Orbital Mechanics Math, Step by Step

Circular orbits, period and escape speed

• Teens
• Community college
• Undergraduate (freshman–senior)

Track: Orbital Mechanics · About 15 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\frac{GMm}{r^2}=\frac{mv^2}{r}$$
   Gravity = mass × centripetal acceleration.
2. $$v_c=\sqrt{\frac{\mu}{r}}$$
   Cancel m and solve for v.
3. $$T=\frac{2\pi r}{v_c}=2\pi\sqrt{\frac{r^3}{\mu}}$$
   Period = circumference / speed.
4. $$\tfrac12 v_{esc}^2-\frac{\mu}{r}=0\;\Rightarrow\; v_{esc}=\sqrt{\frac{2\mu}{r}}=\sqrt2\,v_c$$
   Escape: total energy just reaches zero.

import math
mu, RE = 398600.4418, 6378.137   # km^3/s^2, km
r = RE + 400
v = math.sqrt(mu / r)
T = 2 * math.pi * r / v
print(v, T / 60)   # km/s, min

Auto-checked practice

1. Circular speed at 800 km altitude, in km/s?

   km/s Check Show a hint
   Full solution

   r = 7178.137 km; v = √(398600.4418/7178.137) = 7.4518 km/s.

2. Period of that 800 km orbit, in minutes?

   min Check Show a hint
   Full solution

   T = 2π√(7178.137³/398600.4418) s = 100.87 min.

Practice

1. Find the radius of an orbit whose period is one sidereal day (86 164.09 s).

   Show solution

   r = (μ T²/4π²)^(1/3) = 42164 km (geostationary), where v_c = 3.075 km/s.

References

• NASA, Basics of Space Flight.
• H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann (Elsevier), 2020.

Kepler’s three laws

• Teens
• Community college
• Undergraduate (freshman–senior)

Track: Orbital Mechanics · About 15 min · Numbers verified in build: 2026-10-16

Step by step

1. $$dA=\tfrac12\,|\mathbf r\times d\mathbf r|\;\Rightarrow\;\frac{dA}{dt}=\tfrac12|\mathbf r\times\mathbf v|=\frac{h}{2}$$
   Area swept per unit time is half the specific angular momentum h.
2. $$\dot{\mathbf h}=\dot{\mathbf r}\times\mathbf v+\mathbf r\times\left(-\frac{\mu}{r^3}\mathbf r\right)=\mathbf 0$$
   For a central force h is constant, so dA/dt is constant (2nd law).
3. $$T=\frac{\pi ab}{h/2},\quad h=\sqrt{\mu a(1-e^2)},\quad b=a\sqrt{1-e^2}$$
   Total area of the ellipse divided by the areal rate.
4. $$T=2\pi\sqrt{\frac{a^3}{\mu}}$$
   Third law: depends only on a, not on e.

import math
mu, RE = 398600.4418, 6378.137   # km^3/s^2, km
a = 42164.17
T = 2 * math.pi * math.sqrt(a**3 / mu)
print(T / 3600)   # h

Practice

1. Two orbits have the same a but e = 0 and e = 0.7. Which has the longer period?

   Show solution

   Neither: T depends only on a.

References

• H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann (Elsevier), 2020.
• R. R. Bate, D. D. Mueller, J. E. White, W. W. Saylor, Fundamentals of Astrodynamics, 2nd ed., Dover, 2020.

