Deriving Gravitational Acceleration from Newton’s Inverse-Square Law

Welcome back. You have already fixed the coordinate language for the first model: an Earth-centred, non-rotating Cartesian frame, with a satellite described by its position $\mathbf{r}(t)$ and velocity $\mathbf{v}(t)$. The remaining physical ingredient is a rule that assigns an acceleration to each possible position.

Here we derive that rule carefully. The destination is the compact expression

$$\mathbf{a}(\mathbf{r})=-\frac{GM}{\Vert \mathbf{r}{\Vert }^3}\mathbf{r},$$

for a satellite at position $\mathbf{r}$ under the gravity of a point mass $M$ at the origin. The power $-3$ in the denominator is not a different distance law: combined with the vector $\mathbf{r}$, it produces an acceleration whose magnitude obeys the familiar inverse-square law and whose direction is toward the attracting mass.

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Inverse-square magnitude: a geometric intuition

Newton’s law of universal gravitation says that two point masses $M$ and $m$, separated by a distance $\rho$, attract with force magnitude

$$F=\frac{GMm}{{\rho }^2}.$$

Here $G$ is the universal gravitational constant. In SI units,

$$G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}.$$

The inverse-square dependence means that doubling separation reduces the force to one quarter; tripling separation reduces it to one ninth.

The law is an empirical law of the Newtonian model, rather than something that can be proved solely from a diagram. But spherical geometry gives useful intuition for why point-source effects commonly have this form. At greater distances, the same influence from a source is distributed across larger spherical surfaces, whose areas scale as ${\rho }^2$.

This geometric argument concerns the magnitude only. A simulator also needs the direction, and it must represent that direction consistently in three dimensions.

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From a scalar law to a vector force

The following reading uses a two-body convention that will become important in the next module: it defines a relative displacement first, then lets that displacement determine both the direction and magnitude of each force.

We will now use a convention tailored to the fixed-Earth satellite model.

Let:

• $M$ be the source mass, such as Earth;
• $m$ be the mass of the body being accelerated, such as a satellite;
• $\mathbf{R}$ be the source position;
• $\mathbf{r}$ be the satellite position.

Define the displacement from the source to the satellite as

$$\mathbf{\rho }=\mathbf{r}-\mathbf{R}.$$

Its length is the separation distance:

$$\rho =\Vert \mathbf{\rho }\Vert .$$

The corresponding outward unit vector, pointing from the source toward the satellite, is

$${\mathbf{e}}_{\rho }=\frac{\mathbf{\rho }}{\rho }.$$

That word outward determines the sign. Gravitation is attractive, so the force on the satellite points in the direction opposite to ${\mathbf{e}}_{\rho }$: back toward the source.

Thus the vector force on mass $m$ due to source mass $M$ is

$${\mathbf{F}}_m^{(M)}=-\frac{GMm}{{\rho }^2}{\mathbf{e}}_{\rho }.$$

The scalar coefficient supplies the inverse-square magnitude. The minus sign reverses the outward unit vector, making the force inward.

The equal-and-opposite pair in the diagram is Newton’s third law. In the restricted Earth–satellite model, we retain the force and acceleration of the satellite while neglecting Earth’s extremely small response. That is an approximation, not a claim that Earth feels no gravitational force. When we move to the full two-body problem, both accelerations will be evolved.

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Why the vector formula contains ${\rho }^3$

Substitute the unit-vector definition into the force law:

$${\mathbf{F}}_m^{(M)}=-\frac{GMm}{{\rho }^2}\left(\frac{\mathbf{\rho }}{\rho }\right).$$

Combining the distance factors gives

$${\mathbf{F}}_m^{(M)}=-GMm\frac{\mathbf{\rho }}{{\rho }^3}.$$

This is the form worth remembering. The ${\rho }^3$ denominator appears because we began with:

1. an inverse-square scalar magnitude, ${\rho }^{-2}$;
2. a unit direction vector, $\mathbf{\rho }/\rho$.

It does not mean that gravity’s magnitude decays as ${\rho }^{-3}$. Taking the norm makes the cancellation explicit:

$$\Vert \frac{\mathbf{\rho }}{{\rho }^3}\Vert =\frac{\Vert \mathbf{\rho }\Vert }{{\rho }^3}=\frac{1}{{\rho }^2}.$$

Therefore,

$$\Vert {\mathbf{F}}_m^{(M)}\Vert =\frac{GMm}{{\rho }^2},$$

exactly as Newton’s scalar law requires.

