Nondimensionalising Gravitational Models and Deriving Orbital Timescales

Welcome back. In the previous lesson, you expressed the fixed-Earth orbital model as the first-order initial-value problem

$$\dot{\mathbf{y}}=\left[\begin{array}{c}\mathbf{v}\\ -{\mu }_{\mathrm{E}}\mathbf{r}/\Vert \mathbf{r}{\Vert }^3\end{array}\right],\mathbf{y}=\left[\begin{array}{c}\mathbf{r}\\ \mathbf{v}\end{array}\right].$$

That formulation is ready for an ODE solver, but its numerical values depend heavily on our chosen units. A low-Earth-orbit position might be thousands of kilometres, a velocity around $7.5\mathrm{km}{\mathrm{s}}^{-1}$, and an acceleration around $0.008\mathrm{km}{\mathrm{s}}^{-2}$. In SI units, those same quantities have very different magnitudes again.

This lesson develops a standard applied-mathematics move: nondimensionalisation. We will scale mass, length, position, velocity, and time so that the gravitational equation has a parameter-free canonical form. In doing so, we will derive the natural timescale of an orbit and see exactly how it relates to an orbital period.

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Why rescale a model?

A dimensional model carries both physical structure and arbitrary human conventions. The Earth does not behave differently because we report distance in metres rather than kilometres; only the numerical representation has changed.

Nondimensionalisation separates those two things. We choose characteristic scales that are representative of the problem:

• a characteristic mass $M_c$,
• a characteristic length $L_c$,
• and a derived characteristic time $t_c$.

We then express every variable as “how many characteristic units” it contains. The resulting variables have no physical dimensions and are ideally of order one.

This is useful for three reasons.

1. It exposes the true number of independent parameters. A one-centre gravitational problem with a sensibly chosen mass scale has no remaining physical parameter in its dimensionless equation.

2. It improves numerical interpretation. A dimensionless timestep of $h=0.01$ means “one hundredth of the local gravitational timescale,” regardless of whether the simulated system is an Earth satellite, an asteroid around the Sun, or a compact binary.

3. It makes solutions transferable. One dimensionless trajectory can represent many physical systems after rescaling, provided their models have the same form.

Nondimensionalisation does not itself make a numerical integrator more accurate. It makes magnitudes and error tolerances easier to reason about, while revealing what timestep means physically.

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Scaling the fixed-centre gravitational model

Write the gravitational equation with the source mass shown explicitly:

$$\ddot{\mathbf{r}}=-G{M}_s\frac{\mathbf{r}}{\Vert \mathbf{r}{\Vert }^3}.$$

Here, $M_s$ is the mass of the central source, such as Earth, and the orbiting satellite’s mass has cancelled from the equation. Let

$${\mu }_s=G{M}_s.$$

The units of the gravitational constant are

$$[G]=L^3{M}^{-1}{T}^{-2},$$

so the combination $G{M}_s$ has units

$$[{\mu }_s]=L^3{T}^{-2}.$$

Choose a characteristic mass $M_c$ and length $L_c$. Define dimensionless position and time variables by

$$\mathbf{r}(t)=L_c\mathbf{\rho }(\tau ),t=t_c\tau .$$

The bold vector $\mathbf{\rho }$ is a position measured in units of $L_c$, while $\tau$ is time measured in units of $t_c$.

Apply the chain rule carefully. Since

$$\tau =\frac{t}{t_c},$$

we have

$$\frac{d}{dt}=\frac{1}{t_c}\frac{d}{d\tau }.$$

Therefore,

$$\dot{\mathbf{r}}=\frac{L_c}{t_c}\frac{d\mathbf{\rho }}{d\tau },$$

and

$$\ddot{\mathbf{r}}=\frac{L_c}{t_c^2}\frac{d^2\mathbf{\rho }}{d{\tau }^2}.$$

Substituting these expressions into the physical equation gives

$$\frac{L_c}{t_c^2}\frac{d^2\mathbf{\rho }}{d{\tau }^2}=-G{M}_s\frac{L_c\mathbf{\rho }}{\Vert L_c\mathbf{\rho }{\Vert }^3}.$$

Because

$$\Vert L_c\mathbf{\rho }{\Vert }^3=L_c^3\Vert \mathbf{\rho }{\Vert }^3,$$

we obtain

$$\frac{d^2\mathbf{\rho }}{d{\tau }^2}=-\left(\frac{G{M}_s{t}_c^2}{L_c^3}\right)\frac{\mathbf{\rho }}{\Vert \mathbf{\rho }{\Vert }^3}.$$

At this point, all of the dimensional information is concentrated in one dimensionless coefficient:

$$\frac{G{M}_s{t}_c^2}{L_c^3}.$$

The aim is to choose $t_c$ so that this coefficient becomes simple.

