6888 \hat{K} \ km$
$\bar{V} = -0.7000 \hat{I} + 6.600 \hat{J} -0.6000 \hat{K} \ km/s$
Find:
$a \ (semimajor \ axis)$
$e \ (eccentricity)$
$i \ (inclination)$
$\Omega \ (right \ ascension \ of \ the \ ascending \ node)$
$\omega \ (argument \ of \ perigee)$
$\nu \ (true \ anomaly)$
Solution:
Step 1: Determine magnitudes of $\bar{R}$ and $\bar{V}$
$R = \sqrt{(8228)^{2} + (389.0)^{2} + (6888)^{2}} = 10,738 \ km$
$V = \sqrt{(0.7000)^{2} + (6.600)^{2} + (0.6000)^{2}} = 6.664 \ km/s$
Step 2: Solve for the semimajor axis, $a$
$\varepsilon = \frac{V^{2}}{2} - \frac{\mu}{R}$
$\varepsilon = \frac{(6.664 \ km/s)^{2}}{2} - \frac{3.986 \times 10^{5} \ km^{3}/s^{2}}{10,738 \ km} = -14.916 \ km^{2}/s^{2}$
$\varepsilon = - \frac{\mu}{2a} \Rightarrow a = - \frac{\mu}{2 \varepsilon}$
$a = \frac{3.986 \times 10^{5} \ km^{3}/s^{2}}{2(-14.916 \ km^{2}/s^{2}} = 13360 \ km \rightarrow 1.336 \times 10^{4} \ km$
First Answer:
$a = 1.336 \times 10^{4} \ km$
Step 3: Solve for eccentricity vector, $\bar{e}$, and its magnitude, $e$
$\bar{e} = \frac{1}{\mu} \left[ \left(V^{2} - \frac{\mu}{R} \right) \bar{R} - \left( \bar{R} \cdot \bar{V} \right) \bar{V} \right]$
$\bar{R} \cdot \bar{V} = (8228)(-0.7000) + (389.0)(6.600) + (6888)(-0.600) = -7325 \ km^{2}/s$
$\bar{e} = \frac{1}{3.986 \times 10^{5} \ km^{3}/s^{2}} \left[ \left( \left( 6.664 \ m/s \right)^{2} - \frac{3.986 \times 10^{5} \ km^{3}/s^{2}}{10,738 \ km} \right) \bar{R} - \left(-7325 \ km^{2}/s \right) \bar{V} \right]$
$\bar{e} = (2.5088 \times 10^{-6})[(7.288) \bar{R} - (-7325) \bar{V}]$
$\bar{e} = (1.8248 \times 10^{-5})[8228 \hat{I} + 389.0 \hat{J} + 6888 \hat{K}] - (-0.018377)[-0.7000 \hat{I} + 6.600 \hat{J} - 0.6000 \hat{K}]$
$\bar{e} = [0.15044 \hat{I} + 0.0071125 \hat{J} + 0.12594 \hat{K}] - [0.012864 \hat{I} - 0.12129 \hat{J} + 0.011026 \hat{K}]$
$\bar{e} = 0.1376 \hat{I} + 0.1284 \hat{J} + 0.1149 \hat{K}$
$e = \sqrt{(0.1376)^{2} + (0.1284)^{2} + (0.1149)^{2}} = 0.2205$
Second Answer:
$e = 0.2205$
Step 4: Solve for specific angular momentum vector, $\bar{h}$, and its magnitude, h
$\bar{h} = \bar{R} \times \bar{V}$
$\bar{h} = [(389.0)(-0.600) - (6.600)(6888)] \hat{I} - [(8228)(-0.6000) - (-0.7000)(6888)] \hat{J} + [(8228)(6.600) - (0.700)(389.0)] \hat{K} \ km^{2}/s$
$\bar{h} = [(-233.4)-(45,460.8)] \hat{I} - [(-4936.8) + (4821.6)] \hat{J} + [(54,304.8) + (272.3)] \hat{K} \ km^{2}/s$
$\bar{h} = -45,694.2 \hat{I} + 115.2 \hat{J} + 54,577.1 \hat{K} \ km^{2}/s$
$h = \sqrt{(45,694.2)^{2} + (115.2)^{2} + (54,577.1)^{2}} = 71,180.3 \ km^{2}/s$
Step 5: Solve for the inclination angle, $i$
$i = \cos^{-1} \left(\frac{\hat{K} \cdot \bar{h}}{Kh}\right) \Rightarrow \cos^{-1} \left( \frac{\hat{K} \cdot \bar{h}}{h}\right)$