The two-body problem and the orbit equation

• Undergraduate (freshman–senior)

Track: Orbital Mechanics · About 30 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\ddot{\mathbf r}=-\frac{\mu}{r^3}\mathbf r,\qquad \mu=G(m_1+m_2)$$
   Relative equation of motion.
2. $$\ddot{\mathbf r}\times\mathbf h=-\frac{\mu}{r^3}\,\mathbf r\times(\mathbf r\times\dot{\mathbf r})=\mu\frac{d}{dt}\!\left(\frac{\mathbf r}{r}\right)$$
   Cross with constant h and use BAC-CAB.
3. $$\dot{\mathbf r}\times\mathbf h-\mu\frac{\mathbf r}{r}=\mu\,\mathbf e$$
   Integrate once: e is a constant vector pointing to periapsis.
4. $$\mathbf r\cdot(\dot{\mathbf r}\times\mathbf h)=h^2=\mu r+\mu r e\cos\nu\;\Rightarrow\; r=\frac{h^2/\mu}{1+e\cos\nu}$$
   Dot with r: the conic equation.

import numpy as np
mu = 398600.4418
r = np.array([7000., 1000., 500.]); v = np.array([-1.0, 7.4, 1.0])
h = np.cross(r, v)
e = np.cross(v, h) / mu - r / np.linalg.norm(r)
eps = v @ v / 2 - mu / np.linalg.norm(r)
print(np.linalg.norm(h), np.linalg.norm(e), -mu / (2 * eps))

Practice

1. Show that e < 1 gives a closed orbit and e > 1 an open one.

   Show solution

   The denominator 1 + e cosν stays positive for every ν only if e < 1; for e > 1 it vanishes at cosν = -1/e, so r → ∞.

References

• H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann (Elsevier), 2020.
• D. A. Vallado, Fundamentals of Astrodynamics and Applications, 4th ed., Microcosm Press, 2013.
• R. R. Bate, D. D. Mueller, J. E. White, W. W. Saylor, Fundamentals of Astrodynamics, 2nd ed., Dover, 2020.

Energy and the vis-viva equation

• Community college
• Undergraduate (freshman–senior)

Track: Orbital Mechanics · About 20 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\dot{\mathbf r}\cdot\ddot{\mathbf r}+\frac{\mu}{r^3}\,\mathbf r\cdot\dot{\mathbf r}=0\;\Rightarrow\;\frac{d}{dt}\!\left(\frac{v^2}{2}-\frac{\mu}{r}\right)=0$$
   Dot the equation of motion with velocity.
2. $$\varepsilon=\frac{v_p^2}{2}-\frac{\mu}{r_p},\quad r_p=a(1-e),\quad v_p=\frac{h}{r_p}$$
   Evaluate at periapsis, where v is perpendicular to r.
3. $$\varepsilon=-\frac{\mu}{2a}$$
   Simplify with h² = μa(1-e²).
4. $$v^2=\mu\left(\frac{2}{r}-\frac{1}{a}\right)$$
   Vis-viva.

import math
mu, RE = 398600.4418, 6378.137   # km^3/s^2, km
rp, ra = RE + 300, 42164.17
a = (rp + ra) / 2
vv = lambda r: math.sqrt(mu * (2/r - 1/a))
print(vv(rp), vv(ra), rp*vv(rp), ra*vv(ra))

Auto-checked practice

1. An orbit has a = 10 000 km. Speed when r = 8000 km, in km/s?

   km/s Check Show a hint
   Full solution

   v = √(398600.4418 × (2/8000 − 1/10000)) = 7.7324 km/s.

Practice

1. What is the sign of ε for an escape (parabolic) trajectory?

   Show solution

   Zero (a → ∞). Bound orbits have ε < 0, hyperbolas ε > 0.

References

• H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann (Elsevier), 2020.
• R. R. Bate, D. D. Mueller, J. E. White, W. W. Saylor, Fundamentals of Astrodynamics, 2nd ed., Dover, 2020.