A useful way to diagnose a sign or power error is to test a simple position. Put the source at the origin and the satellite on the positive $x$-axis:

$$\mathbf{r}=\left[\begin{array}{c}{r}_0\\ 0\\ 0\end{array}\right],r_0>0.$$

The acceleration must point toward the origin, in the negative $x$-direction. Any formula that produces a positive $x$-component has the direction wrong.

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Dividing by inertial mass: gravitational acceleration

Newton’s second law is

$$\mathbf{F}=m\mathbf{a}.$$

Insert the gravitational force and divide by the accelerated body’s mass $m$:

$$m\mathbf{a}=-GMm\frac{\mathbf{\rho }}{{\rho }^3},$$ $$\mathbf{a}=-GM\frac{\mathbf{\rho }}{{\rho }^3}.$$

Equivalently, in terms of the two absolute positions,

$$\boxed{\mathbf{a}(\mathbf{r})=-GM\frac{\mathbf{r}-\mathbf{R}}{\Vert \mathbf{r}-\mathbf{R}{\Vert }^3}}$$

This is the gravitational acceleration generated by a point mass $M$ at $\mathbf{R}$, evaluated at $\mathbf{r}$.

The cancellation of $m$ means that, in this model, all test bodies at the same location have the same gravitational acceleration, irrespective of their own masses. A $1$-kg and a $1000$-kg satellite released from the same position with the same velocity follow the same trajectory, provided both are treated as test particles and non-gravitational effects are omitted.

It is often helpful to name this position-dependent acceleration field:

$${\mathbf{g}}_M(\mathbf{r})=-GM\frac{\mathbf{r}-\mathbf{R}}{\Vert \mathbf{r}-\mathbf{R}{\Vert }^3}.$$

Then a test satellite obeys simply

$$\mathbf{a}={\mathbf{g}}_M(\mathbf{r}).$$

Near Earth’s surface, one often approximates $\mathbf{g}$ as a constant downward vector. Orbital motion is different: distance and direction both change substantially around the orbit, so we retain the full position-dependent expression.

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The Earth-centred formula

For the initial satellite model, Earth is fixed at the origin. Therefore,

$$\mathbf{R}=0,\mathbf{\rho }=\mathbf{r}.$$

The acceleration becomes

$$\boxed{\mathbf{a}(\mathbf{r})=-G{M}_{\mathrm{E}}\frac{\mathbf{r}}{\Vert \mathbf{r}{\Vert }^3}}$$

where $M_{\mathrm{E}}$ is Earth’s mass.

In orbital mechanics, it is conventional and computationally convenient to combine the constants into Earth’s standard gravitational parameter:

$${\mu }_{\mathrm{E}}=G{M}_{\mathrm{E}}.$$

The governing acceleration law is then

$$\boxed{\mathbf{a}(\mathbf{r})=-{\mu }_{\mathrm{E}}\frac{\mathbf{r}}{\Vert \mathbf{r}{\Vert }^3}}$$

The parameter ${\mu }_{\mathrm{E}}$ has units

$$[{\mu }_{\mathrm{E}}]=\mathrm{lengt}{\mathrm{h}}^3\mathrm{tim}{\mathrm{e}}^{-2}.$$

That gives a valuable dimensional check:

$$\left[{\mu }_{\mathrm{E}}\frac{\mathbf{r}}{\Vert \mathbf{r}{\Vert }^3}\right]=\frac{\mathrm{lengt}{\mathrm{h}}^3}{\mathrm{tim}{\mathrm{e}}^2}\frac{\mathrm{length}}{\mathrm{lengt}{\mathrm{h}}^3}=\frac{\mathrm{length}}{\mathrm{tim}{\mathrm{e}}^2},$$

which is indeed acceleration.

In Cartesian components, if

$$\mathbf{r}=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],r=\sqrt{x^2+y^2+z^2},$$

then

$$\mathbf{a}=\left[\begin{array}{c}-{\mu }_{\mathrm{E}}x/r^3\\ -{\mu }_{\mathrm{E}}y/r^3\\ -{\mu }_{\mathrm{E}}z/r^3\end{array}\right].$$

Each component is coupled through the common distance $r$. This is why orbital gravity is a nonlinear vector system: the denominator depends on the current position itself.

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Three checks before trusting the formula

Before putting this acceleration into an integrator, make these checks habitual.