$\hat{K} \cdot \bar{h} = h_{K} = 54,577.1$
$i = \cos^{-1} \left( \frac{54,577.1}{71,180.3} \right) = \cos^{-1}(0.76674)$
$i = 39.94^{\circ} \ or \ (360^{\circ} - 39.94^{\circ}) = 320.1^{\circ}$
Resolving this ambiguity requires following the definition of inclination. Because $0^{\circ} < i < 360^{\circ}$:
$i = 39.94^{\circ}$
Third Answer:
$i = 39.94^{\circ}$
Step 6: Solve for the ascending node vector, $\bar{n}$, and its magnitude, $n$
$\bar{n} = \hat{K} \times \bar{h}$
$\bar{n} = -115.2 \hat{I} -45,694.2 \hat{J} + 0 \hat{K}$
$n = \sqrt{(115.2)^{2} + (45,694.2)^{2} + (0)^{2}} = 45,694.3 \ km^{2}/s$
Step 7: Solve for right ascension of the ascending node, $\Omega$. Do a quadrant check.
$\Omega = \cos^{-1} \left( \frac{\hat{I} \cdot \bar{n}}{In} \right) \Rightarrow \cos^{-1} \left( \frac{\hat{I} \cdot \bar{n}}{n} \right)$
$\hat{I} \cdot \bar{n} = n_{I} = -115.2$
$\Omega = \cos^{-1} \left( \frac{-115.2}{45,694.3} \right) = \cos^{-1}(-0.0025211)$
$\Omega = 90.14^{\circ} \ or \ (360^{\circ} - 90.14^{\circ}) = 269.9^{\circ}$
$If \ n_{J} \geq 0, \ then \ 0^{\circ} \leq \Omega \leq 180^{\circ}$
$If \ n_{J} < 0, \ then \ 180^{\circ} < \Omega < 360^{\circ}$
$n_{J} = -45,694.2, \ n_{J} < 0$
$\Omega = 269.9^{\circ}$
Fourth Answer:
$\Omega = 269.9^{\circ}$
Step 8: Solve for the argument of perigee, $\omega$. Do a quadrant check.
$\omega = \cos^{-1} \left( \frac{\bar{n} \cdot \bar{e}}{ne} \right)$
$\bar{n} \cdot \bar{e} = (-115.2)(0.1376) + (-45,694.2)(0.1284) + (0)(0.1149) = -5882.99$
$\omega = \cos^{-1} \left[ \frac{-5882.99}{(45694.3)(0.2205)} \right] = \cos^{-1}(-0.58389)$
$\omega = 125.7^{\circ} \ or \ (360^{\circ} - 125.7^{\circ}) = 234.3^{\circ}$
$If \ e_{K} \geq 0, \ then \ 0^{\circ} \leq \omega \leq 180^{\circ}$
$If \ e_{K} < 0, \ then \ 180^{\circ} < \omega < 360^{\circ}$
$e_{K} = 0.1149, \ e_{K} > 0$
$\omega = 125.7^{\circ}$
Fifth Answer:
$\omega = 125.7^{\circ}$
Step 9: Solve for true anomaly, $\nu$. Do a quadrant check.
$\nu = \cos^{-1} \left( \frac{\bar{e} \cdot \bar{R}}{eR} \right)$
$\bar{e} \cdot \bar{R} = (0.1376)(8228) + (0.1284)(389.0) + (0.1149)(6888) = 1974.05$
$\nu = \cos^{-1} \left[ \frac{1974.05}{(0.2205)(10,738)} \right] = \cos^{-1}(0.83373)$
$\nu = 33.52^{\circ} \ or \ (360^{\circ} - 33.52^{\circ}) = 326.48^{\circ}$
$If \ (\bar{R} \cdot \bar{V}) \geq 0, \ then \ 0^{\circ} \leq \nu \ leq 180^{\circ}$
$If \ (\bar{R} \cdot \bar{V} < 0, \ then \ 180^{\circ} < \nu < 360^{\circ}$
$(\bar{R} \cdot \bar{V}) = -7325, \ (\bar{R} \cdot \bar{V}) < 0$
$\nu = 326.5^{\circ}$
Sixth Answer:
$\nu = 326.5^{\circ}$
Whew!!!