Orbital elements from a state vector

• Undergraduate (freshman–senior)
• Graduate

Track: Orbital Mechanics · About 30 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\mathbf h=\mathbf r\times\mathbf v,\quad \mathbf n=\hat{\mathbf k}\times\mathbf h,\quad \mathbf e=\frac{\mathbf v\times\mathbf h}{\mu}-\frac{\mathbf r}{r}$$
   Three key vectors.
2. $$a=-\frac{\mu}{2\varepsilon},\quad \cos i=\frac{h_z}{h},\quad \cos\Omega=\frac{n_x}{n}\;(n_y<0\Rightarrow\Omega=360^\circ-\Omega)$$
   Size, tilt, node (with quadrant check).
3. $$\cos\omega=\frac{\mathbf n\cdot\mathbf e}{ne}\;(e_z<0\Rightarrow 360^\circ-\omega),\quad \cos\nu=\frac{\mathbf e\cdot\mathbf r}{er}\;(\mathbf r\cdot\mathbf v<0\Rightarrow 360^\circ-\nu)$$
   Periapsis and current position.

import numpy as np
mu = 398600.4418
r = np.array([7000., 1000., 500.]); v = np.array([-1.0, 7.4, 1.0])
h = np.cross(r, v); n = np.cross([0, 0, 1], h)
e = np.cross(v, h) / mu - r / np.linalg.norm(r)
i = np.degrees(np.arccos(h[2] / np.linalg.norm(h)))
W = np.degrees(np.arccos(n[0] / np.linalg.norm(n))); W = 360 - W if n[1] < 0 else W
print(i, W)

Practice

1. Why are Ω and ω undefined for an equatorial circular orbit?

   Show solution

   n = 0 (no node line) and e = 0 (no periapsis); use true longitude instead.

References

• D. A. Vallado, Fundamentals of Astrodynamics and Applications, 4th ed., Microcosm Press, 2013.
• H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann (Elsevier), 2020.

Kepler’s equation and Newton–Raphson

• Undergraduate (freshman–senior)
• Graduate

Track: Orbital Mechanics · About 30 min · Numbers verified in build: 2026-10-16

Step by step

1. $$M=n(t-t_p),\qquad n=\sqrt{\mu/a^3}$$
   Mean anomaly grows uniformly in time.
2. $$M=E-e\sin E$$
   Kepler’s equation.
3. $$E_{k+1}=E_k-\frac{E_k-e\sin E_k-M}{1-e\cos E_k}$$
   Newton–Raphson: f/f′.
4. $$\tan\frac{\nu}{2}=\sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2}$$
   Convert to true anomaly.

import math
def kepler_E(M, e, tol=1e-10):
    E = math.pi if e > 0.8 else M
    while True:
        dE = (E - e*math.sin(E) - M) / (1 - e*math.cos(E))
        E -= dE
        print(E, abs(dE))
        if abs(dE) < tol:
            return E
kepler_E(3.6029, 0.37255)

Auto-checked practice

1. Solve M = E − e sin E for E (radians) with M = 1 rad, e = 0.2.

   rad Check Show a hint
   Full solution

   Newton–Raphson converges in a few steps to E = 1.185324 rad.

Practice

1. Why must M and E be in radians inside the iteration?

   Show solution

   Because the E term and the e sin E term must share units; e sin E is dimensionless (radians).

References

• D. A. Vallado, Fundamentals of Astrodynamics and Applications, 4th ed., Microcosm Press, 2013.
• R. H. Battin, An Introduction to the Mathematics and Methods of Astrodynamics, rev. ed., AIAA, 1999.

Numerical propagation with RK4

• Undergraduate (freshman–senior)
• Graduate

Track: Orbital Mechanics · About 35 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\mathbf y=\begin{bmatrix}\mathbf r\\ \mathbf v\end{bmatrix},\qquad \dot{\mathbf y}=\mathbf f(\mathbf y)=\begin{bmatrix}\mathbf v\\ -\mu\mathbf r/r^3\end{bmatrix}$$
   Second-order ODE becomes a 6-state first-order system.
2. $$\mathbf k_1=\mathbf f(\mathbf y_n),\ \mathbf k_2=\mathbf f(\mathbf y_n+\tfrac{\Delta t}{2}\mathbf k_1),\ \mathbf k_3=\mathbf f(\mathbf y_n+\tfrac{\Delta t}{2}\mathbf k_2),\ \mathbf k_4=\mathbf f(\mathbf y_n+\Delta t\,\mathbf k_3)$$
   Four slope samples.
3. $$\mathbf y_{n+1}=\mathbf y_n+\frac{\Delta t}{6}\left(\mathbf k_1+2\mathbf k_2+2\mathbf k_3+\mathbf k_4\right)$$
   Weighted average (Simpson-like).