1. Direction check

The acceleration must be antiparallel to the source-to-satellite displacement. Their dot product is negative:

$$\mathbf{a}\cdot \mathbf{\rho }=-\frac{GM}{\rho }.$$

Since $G$, $M$, and $\rho$ are positive for separated bodies, the result is negative. This confirms that gravity pulls inward.

2. Magnitude and scaling check

The norm satisfies

$$\Vert \mathbf{a}\Vert =\frac{GM}{{\rho }^2}.$$

For a scaled position $\lambda \mathbf{\rho }$, with $\lambda >0$,

$$\mathbf{a}(\lambda \mathbf{\rho })={\lambda }^{-2}\mathbf{a}(\mathbf{\rho }).$$

The direction is unchanged along a radial line, while the magnitude follows the inverse-square rule.

3. Translation check

The general formula depends on positions only through their difference:

$$\mathbf{r}-\mathbf{R}.$$

If the same constant displacement is added to both positions, their relative displacement and the acceleration do not change. This is exactly what should happen: the force depends on separation, not on an arbitrary coordinate origin.

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A direct computational translation

For the Earth-centred restricted problem, the force calculation is small enough to inspect line by line:

import numpy as np

def gravitational_acceleration(position, mu):
    r2 = np.dot(position, position)

    if r2 == 0.0:
        raise ValueError("Acceleration is undefined at the point-mass source.")

    return -mu * position / (r2 * np.sqrt(r2))

The expression r2 * np.sqrt(r2) is $r^3$, computed from $r^2$. The returned array has the same three components as position.

For a satellite initially at positive $x$,

position = np.array([7000.0, 0.0, 0.0])  # km

the returned acceleration must have:

• a negative first component;
• zero second and third components, up to floating-point roundoff;
• units of $\mathrm{km}{\mathrm{s}}^{-2}$ if $\mu$ is supplied in $\mathrm{k}{\mathrm{m}}^3{\mathrm{s}}^{-2}$.

The unit system is a property of the whole calculation. If position is in kilometres, then $\mu$ must use kilometres cubed per second squared; combining kilometres with an SI value of $\mu$ expressed in metres cubed per second squared introduces a factor-of-$10^9$ error.

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Scope and singularity of the point-mass model

The formula

$$\mathbf{a}=-\mu \frac{\mathbf{r}}{\Vert \mathbf{r}{\Vert }^3}$$

is undefined at $\mathbf{r}=0$. Its magnitude tends without bound as $r$ tends to zero. This is a mathematical singularity of the point-mass idealisation.

For a satellite outside a nearly spherical Earth, the point-mass field is an excellent first model: Earth’s external gravitational field behaves as though its mass were concentrated at its centre. But the formula does not describe motion through Earth’s interior, a collision with Earth’s surface, atmospheric drag, or the gravitational influence of Earth’s oblateness, Moon, Sun, and other bodies. Those exclusions are consistent with the modelling brief you established in the first lesson.

For now, the useful operational condition is simple:

$$\Vert \mathbf{r}(t)\Vert >R_{\mathrm{E}}$$

throughout a physically meaningful Earth-orbit simulation, where $R_{\mathrm{E}}$ is Earth’s radius in the chosen distance units.

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Key takeaways

Newton’s inverse-square law gives the gravitational force magnitude:

$$F=\frac{GMm}{{\rho }^2}.$$

Defining the displacement from source to satellite as

$$\mathbf{\rho }=\mathbf{r}-\mathbf{R}$$

makes the attractive direction unambiguous. The vector force and acceleration are

$${\mathbf{F}}_m^{(M)}=-GMm\frac{\mathbf{\rho }}{{\rho }^3},$$ $$\mathbf{a}(\mathbf{r})=-GM\frac{\mathbf{r}-\mathbf{R}}{\Vert \mathbf{r}-\mathbf{R}{\Vert }^3}.$$

With Earth fixed at the origin, this reduces to

$$\mathbf{a}(\mathbf{r})=-{\mu }_{\mathrm{E}}\frac{\mathbf{r}}{\Vert \mathbf{r}{\Vert }^3}.$$

The vector numerator supplies radial direction; together with the cubic denominator, it yields an inverse-square acceleration magnitude. This is the central physical function your simulator will repeatedly evaluate.

Next, we will place this acceleration law into a formal initial-value problem and convert the resulting second-order vector equation into the first-order state-space form used by numerical ODE solvers.