Please note: This example is based on an example found in this (fantabulous!!—that's a word, right??) book:
Sellers, Jerry Jon, et al. Understanding Space: An Introduction to Astronautics. New York: McGraw-Hill, 1994.
\[a = - \frac{\mu}{2 \varepsilon}\]
Where:
$a = semimajor \ axis \ (km)$
$\mu = gravitational \ parameter = 3.986 \times 10^{5} \ (km^{3}/s^{2})$
$\varepsilon = spacecraft's \ specific \ mechanical \ energy \ (km^{2}/s^{2})$
\[\bar{e} = \frac{1}{ \mu } \Big[\Big(V^{2} - \frac{\mu}{R}\Big)\bar{R} - \Big(\bar{R} \cdot \bar{V}\Big)\bar{V}\Big]\]
Where:
$\bar{e} = eccentricity \ vector \ (unitless)$
$\mu = gravitational \ parameter = 3.986 \times 10^{5} \ (km^{3}/s^{2})$
$V = magnitude \ of \ \bar{V} \ (km/s)$
$R = magnitude \ of \ \bar{R} \ (km)$
$\bar{R} = position \ vector \ (km)$
$\bar{V} = velocity \ vector \ (km/s)$
\[i = \cos^{-1} \Big(\frac{\hat{K} \cdot \bar{h}}{Kh}\Big)\]
Where:
$i = inclination \ (deg \ or \ rad)$
$\hat{K} = unit \ vector \ through \ the \ North \ Pole$
$\bar{h} = specific \ angular \ momentum \ vector \ (km^{2}/s)$
$K = magnitude \ of \ \hat{K} = 1$
$h = magnitude \ of \ \bar{h} \ (km^{2}/s)$
\[\bar{n} = \hat{K} \times \bar{h}\]
Where:
$\bar{n} = ascending \ node \ vector \ (km^{2}/s)$
$\hat{K} = unit \ vector \ through \ the \ North \ Pole$
$\bar{h} = specific \ angular \ momentum \ vector \ (km^{2}/s)$
\[\Omega = \cos^{-1} \Big(\frac{\hat{I} \cdot \bar{n}}{In}\Big)\]
Where:
$\Omega = right \ ascension \ of \ the \ ascending \ node \ (deg \ or \ rad)$
$\hat{I} = unit \ vector \ in \ the \ principal \ direction$
$\bar{n} = ascending \ node \ vector \ (km^{2}/s)$
$I = magnitude \ of \ \hat{I} = 1$
$n = magnitude \ of \ \bar{n} \ (km^{2}/s)$
$If \ n_{j} \geq 0, \ then \ 0 \leq \Omega \leq 180^{\circ}$
$If \ n_{j} < 0, \then \ 180^{\circ} < \Omega < 360^{\circ}$
\[\omega = \cos^{-1} \Big(\frac{\bar{n} \cdot \bar{e}}{ne} \Big)\]
Where:
$\omega = argument \ of \ perigee \ (deg \ or \ rad)$
$\bar{n} = ascending \ node \ vector \ (km^{2}/s)$
$\bar{e} = eccentricity \ vector \ (unitless)$
$n = magnitude \ of \ \bar{n} \ (km^{2}/s)$
$e = magnitude \ of \ \bar{e} \ (unitless)$
$If \ e_{k} \geq 0, \ then \ 0^{\circ} \leq \omega \leq 180^{\circ}$
$If \ e_{k} < 0, \ then \ 180^{\circ} < \omega < 360^{\circ}$
\[\nu = \cos^{-1} \Big(\frac{\bar{e} \cdot \bar{R}}{eR} \Big)\]
Where:
$\nu = true \ anomaly \ (deg \ or \ rad)$
$\bar{e} = eccentricity \ vector \ (unitless)$
$\bar{R} = position \ vector \ (km)$
$e = magnitude \ of \ \bar{e} \ (unitless)$
$R = magnitude \ of \ \bar{R} \ (km)$
$If \ \left(\bar{R} \cdot \bar{V} \right) \geq 0 \ \left(\phi \geq 0 \right), \ then \ 0^{\circ} \leq \nu \leq 180^{\circ}$
$If \ \left(\bar{R} \cdot \bar{V} \right) < 0 \ \left(\phi < 0 \right), \ then \ 180^{\circ} < \nu < 360^{\circ}$
\[\Delta N = 360^{\circ} - longitude \ between \ successive \ ascending \ nodes\]
Where:
$\Delta N = nodal \ displacement \ (deg)$
\[P = \frac{\Delta N}{15^{\circ} /hr} \ (for \ direct \ orbits)\]
Where:
$P = period \ (hours)$
$\Delta N = nodal \ displacement \ (deg)$
\[a = \sqrt[3]{\mu (P/2\pi)^{2}}\]
Where:
$a = semimajor \ axis \ (km)$
$\mu = gravitational \ parameter \approx 3.986 \times 10^{5} \ (km^{3}/s^{2})$
$P = period \ (seconds)$
$\pi = 3.14159... \ (unitless)$
For a direct (prograde) orbit, $(0 < i < 90^{\circ})$, the latitude of the northernmost or southernmost point on the ground track (referred to as the "maximum latitude") is equal to the orbit's inclination.