import numpy as np
mu = 398600.4418
def f(y): r = y[:3]; return np.r_[y[3:], -mu * r / np.linalg.norm(r)**3]
def rk4(y, dt):
    k1 = f(y); k2 = f(y + dt/2*k1); k3 = f(y + dt/2*k2); k4 = f(y + dt*k3)
    return y + dt/6*(k1 + 2*k2 + 2*k3 + k4)

Practice

1. Halving Δt reduces RK4 global error by roughly what factor?

   Show solution

   About 2⁴ = 16.

References

• D. A. Vallado, Fundamentals of Astrodynamics and Applications, 4th ed., Microcosm Press, 2013.
• MIT OpenCourseWare, 16.346 Astrodynamics (CC BY-NC-SA 4.0).

Hohmann transfer

• Community college
• Undergraduate (freshman–senior)

Track: Orbital Mechanics · About 25 min · Numbers verified in build: 2026-10-16

Step by step

1. $$a_t=\frac{r_1+r_2}{2}$$
   Transfer ellipse touches both circles.
2. $$\Delta v_1=\sqrt{\mu\left(\frac{2}{r_1}-\frac{1}{a_t}\right)}-\sqrt{\frac{\mu}{r_1}}$$
   Perigee burn (vis-viva minus circular).
3. $$\Delta v_2=\sqrt{\frac{\mu}{r_2}}-\sqrt{\mu\left(\frac{2}{r_2}-\frac{1}{a_t}\right)}$$
   Apogee burn.
4. $$t_{t}=\pi\sqrt{\frac{a_t^3}{\mu}}$$
   Half the transfer-orbit period.

import math
mu, RE = 398600.4418, 6378.137          # km^3/s^2, km
r1, r2 = RE + 300, 42164.17
at = (r1 + r2) / 2
vv = lambda r, a: math.sqrt(mu * (2/r - 1/a))
dv1 = vv(r1, at) - math.sqrt(mu/r1)
dv2 = math.sqrt(mu/r2) - vv(r2, at)
tof = math.pi * math.sqrt(at**3 / mu)
print(dv1, dv2, dv1 + dv2, tof/3600)   # km/s, km/s, km/s, h

Auto-checked practice

1. Total Hohmann Δv from a 400 km to a 1000 km circular orbit, in km/s?

   km/s Check Show a hint
   Full solution

   Δv₁ = 0.1608, Δv₂ = 0.1575; total 0.3183 km/s.

Practice

1. Why do real GEO missions need more than this total?

   Show solution

   Launch sites are not on the equator, so a plane change is also needed (see Plane changes).

References

• H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann (Elsevier), 2020.
• R. R. Bate, D. D. Mueller, J. E. White, W. W. Saylor, Fundamentals of Astrodynamics, 2nd ed., Dover, 2020.
• NASA, Basics of Space Flight.

Bi-elliptic transfer

• Undergraduate (freshman–senior)
• Graduate

Track: Orbital Mechanics · About 20 min · Numbers verified in build: 2026-10-16

Step by step

1. $$a_1=\frac{r_1+r_b}{2},\qquad a_2=\frac{r_2+r_b}{2}$$
   Two transfer ellipses.
2. $$\Delta v=\big|v_{p1}-v_{c1}\big|+\big|v_{a2}-v_{a1}\big|+\big|v_{c2}-v_{p2}\big|$$
   Burn at r₁, at r_b, and at r₂, each from vis-viva.

