Luna Vector — 3D Mass Driver Flight Simulator

SIM TIME—2.34 km/s

Scroll or double-click to zoom · Earth-centred

TRAJECTORY SPEED · KM/S

1.076.0611.04

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01 · LIVE

EQUATIONS OF MOTION

\ddot{\mathbf r}=-\mu_E\frac{\mathbf r}{r^3}+\mathbf a_M+\mathbf a_\odot+\mathbf a_{J_2}

r=388882\,\mathrm{km}\quad |\mathbf a_E|=0.0026\,\mathrm{m\,s^{-2}}\quad |\mathbf a_M|=1.62e+0\,\mathrm{m\,s^{-2}}

Earth’s pull rises sharply as range falls. The lunar value is differential gravity: the Moon’s pull on the payload minus its pull on Earth.

TERM GUIDE

$\mathbf r$: Payload position measured from Earth’s centre.
$\ddot{\mathbf r}$: Total acceleration: how the velocity vector changes.
$\mu_E$: Earth’s gravitational parameter, GM.
$\mathbf a_M$: Moon’s differential gravitational acceleration.
$\mathbf a_\odot$: Differential acceleration caused by the Sun.
$\mathbf a_{J_2}$: Correction for Earth’s equatorial bulge.

Derive from first principles

01 · Begin with Newton’s second law
$m\ddot{\mathbf r}=\sum_i\mathbf F_i\quad\Longrightarrow\quad\ddot{\mathbf r}=\sum_i\mathbf a_i$
The payload mass cancels from gravitational acceleration, so objects follow the same ideal trajectory regardless of payload mass.
02 · Insert Earth’s inverse-square force
$\mathbf F_E=-\frac{GM_E m}{r^2}\hat{\mathbf r}\quad\Longrightarrow\quad\mathbf a_E=-\mu_E\frac{\mathbf r}{r^3}$
The force magnitude is GMm/r² and its direction points back toward Earth’s centre. Writing the unit vector as r/r produces the vector form.
03 · Express lunar gravity relative to Earth
$\mathbf a_M=\mu_M\!\left[\frac{\mathbf r_M-\mathbf r}{\lVert\mathbf r_M-\mathbf r\rVert^3}-\frac{\mathbf r_M}{r_M^3}\right]$
A geocentric frame accelerates toward the Moon too. Subtracting Earth’s lunar acceleration leaves the Moon’s differential, or tidal, effect on the payload.
04 · Add the remaining perturbations
$\ddot{\mathbf r}=\mathbf a_E+\mathbf a_M+\mathbf a_\odot+\mathbf a_{J_2}$
Solar gravity enters with the same third-body subtraction. J₂ is the leading correction obtained by expanding Earth’s non-spherical gravitational potential.

02 · INPUT

LAUNCH STATE

\mathbf v_{0,\mathrm{ECI}}=\mathbf v_M+\boldsymbol\omega_M\times\boldsymbol\rho+v_0\,\hat{\mathbf u}(A,h)

|\mathbf v_M|=1.009\,\mathrm{km/s}\quad v_0=2.70\,\mathrm{km/s}\quad (A,h)=(84.9^\circ,56.0^\circ)

This is fixed after launch. Change muzzle velocity, azimuth, elevation, or the lunar site to build a different initial state.

TERM GUIDE

$\mathbf v_{0,\mathrm{ECI}}$: Payload’s initial velocity in the inertial frame.
$\mathbf v_M$: Moon’s orbital velocity around Earth at launch.
$\boldsymbol\omega_M\!\times\!\boldsymbol\rho$: Velocity of the rotating lunar launch site.
$v_0$: Mass-driver muzzle speed relative to the Moon.
$\hat{\mathbf u}(A,h)$: Unit launch direction from azimuth A and elevation h.

Derive from first principles

01 · Place the launch site on the Moon
$\mathbf r_0=\mathbf r_M+\boldsymbol\rho$
The payload starts at the Moon’s centre position plus the launch-site offset ρ from that centre.
02 · Differentiate the rotating offset
$\frac{d\boldsymbol\rho}{dt}\bigg|_{\mathrm{ECI}}=\boldsymbol\omega_M\times\boldsymbol\rho$
For an offset fixed to a synchronously rotating Moon, the transport theorem converts lunar rotation into inertial surface velocity.
03 · Build the rail direction in the local frame
$\hat{\mathbf u}=\cos h\,(\cos A\,\hat{\mathbf n}+\sin A\,\hat{\mathbf e})+\sin h\,\hat{\mathbf z}$
Azimuth A selects a horizontal direction between local north and east; elevation h tilts that direction toward the local zenith.
04 · Add all inertial velocity contributions
$\mathbf v_{0,\mathrm{ECI}}=\mathbf v_M+\boldsymbol\omega_M\times\boldsymbol\rho+v_0\hat{\mathbf u}$
Moon translation, launch-site rotation, and rail-relative muzzle velocity are vectors in the same inertial frame, so they add directly.

03 · STEP 0

RUNGE–KUTTA PROPAGATION

\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)

\Delta t=30\,\mathrm{s}\quad |\mathbf v|=2.337\,\mathrm{km/s}\quad |\Delta\mathbf r|\approx 70.1\,\mathrm{km/step}

RK4 samples the acceleration four times inside every step, then combines those slopes. A smaller Δt costs more work but reduces numerical error.