For an indirect (retrograde) orbit, $(90^{\circ} < i < 180^{\circ})$, subtract the maximum latitude from $180^{\circ}$ to get the inclination.
\[ \Delta V = |V_{selected} - V_{present}|\]
Where:
$\Delta V = velocity \ change \ to \ go \ from \ one \ orbit \ to \ another \ (km/s)$
$V_{selected} = velocity \ in \ desired \ orbit \ at \ present \ orbit \ radius \ (km/s)$
$V_{present} = velocity \ in \ present \ orbit \ (km/s)$
\[2a_{transfer} = R_{orbit1} + R_{orbit2}\]
Where:
$2a_{transfer} = major \ axis \ of \ transfer \ orbit \ (km)$
$R_{orbit1} = radius \ of \ first \ orbit \ (km)$
$R_{orbit2} = radius \ of \ second \ orbit \ (km)$
\[\varepsilon_{transfer} = - \frac{\mu}{2a_{transfer}}\]
Where:
$\varepsilon_{transfer} = specific \ mechanical \ energy \ of \ tranfer \ orbit \ (km^{2}/s^{2})$
$\mu = gravitational \ parameter \approx 3.986 \times 10^{5} \ (km^{3}/s^{s}) \ for \ Earth$
$2a_{transfer} = major \ axis \ of \ transfer \ orbit$
\[TOF = \frac{P}{2} = \pi \sqrt{\frac{a_{transfer}^{3}}{\mu}}\]
Where:
$TOF = spacecraft's \ time \ of \ flight \ (s)$
$P = orbital \ period \ (s)$
$\pi = 3.14159... \ (unitless)$
$a = semimajor \ axis \ of \ transfer \ orbit \ (km)$
$\mu = gravitational \ parameter \ (km^{3}/s^{2})$
\[\Delta V_{simple} = 2V_{initial} \sin \Big(\frac{\theta}{2}\Big)\]
Where:
$\Delta V_{simple} = change \ in \ velocity \ for a \ simple \ plane \ change \ (km/s)$
$V_{initial} = V_{final} = velocities \ in \ the \ initial \ and \ final \ orbits \ (km/s)$
$\theta = plane-change \ angle \ (deg \ or \ rad)$
\[\Delta V_{combined} = \sqrt{(|\overline{V_{initial}}|)^{2} + (|\overline{V_{final}}|)^{2} - 2|\overline{V_{initial}}||\overline{V_{final}}| \cos \theta}\]
Where:
$\Delta V_{combined} = change \ in \ velocity \ for \ combined \ plane \ change \ (km/s)$
$|\overline{V_{initial}}| = magnitude \ of \ velocity \ in \ initial \ orbit \ (km/s)$
$|\overline{V_{final}}| = magnitude \ of \ velocity \ in \ final \ orbit \ (km/s)$
$\theta = plane-change \ angle \ (deg \ or \ rad)$
\[\omega = \sqrt{\frac{\mu}{a^{3}}}\]
Where:
$\omega = spacecraft's \ angular \ momentum \ (rad/s)$
$\mu = gravitational \ parameter \approx 3.986 \times 10^{5} \ for \ Earth \ (km^{3}/s^{2})$
$a = semimajor \ axis \ (km)$
\[\alpha_{lead} = \omega_{target}TOF\]
Where:
$\alpha_{lead} = angle \ by \ which \ the \ interceptor \ must \ lead \ the \ target \ (rad)$
$\omega_{target} = target \ spacecraft's \ angular \ velocity \ (rad/s)$
$TOF = time \ of \ flight \ (s)$
\[\phi_{final} = \pi - \alpha_{lead}\]
Where:
$\phi_{final} = phase \ angle \ between \ interceptor \ and \ target \ as \ transfer \ begins \ (rad)$
$\pi = 3.14159... \ (unitless)$
$\alpha_{lead} = angle \ by \ which \ the \ interceptor \ must \ lead \ the \ target \ (rad)$