import math
mu, RE = 398600.4418, 6378.137   # km^3/s^2, km
vv = lambda r, a: math.sqrt(mu * (2/r - 1/a)); vc = lambda r: math.sqrt(mu / r)
r1 = RE + 300; r2, rb = 20*r1, 40*r1
a1, a2 = (r1+rb)/2, (r2+rb)/2
dv = (vv(r1,a1)-vc(r1)) + (vv(rb,a2)-vv(rb,a1)) + (vv(r2,a2)-vc(r2))
print(abs(dv))

Practice

1. For LEO to GEO (r₂/r₁ ≈ 6.3), should you use a bi-elliptic transfer?

   Show solution

   No: below a ratio of about 11.94 Hohmann is always cheaper.

References

• D. A. Vallado, Fundamentals of Astrodynamics and Applications, 4th ed., Microcosm Press, 2013.
• H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann (Elsevier), 2020.

Plane changes

• Undergraduate (freshman–senior)

Track: Orbital Mechanics · About 20 min · Numbers verified in build: 2026-10-16

Step by step

1. $$\Delta v=\sqrt{v_1^2+v_2^2-2v_1v_2\cos\Delta i}$$
   Law of cosines on the two velocity vectors.
2. $$v_1=v_2=v\;\Rightarrow\;\Delta v=2v\sin\frac{\Delta i}{2}$$
   Pure plane change at constant speed.

import math
va, vg, di = 1.6078, 3.0747, math.radians(28.5)
comb = math.sqrt(va**2 + vg**2 - 2*va*vg*math.cos(di))
sep = (vg - va) + 2*vg*math.sin(di/2)
print(comb, sep)

Auto-checked practice

1. A pure 10° plane change at 7.5 km/s costs how much Δv, in km/s?

   km/s Check Show a hint
   Full solution

   Δv = 2 × 7.5 × sin 5° = 1.3073 km/s.

Practice

1. Where on an elliptical orbit is a plane change cheapest?

   Show solution

   At apoapsis (lowest speed), and it must be done at a node.

References

• H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann (Elsevier), 2020.
• D. A. Vallado, Fundamentals of Astrodynamics and Applications, 4th ed., Microcosm Press, 2013.

The rocket equation

• Teens
• Community college
• Undergraduate (freshman–senior)

Track: Orbital Mechanics · About 15 min · Numbers verified in build: 2026-10-16

Step by step

1. $$m\,dv=-v_e\,dm$$
   Momentum balance over a short burn (no gravity or drag).
2. $$\int_{0}^{\Delta v}dv=-v_e\int_{m_0}^{m_f}\frac{dm}{m}$$
   Separate variables and integrate.
3. $$\Delta v=v_e\ln\frac{m_0}{m_f},\qquad v_e=I_{sp}\,g_0$$
   Tsiolkovsky; g₀ = 9.80665 m/s².

import math
dv, isp, g0 = 3.8926, 450, 9.80665e-3   # km/s, s, km/s^2
mr = math.exp(dv / (isp * g0))
print(mr, 1 - 1/mr)

Auto-checked practice

1. Mass ratio m₀/m_f for Δv = 2 km/s with I_sp = 300 s?

   Check Show a hint
   Full solution

   m₀/m_f = exp(2/2.942) = 1.974.

Practice

1. Doubling Δv at fixed I_sp does what to the mass ratio?

   Show solution

   It is squared, not doubled: e^(2x) = (e^x)².

References

• H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann (Elsevier), 2020.
• NASA, Basics of Space Flight.