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The characteristic gravitational timescale

Let the mass scale define

$${\mu }_c=G{M}_c.$$

Choose the time scale

$$\boxed{t_c=\sqrt{\frac{L_c^3}{G{M}_c}}.}$$

This is dimensionally a time because

$$\left[\frac{L_c^3}{G{M}_c}\right]=T^2.$$

Substituting this choice gives

$$\frac{G{M}_s{t}_c^2}{L_c^3}=\frac{M_s}{M_c}.$$

Define the dimensionless source-mass ratio

$$\beta =\frac{M_s}{M_c}.$$

The nondimensional gravitational equation is therefore

$$\boxed{\frac{d^2\mathbf{\rho }}{d{\tau }^2}=-\beta \frac{\mathbf{\rho }}{\Vert \mathbf{\rho }{\Vert }^3}.}$$

For the current fixed-Earth model, the natural choice is simply

$$M_c=M_s=M_{\mathrm{E}}.$$

Then $\beta =1$, and the equation becomes

$$\boxed{\frac{d^2\mathbf{\rho }}{d{\tau }^2}=-\frac{\mathbf{\rho }}{\Vert \mathbf{\rho }{\Vert }^3}.}$$

This is the canonical dimensionless point-mass gravity equation. There is no $G$, no Earth mass, no kilometre, and no second in it. Those quantities have not been discarded; they are stored in the conversion factors $L_c$ and $t_c$.

The same idea will scale naturally into the next module. Once both bodies move freely, the relative two-body equation contains

$$\mu =G(m_1+m_2).$$

There, choosing

$$M_c=m_1+m_2$$

produces exactly the same parameter-free dimensionless relative-motion equation.

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The timescale is not quite the orbital period

The scale

$$t_c=\sqrt{\frac{L_c^3}{G{M}_c}}$$

is often called a dynamical time or gravitational timescale. It is the time required for gravity to change motion substantially across a distance comparable with $L_c$.

For a circular orbit of radius $L_c$, the physical angular frequency follows from balancing centripetal and gravitational acceleration:

$${\omega }_c^2{L}_c=\frac{G{M}_c}{L_c^2}.$$

Hence,

$${\omega }_c=\sqrt{\frac{G{M}_c}{L_c^3}}=\frac{1}{t_c}.$$

The corresponding circular-orbit period is

$$\boxed{P_c=\frac{2\pi }{{\omega }_c}=2\pi \sqrt{\frac{L_c^3}{G{M}_c}}=2\pi t_c.}$$

So $t_c$ is not the full period. It is the period divided by $2\pi$.

This distinction becomes transparent in dimensionless form. Consider the unit-radius circular solution

$$\mathbf{\rho }(\tau )=\left[\begin{array}{c}\cos \tau \\ \sin \tau \\ 0\end{array}\right].$$

Its norm is one, so the right-hand side of the dimensionless equation is simply $-\mathbf{\rho }$. Also,

$$\frac{d^2\mathbf{\rho }}{d{\tau }^2}=-\mathbf{\rho }.$$

Thus it solves the equation exactly, and it returns to its starting position after dimensionless time

$$\tau =2\pi .$$

Converting back to dimensional time gives

$$t=2\pi t_c=P_c.$$

There is another legitimate convention: use $P_c$, rather than $t_c$, as the unit of time. Then a circular orbit has dimensionless period one, but the equation becomes

$$\frac{d^2\mathbf{\rho }}{d{\hat{\tau }}^2}=-4{\pi }^2\frac{\mathbf{\rho }}{\Vert \mathbf{\rho }{\Vert }^3}.$$

Neither convention is more correct. The $t_c$ convention makes the force coefficient equal to one; the period convention makes a reference circular orbit have period one. The key discipline is to state which convention you use and apply it consistently.

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A concrete Earth-orbit scaling

Suppose the reference length is a low-Earth-orbit radius:

$$L_c=7000\mathrm{km}.$$

Use Earth’s gravitational parameter in compatible units:

$${\mu }_{\mathrm{E}}=398600\mathrm{k}{\mathrm{m}}^3{\mathrm{s}}^{-2}.$$

Taking $M_c=M_{\mathrm{E}}$, the characteristic time is

$$t_c=\sqrt{\frac{L_c^3}{{\mu }_{\mathrm{E}}}}\approx 927.6\mathrm{s}.$$

The corresponding velocity scale is

$$v_c=\frac{L_c}{t_c}=\sqrt{\frac{{\mu }_{\mathrm{E}}}{L_c}}\approx 7.55\mathrm{km}{\mathrm{s}}^{-1}.$$

That is precisely the circular speed at $7000\mathrm{km}$. The circular period is

$$P_c=2\pi t_c\approx 5828\mathrm{s}\approx 97.1\min .$$

A reference circular initial condition becomes unusually clean:

$${\mathbf{\rho }}_0=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],{\mathbf{u}}_0=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right].$$

Here $\mathbf{u}$ is dimensionless velocity, defined by

$$\mathbf{u}=\frac{t_c}{L_c}\mathbf{v}.$$

The state says: begin one characteristic length from the origin and move perpendicular to the radial direction with one characteristic velocity. No physical constants are needed by the dimensionless right-hand side.

The illustrative velocity from the previous lesson,

$$7.5\mathrm{km}{\mathrm{s}}^{-1},$$

would correspond at this radius to a dimensionless tangential speed of approximately

$$u_y\approx \mathrm{0.994.}$$

That small difference from one is physically meaningful: it is close to, but not exactly, the circular reference state.