\[wait \ time = \frac{\phi_{final} - \phi_{initial}}{\omega_{target} - \omega_{interceptor}}\]
Where:
$wait \ time = time \ until \ interceptor \ initiates \ rendezvous \ (s)$
$\phi_{final}, \ \phi_{initial} = final \ and \ initial \ phase \ angles \ (rad)$
$\omega_{target}, \ \omega_{interceptor} = angular \ velocities \ of \ target \ and \ interceptor \ (rad/s)$
\[a_{phasing} = \sqrt[3]{ \mu \Big(\frac{\phi_{travel}}{2 \pi \omega_{target}}\Big)^{2}} \]
Where:
$a_{phasing} = semimajor \ axis \ of \ the \ phasing \ orbit \ (km)$
$\mu = gravitational \ parameter \approx 3.986 \times 10^{5} \ (km^{3}/s^{2}) \ for \ Earth$
$\phi_{travel} = angular \ distance \ traveled \ by \ target \ to \ reach \ rendezvous \ (rad)$
$\pi = 3.14159... \ (unitless)$
$\omega_{target} = target's \ angular \ velocity \ (rad/s)$
\[R_{SOI} = a_{planet} \Big( \frac{m_{planet}}{m_{Sun}} \Big)^{2/5}\]
Where:
$R_{SOI} = radius \ of \ planet's \ sphere \ of \ influence \ (km)$
$a_{planet} = semimajor \ axis \ of \ the \ planet's \ orbit \ around \ the \ Sun \ (km)$
$m_{planet} = mass \ of \ planet \ (kg)$
$m_{Sun} = mass \ of \ the \ Sun = 1.989 \times 10^{30} \ (kg)$
\[V_{Earth} = \sqrt{2 \Big( \frac{\mu_{Sun}}{R_{to Earth}} + \varepsilon_{Earth} \Big)} \]
Where:
$V_{Earth} = Earth's \ orbital \ velocity \ with \ respect \ to \ the \ Sun \ (km/s)$
$\mu_{Sun} = gravitational \ parameter \ of \ the \ Sun \approx 1.327 \times 10^{11} \ (km^{3}/s^{2})$
$R_{to Earth} = distance \ from \ Sun \ to \ Earth \approx 1.496 \times 10^{8} \ (km)$
$\varepsilon_{Earth} = specific \ mechanical \ energy \ of \ Earth's \ orbit \ (km^{2}/s^{2})$
\[\varepsilon_{transfer} = - \frac{\mu_{Sun}}{2a_{transfer}}\]
Where:
$\varepsilon_{transfer} = spacecraft's \ specific \ mechanical \ energy \ in \ heliocentric \ transfer \ orbit \ (km^{2}/s^{2})$
$\mu_{Sun} = gravitational \ parameter \ of \ the \ Sun \approx 1.327 \times 10^{11} \ (km^{3}/s^{2})$
$a_{transfer} = semimajor \ axis \ of \ the \ transfer \ orbit \ (km)$
\[a_{transfer} = \frac{R_{to Earth} + R_{to Target}}{2}\]
Where:
$a_{transfer} = semimajor \ axis \ of \ the \ transfer \ orbit \ (km)$
$R_{to Earth} = radius \ from \ the \ Sun \ to Earth \approx 1.496 \times 10^{8} \ (km)$
$R_{to Target} = radius \ from \ the \ Sun \ to \ the \ target \ planet \ (km)$
\[V_{transfer at Earth} = \sqrt{2 \Big( \frac{\mu_{Sun}}{R_{toEarth}} + \varepsilon_{transfer}\Big)}\]
Where:
$V_{transfer at Earth} = velocity \ spacecraft \ needs \ at \ Earth's \ radius \ from \ Sun \ to \ transfer \ to \ target \ (km/s)$
$\mu_{Sun} = gravitational \ parameter \ of \ the \ Sun \approx 1.327 \times 10^{11} \ (km^{3}/s^{2})$
$R_{to Earth} = radius \ from \ the \ Sun \ to Earth \approx 1.496 \times 10^{8} \ (km)$