TERM GUIDE

$\mathbf y_n$: Position and velocity together at step n.
$\Delta t$: Time advanced by one integration step.
$\mathbf k_1$: Slope evaluated at the start of the step.
$\mathbf k_2,\mathbf k_3$: Two slope estimates at the step midpoint.
$\mathbf k_4$: Slope evaluated at the end of the step.
$\tfrac{1}{6}(1,2,2,1)$: Weights used to average the four slopes.

Derive from first principles

01 · Write orbital motion as a first-order system
$\mathbf y=\begin{bmatrix}\mathbf r\\\mathbf v\end{bmatrix},\qquad\dot{\mathbf y}=\mathbf f(t,\mathbf y)=\begin{bmatrix}\mathbf v\\\mathbf a(t,\mathbf r,\mathbf v)\end{bmatrix}$
Position and velocity are combined into one state vector, allowing the equations of motion to be treated as a standard first-order ODE.
02 · Use the exact integral statement
$\mathbf y(t+\Delta t)=\mathbf y(t)+\int_t^{t+\Delta t}\mathbf f(\tau,\mathbf y(\tau))\,d\tau$
Advancing the state means accumulating its derivative across the step. The difficulty is that the path inside this integral is not yet known.
03 · Sample four representative slopes
$\begin{aligned}\mathbf k_1&=\mathbf f(t,\mathbf y)\\\mathbf k_2&=\mathbf f(t+\tfrac{\Delta t}{2},\mathbf y+\tfrac{\Delta t}{2}\mathbf k_1)\\\mathbf k_3&=\mathbf f(t+\tfrac{\Delta t}{2},\mathbf y+\tfrac{\Delta t}{2}\mathbf k_2)\\\mathbf k_4&=\mathbf f(t+\Delta t,\mathbf y+\Delta t\,\mathbf k_3)\end{aligned}$
Two midpoint estimates correct each other, while the first and last estimates capture how the slope changes across the interval.
04 · Approximate the integral to fourth order
$\mathbf y_{n+1}=\mathbf y_n+\frac{\Delta t}{6}(\mathbf k_1+2\mathbf k_2+2\mathbf k_3+\mathbf k_4)$
These weights cancel the lower-order Taylor-series errors, leaving a local truncation error proportional to Δt⁵ and a global error proportional to Δt⁴.

04 · LIVE

ROTATING TARGET FRAME

\mathbf r_{\mathrm{ECEF}}=R_3(-\theta_{\mathrm{GST}})\,R_{\mathrm{EQJ}\rightarrow\mathrm{EQD}}\,\mathbf r_{\mathrm{ECI}}

\Delta\theta_E=0.00^\circ\quad v_{\mathrm{target}}=0.294\,\mathrm{km/s}\quad h=382511\,\mathrm{km}

The target keeps moving beneath the inertial trajectory. Its surface speed depends on latitude, so equatorial destinations move fastest.

TERM GUIDE

$\mathbf r_{\mathrm{ECI}}$: Position in the non-rotating Earth-centred frame.
$R_{\mathrm{EQJ}\rightarrow\mathrm{EQD}}$: Precession and nutation from J2000 to the current date.
$\theta_{\mathrm{GST}}$: Greenwich sidereal angle: Earth’s current rotation.
$R_3(-\theta)$: Rotation about Earth’s polar, or z, axis.
$\mathbf r_{\mathrm{ECEF}}$: Earth-fixed position used for latitude and longitude.

Derive from first principles

01 · Start with inertial components
$\mathbf r_{\mathrm{ECI}}=x\,\hat{\mathbf i}_{J2000}+y\,\hat{\mathbf j}_{J2000}+z\,\hat{\mathbf k}_{J2000}$
The propagated vector is expressed in a non-rotating J2000 equatorial basis, which is convenient for integrating forces.
02 · Move the equator and equinox to the date
$\mathbf r_{\mathrm{EQD}}=R_{\mathrm{EQJ}\rightarrow\mathrm{EQD}}\mathbf r_{\mathrm{ECI}}$
Precession slowly moves Earth’s spin axis and nutation adds shorter oscillations. This rotation aligns the basis with Earth’s orientation at the flight epoch.
03 · Rotate into Earth’s spinning basis
$\mathbf r_{\mathrm{ECEF}}=R_3(-\theta_{\mathrm{GST}})\mathbf r_{\mathrm{EQD}}$
Earth-fixed axes turn eastward through the Greenwich sidereal angle. Coordinates therefore rotate by the opposite angle, giving the minus sign.
04 · Read geographic coordinates
$\lambda=\operatorname{atan2}(y_{\mathrm{ECEF}},x_{\mathrm{ECEF}}),\qquad\varphi\ \text{from the oblate-Earth geodetic solution}$
Longitude follows directly from the equatorial components. Latitude is solved against Earth’s reference ellipsoid rather than treating Earth as a perfect sphere.

✓Target corridor reachedEntry interface crossed 1 km from the selected target.ENTRY

ENTRY SOLUTION51.50°N · 0.14°W

Flight time2.59 dTo terminal event

Target miss1.2 kmEarth frame

Entry speed11.04 km/sBefore interface

Entry angle-58.4°Local horizon

Grid energy4.29 GJ85% efficiency

Rail acceleration37.2 g10 km rail

Moon moves33.8°During flight

Earth rotates215.0°Sidereal

Earth-centered inertial view