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Dimensionless state space and code

The scaling should be applied to the entire state, not position alone:

$$\mathbf{y}=\left[\begin{array}{c}\mathbf{r}\\ \mathbf{v}\end{array}\right],\mathbf{Y}=\left[\begin{array}{c}\mathbf{\rho }\\ \mathbf{u}\end{array}\right]=\left[\begin{array}{c}\mathbf{r}/L_c\\ t_c\mathbf{v}/L_c\end{array}\right].$$

In terms of dimensionless time, the first-order state equation is

$$\boxed{\frac{d}{d\tau }\left[\begin{array}{c}\mathbf{\rho }\\ \mathbf{u}\end{array}\right]=\left[\begin{array}{c}\mathbf{u}\\ -\beta \mathbf{\rho }/\Vert \mathbf{\rho }{\Vert }^3\end{array}\right].}$$

For the natural single-source choice $M_c=M_s$, set $\beta =1$.

import numpy as np

def gravity_rhs_nondimensional(tau, state, beta=1.0):
    """
    Dimensionless gravitational right-hand side.

    state = [rho_x, rho_y, rho_z, u_x, u_y, u_z]
    tau and state are dimensionless.
    """
    rho = state[:3]
    u = state[3:]

    rho2 = np.dot(rho, rho)

    if rho2 == 0.0:
        raise ValueError("Point-mass gravity is undefined at rho = 0.")

    acceleration = -beta * rho / (rho2 * np.sqrt(rho2))

    return np.concatenate((u, acceleration))

Notice what has disappeared from the force function: there is no mu argument. The dimensional gravitational parameter was used once, during construction of the time scale

$$t_c=\sqrt{\frac{L_c^3}{{\mu }_s}}.$$

Conversion functions can make the boundary between physical inputs and dimensionless integration explicit:

def state_to_nondimensional(state, length_scale, time_scale):
    nd_state = np.asarray(state, dtype=float).copy()
    nd_state[:3] /= length_scale
    nd_state[3:] *= time_scale / length_scale
    return nd_state

def state_to_dimensional(nd_state, length_scale, time_scale):
    state = np.asarray(nd_state, dtype=float).copy()
    state[:3] *= length_scale
    state[3:] *= length_scale / time_scale
    return state

If an integrator advances $\tau$ by a dimensionless timestep $h$, the associated physical timestep is

$$\Delta t=t_{c}h.$$

For the low-Earth-orbit scaling above:

$$h=0.01$$

corresponds to about

$$9.28\mathrm{s}.$$

A useful implementation check is that the unit circular state

state0 = np.array([1.0, 0.0, 0.0, 0.0, 1.0, 0.0])

should produce the derivative

np.array([0.0, 1.0, 0.0, -1.0, 0.0, 0.0])

up to floating-point representation. The first three entries are the velocity, and the final three are the inward unit acceleration.

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Choosing useful characteristic scales

There is no universal best $L_c$. Its choice should reflect the investigation.

Investigation	Sensible $L_c$	Sensible $M_c$
One Earth satellite	Reference orbital radius	Earth mass
Planet around the Sun	One astronomical unit or semi-major axis	Solar mass
Binary-star relative orbit	Typical stellar separation	Sum of stellar masses
Compact multi-body system	Characteristic system size	Total system mass

A good scale makes typical dimensionless position and velocity components reasonably close to one. If $\Vert \mathbf{\rho }\Vert$ ranges from $10^{-9}$ to $10^9$, the scaling has not matched the geometry of the problem very well, even though the mathematics is still valid.

Two practical cautions matter:

• Maintain unit consistency before scaling. If $L_c$ is in kilometres, then $\mu$ must use kilometres and the corresponding time unit. Do not combine kilometres with ${\mathrm{m}}^3{\mathrm{s}}^{-2}$.

• Do not confuse scale choice with model choice. Rescaling cannot repair omitted physics such as atmospheric drag, oblateness, relativity, or perturbations from the Moon. It only rewrites the model already chosen.

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Key takeaways

Nondimensionalisation rewrites the gravitational model using characteristic mass and length scales:

$$M_c,L_c.$$

Choosing

$$t_c=\sqrt{\frac{L_c^3}{G{M}_c}}$$

gives the dimensionless equation

$$\frac{d^2\mathbf{\rho }}{d{\tau }^2}=-\frac{M_s}{M_c}\frac{\mathbf{\rho }}{\Vert \mathbf{\rho }{\Vert }^3}.$$

For a fixed central body, choose $M_c=M_s$. The equation then has the universal parameter-free form

$${\mathbf{\rho }}^{\prime \prime }=-\frac{\mathbf{\rho }}{\Vert \mathbf{\rho }{\Vert }^3}.$$

The characteristic time $t_c$ is a gravitational dynamical time. A circular orbit at radius $L_c$ has period

$$P_c=2\pi t_c.$$

You now have a clean, scale-independent version of the orbital initial-value problem, along with a disciplined conversion boundary between physical inputs and numerical integration.

Next, we move from a fixed Earth to the genuine two-body problem: two masses exerting equal and opposite forces and both moving in an inertial frame.