$\varepsilon_{transfer} = spacecraft's \ specific \ mechanical \ energy \ in \ heliocentric \ transfer \ orbit \ (km^{2}/s^{2})$
\[V_{ \infty Earth} = |V_{transfer at Earth} - V_{Earth}|\]
Where:
$V_{ \infty Earth} = spacecraft's \ velocity \ at \ infinity \ with \ respect \ to \ Earth \ (km/s)$
$V_{transfer at Earth} = velocity \ spacecraft \ needs \ at \ Earth's \ radius \ from \ Sun \ to \ transfer \ to \ target \ (km/s)$
$V_{Earth} = Earth's \ orbital \ velocity \ with \ respect \ to \ the \ Sun \ (km/s)$
\[V_{transfer at target} = \sqrt{2 \Big( \frac{\mu_{Sun}}{R_{to target}} + \varepsilon_{transfer} \Big)}\]
Where
$V_{transfer at target} = spacecraft's \ velocity \ on \ the \ transfer \ orbit \ just \ outside \ the \target's \ SOI \ (km/s)$
$\mu_{Sun} \approx 1.327 \times 10^{11} \ (km^{3}/s^{2})$
$R_{to target} = distance \ from \ the \ Sun \ to \ the \ target \ planet \ (km)$
$\varepsilon_{transfer} = specific \ mechanical \ energy \ of \ the \ transfer \ orbit \ (km^{2}/s^{2})$
\[\varepsilon_{target} = - \frac{\mu_{Sun}}{2a_{target}}\]
Where:
$\varepsilon_{target} = target \ planet's \ specific \ mechanical \ energy \ with \ respect \ to \ the \ Sun \ (km^{2}/s^{2})$
$\mu_{Sun} \approx 1.327 \times 10^{11} \ (km^{3}/s^{2})$
$a_{target} = target \ planet's \ semimajor \ axis \ (km)$
\[V_{target} = \sqrt{2 \Big( \frac{\mu_{Sun}}{R_{to target}} + \varepsilon_{target} \Big)}\]
Where:
$V_{target} = target \ planet's \ velocity \ around \ the \ Sun \ (km/s)$
$\mu_{Sun} \approx 1.327 \times 10^{11} \ (km^{3}/s^{2})$
$R_{totarget} = distance \ from \ the \ Sun \ to \ the \ target \ planet \ (km)$
$\varepsilon_{target} = target \ planet's \ specific \ mechanical \ energy \ (km^{3}/s^{2})$
\[V_{ \infty target} = |V_{transfer at target} - V_{target}|\]
Where:
$V_{ \infty target} = spacecraft's \ velocity \ at \ infinity \ with \ respect \ to \ target \ (km/s)$
$V_{transfer at target} = velocity \ spacecraft \ has \ just \ outside \ target \ planet's \ SOI \ (km/s)$
$V_{target} = target's \ orbital \ velocity \ with \ respect \ to \ the \ Sun \ (km/s)$
\[n = \frac{angle}{time} = \frac{2 \pi}{P} = \sqrt{\frac{\mu}{a^{3}}}\]
Where:
$n = spacecraft's \ mean \ motion \ (rad/s)$
$\pi = 3.14159... \ (unitless)$
$P = orbital \ period \ (s)$
$\mu = central \ body's \ gravitational \ parameter \approx 3.986 \times 10^{5} \ (km^{3}/s^{2}) \ for \ Earth$
$a = semimajor \ axis \ (km)$
\[TOF = \frac{u_{future} - u_{initial}}{n}\]
Where:
$TOF = time \ of \ flight$
$u_{future} = spacecraft's \ future \ argument \ of \ latitude \ (rad)$
$u_{initial} \ spacecraft's \ initial \ argument \ of \ latitude \ (rad)$
$n = spacecraft's \ mean \ motion \ (rad/s)$
\[M = nT\]
Where:
$M = mean \ anomaly \ (rad)$