J2 perturbations and sun-synchronous orbits

• Graduate

Track: Orbital Mechanics · About 40 min · Numbers verified in build: 2026-10-16

Step by step

1. $$U=-\frac{\mu}{r}\left[1-J_2\left(\frac{R_\oplus}{r}\right)^2\frac{3\sin^2\phi-1}{2}\right]$$
   Potential with the J2 zonal term (φ = latitude).
2. $$\dot\Omega=-\frac{3}{2}\,n\,J_2\left(\frac{R_\oplus}{p}\right)^2\cos i,\qquad p=a(1-e^2)$$
   Secular node rate after orbit-averaging (Lagrange equations).
3. $$\dot\Omega_{SSO}=\frac{360^\circ}{365.2422\ \text{d}}\approx 0.9856^\circ/\text{d}\;\Rightarrow\;\cos i=\frac{\dot\Omega_{SSO}}{-\tfrac32 nJ_2(R_\oplus/p)^2}$$
   Sun-synchronous: the node follows the Sun once a year.

import math
mu, RE = 398600.4418, 6378.137   # km^3/s^2, km
J2 = 1.08262668e-3
a = RE + 700; n = math.sqrt(mu / a**3)
target = 2*math.pi / (365.2422*86400)
print(math.degrees(math.acos(-target / (1.5*n*J2*(RE/a)**2))))

Auto-checked practice

1. Inclination (degrees) of a circular sun-synchronous orbit at 800 km?

   ° Check Show a hint
   Full solution

   cos i is negative and i = 98.60°, slightly higher than at 700 km.

Practice

1. Why are sun-synchronous orbits slightly retrograde (i > 90°)?

   Show solution

   The node must move east (positive rate); the formula has -cos i, so cos i < 0.

References

• D. A. Vallado, Fundamentals of Astrodynamics and Applications, 4th ed., Microcosm Press, 2013.
• N. K. Pavlis et al., "The development and evaluation of the Earth Gravitational Model 2008 (EGM2008)", J. Geophys. Res. 117, B04406 (2012). doi:10.1029/2011JB008916
• R. H. Battin, An Introduction to the Mathematics and Methods of Astrodynamics, rev. ed., AIAA, 1999.

Lambert’s problem

• Graduate
• PhD / research

Track: Orbital Mechanics · About 60 min · Numbers verified in build: 2026-10-16

Step by step

1. $$A=\sin\Delta\nu\sqrt{\frac{r_1r_2}{1-\cos\Delta\nu}},\qquad y(z)=r_1+r_2+A\,\frac{zS(z)-1}{\sqrt{C(z)}}$$
   Geometry constant A and auxiliary y(z) (Stumpff C, S).
2. $$\sqrt{\mu}\,\Delta t=\left(\frac{y}{C}\right)^{3/2}S+A\sqrt{y}$$
   Time-of-flight equation: solve for z by root finding.
3. $$f=1-\frac{y}{r_1},\quad g=A\sqrt{\frac{y}{\mu}},\quad \dot g=1-\frac{y}{r_2},\quad \mathbf v_1=\frac{\mathbf r_2-f\mathbf r_1}{g},\quad \mathbf v_2=\frac{\dot g\,\mathbf r_2-\mathbf r_1}{g}$$
   Lagrange coefficients give the velocities.

# universal-variable Lambert (Curtis Alg. 5.2): see scripts/learn_calc.py lambert()
from learn_calc import lambert, propagate
v1, v2 = lambert([7000, 0, 0], [0, 8000, 0], 3600)
print(v1, v2)

Practice

1. Research: what does Izzo (2015) change compared with the universal-variable method?

   Show solution

   A single non-dimensional variable x with a good initial guess and Householder (third-order) iterations, handling multi-revolution cases robustly in very few iterations.

References

• D. Izzo, "Revisiting Lambert’s problem", Celestial Mechanics and Dynamical Astronomy 121, 1–15 (2015). doi:10.1007/s10569-014-9587-y
• R. H. Battin, An Introduction to the Mathematics and Methods of Astrodynamics, rev. ed., AIAA, 1999.
• D. A. Vallado, Fundamentals of Astrodynamics and Applications, 4th ed., Microcosm Press, 2013.
• H. D. Curtis, Orbital Mechanics for Engineering Students, 4th ed., Butterworth-Heinemann (Elsevier), 2020.

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