$n = mean \ motion \ (rad/s)$
$T = time \ since \ last \ perigee \ passage \ (s)$
\[M_{future} - M_{initial} = n \left( t_{future} - t_{initial} \right) - 2k \pi \]
Where:
$M_{future} = mean \ anomaly \ when \ spacecraft \ is \ in \ the \ future \ position \ (rad)$
$M_{initial} = mean \ anomaly \ of \ spacecraft's \ initial \ position \ (rad)$
$t_{future} - t_{initial} = TOF \ between \ two \ points \ in \ the \ orbit \ (s)$
$k = the \ number \ of \ times \ the \ spacecraft \ passes \ perigee \ during \ the \ TOF$
$\pi = 3.14159... \ (unitless)$
\[M = E - e \sin E\]
Where:
$M = mean \ anomaly \ (rad)$
$E = eccentric \ anomaly \ (rad)$
$e = eccentricity \ (unitless)$
\[\cos E = \frac{e + \cos \nu}{1 + e \cos \nu}\]
Where:
$E = eccentric \ anomaly \ (rad)$
$e = eccentricy \ (unitless)$
$\nu = true \ anomaly \ (rad)$
\[\cos \nu = \frac{\cos E - e}{1 - e \cos E}\]
Where:
$\nu = true \ anomaly \ (rad)$
$E = eccentric \ anomaly \ (rad)$
$e = eccentricy \ (unitless)$
The desired orbital inclination must be equal to, or greater than, the launch site's latitude, $L_{0}$.
A launch window can only exist for the following conditions:
\[L_{0} \leq i \ (direct \ (prograde) \ orbits)\]
\[L_{0} \leq 180^{\circ} - i \ (indirect \ (retrograde) \ orbits)\]
\[ \sin \gamma = \frac{ \cos \alpha}{ \cos L_{0}}\]
Where:
$\gamma = launch-direction \ auxiliary \ angle \ (deg \ or \ rad)$
$\alpha = inclination \ auxiliary \ angle \ (deg \ or \ rad)$
$L_{0} = launch \ site \ latitude \ (deg \ or \ rad)$
\[ \cos \delta = \frac{ \cos \gamma}{ \sin \alpha}\]
Where:
$\delta = launch-window \ location \ angle \ (deg \ or \ rad)$
$\gamma = launch-direction \ auxiliary \ angle \ (deg \ or \ rad)$
$\alpha = inclination \ auxiliary \ angle \ (deg \ or \ rad)$
\[LWST_{AN} = \Omega + \delta \]
Where:
$LWST_{AN} = launch-window \ sidereal \ time \ at \ ascending \ node$
$\Omega = right \ ascension \ of \ the \ ascending \ node \ (deg \ or \ rad)$
$\delta = launch-window \ location \ angle \ (deg \ or \ rad)$
\[LWST_{DN} = \Omega + \left( 180^{\circ} - \delta \right) \]
Where:
$LWST_{DN} = launch-window \ sidereal \ time \ at \ descending \ node$
$\Omega = right \ ascension \ of \ the \ ascending \ node \ (deg \ or \ rad)$
$\delta = launch-window \ location \ angle \ (deg \ or \ rad)$
\[\bar{q} = \frac{\rho V^{2}}{2}\]
Where:
$\bar{q} = dynamic \ pressure \ on \ the \ spacecraft \ (N/m^{2})$
$\rho = atmospheric \ density \ (kg/m^{3})$
$V = spacecraft's \ velocity \ (m/s)$
\[F_{drag} = \bar{q} C_{D} A = \frac{1}{2} \rho V^{2} C_{D} A \]
Where:
$F_{drag} = drag \ force \ on \ spacecraft \ (N)$
$\bar{q} = dynamic \ pressure \ on \ the \ spacecraft \ (N/m^{2})$
$C_{D} = drag \ coefficient \ (unitless)$
$A = spacecraft's \ cross-sectional \ area \ (m^{2})$
$\rho = atmospheric \ density \ (kg/m^{3})$
$V = spacecraft's \ velocity \ (m/s)$
\[\bar{a} = \Big( - \bar{q} \frac{C_{D}A}{m} \cos \gamma \Big) \hat{X} + \Big( \bar{q} \frac{C_{D}A}{m} \sin \gamma \Big) \hat{Z} \]
Where:
$\bar{a} = spacecraft's \ acceleration \ (m/s^{2})$
$\bar{q} = dynamic \ pressure \ on \ the \ spacecraft \ (N/m^{2})$
$C_{D} = drag \ coefficient \ (unitless)$
$A = spacecraft's \ cross-sectional \ area \ (m^{2})$
$m = spacecraft's \ mass \ (kg)$
$\gamma = spacecraft's \ flight-path \ angle \ (deg)$
\[BC = \frac{m}{C_{D}A}\]
Where:
$BC = spacecraft's \ ballistic \ coefficient \ (kg/m^{2})$
$m = spacecraft's \ mass \ (kg)$
$C_{D} = drag \ coefficient \ (unitless)$
$A = spacecraft's \ cross-sectional \ area \ (m^{2})$
\[a_{max} = \frac{V_{re-entry}^{2} \beta \sin \gamma}{2e}\]
Where:
$a_{max} = spacecraft's \ maximum \ deceleration \ (m/s^{2})$
$V_{re-entry} = spacecraft's \ re-entry \ velocity \ (m/s)$
$\beta = atmospheric \ scale \ height = 0.000139 \ (m^{-1}) \ for \ Earth$
$\gamma = flight \ path \ angle \ (deg)$
$e = base \ of \ the \ natural \ logarithm = 2.7182...$
\[h_{max} = \frac{1}{\beta} \ln \Big( \frac{\rho_{0}}{BC \beta \sin \gamma} \Big) \]
Where:
$h_{max} = altitude \ of \ spacecraft's \ maximum \ acceleration \ (m)$
$\beta = atmospheric \ scale \ height = 0.000139 \ (m^{-1}) \ for \ Earth$
$\rho_{0} = atmospheric \ density \ at \ sea \ level = 1.225 \ (kg/m^{3})$
$BC = spacecraft's \ ballistic \ coefficient \ (kg/m^{2})$
$\gamma = flight \ path \ angle \ (deg)$
\[\dot{q} \cong 1.83 \times 10^{-4} V^{3} \sqrt{ \frac{ \rho}{r_{nose}}}\]
Where:
$\dot{q} = spacecraft's \ heating \ rate \ (W/m^{2})$
$V = spacecraft's \ velocity \ (m/s)$
$\rho = air \ density \ (km/m^{3})$
$r_{nose} = spacecraft's \ nose \ radius \ (m)$
\[h_{\dot{q}max} = \frac{1}{\beta} \ln \Big( \frac{\rho_{0}}{3BC \beta \sin \gamma} \Big) \]
Where:
$h_{\dot{q}max} = altitude \ of \ spacecraft's \ maximum \ heating \ rate \ (m)$
$\beta = atmospheric \ scale \ height = 0.000139 \ (m^{-1}) \ for \ Earth$
$\rho_{0} = atmospheric \ density \ at \ sea \ level = 1.225 \ (kg/m^{3})$
$BC = spacecraft's \ ballistic \ coefficient \ (kg/m^{2})$
$\gamma = flight \ path \ angle \ (deg)$
\[V_{\dot{q} max} \approx 0.846 V_{re-entry}\]
Where:
$V_{\dot{q} max} = spacecraft's \ velocity \ when \ it \ reaches \ maximum \ heating \ rate \ (m/s)$
$V_{re-entry} = spacecraft's \ velocity \ at \ re-entry \ (m/s)$
\[q_{A} = \sigma \varepsilon T^{4}\]
Where:
$q_{A} = object's \ emitted \ power \ per \ unit \ area \ (W/m^{2})$
$\sigma = Boltzmann \ constant = 5.67 \times 10^{-8} \ (W/m^{2}K^{4})$
$\varepsilon = object's \ emissivity \ (0 < \varepsilon < 1) \ (unitless)$
$T = object's \ temperature \ (K)$