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In a spherically dominated Newtonian gravitational field, the circular orbit speed satisfies $v_c^2=GM(<r)/r$. A flat galaxy rotation curve therefore implies $M(<r)=v_0^{2}r/G$. Differentiating the enclosed mass gives $\rho =(4\pi r^2{)}^{-1}dM/dr$, yielding the inverse-square density profile. This is a halo-region approximation, not a mass inversion for a flattened stellar disk. A flat curve extending to infinite radius would imply infinite mass, so a real halo profile must eventually change or truncate.

The product $\mu =GM$ of the gravitational constant and a central mass sets the strength of the exterior Newtonian gravitational field. Its dimensions are length cubed divided by time squared. In the exact relative two-body problem, the parameter is $G(M_1+M_2)$; using $GM$ for one source assumes the test particle approximation.

The combination is useful because orbital and falling-body observations constrain $\mu$ directly. A gravitational circular orbit of radius $a$ has oscillation period $T=2\pi \sqrt{a^3/\mu }$, so measuring $a$ and $T$ gives $\mu =4{\pi }^2{a}^3/T^2$. That measurement alone does not separately determine $G$ and $M$. Similarly, radial free fall from rest has a natural time scale $\sqrt{r_0^3/\mu }$.

Other interactions require a different model. Inverse-square radiation pressure can reduce the effective attraction for a dust Kepler orbit; calling that effective coefficient a gravitational parameter must not hide the added force.

A passing star perturbs a planetary or cometary system through its differential Newtonian gravitational field. A distant encounter is predominantly tidal; a sufficiently close one can substantially change specific orbital energy and angular momentum, eject a comet, or shift its periapsis into or out of the planetary region.

Treating a moving body as a test particle neglects its effect on the externally prescribed field. For a small meteorite moving near a planet, $m/M\ll 1$ permits neglect of the planet's recoil and replacement of the relative two-body gravitational parameter $G(M+m)$ by $GM$. This is separate from neglecting drag or assuming a Newtonian gravitational field.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.

To isolate a tidal force, subtract the gravitational acceleration shared by the body as a whole. Let a reference test particle follow $\ddot{\mathbf{R}}=\mathbf{g}(\mathbf{R})$, and let another have position $\mathbf{R}+\mathbf{x}$. In a translating accelerated reference frame with parallel, nonrotating axes,for a source point mass, when $|\mathbf{x}|/R\ll 1$. Here $\mu =GM$ is its standard gravitational parameter and the last term uses Big O notation. This Taylor expansion removes the common inward pull and keeps the spatial variation. The near side has an excess pull toward the source; the far side has a deficit of inward pull, which is an outward relative acceleration. Both sides can stretch apart while both still accelerate toward the source.^[ref]

For a material element of mass $m$, the residual external force in this translating accelerated reference frame is $m[\mathbf{g}(\mathbf{R}+\mathbf{x})-\mathbf{g}(\mathbf{R})]$. Actual deformation also depends on elasticity, pressure and self-gravity. A rigid body approximation suppresses deformation but can still have internal forces that resist the tidal force. The point-mass tidal stretching and squeezing pattern describes freely falling, noninteracting test particles, rather than promising the same shape for every solid, star or ocean. A rotating reference frame introduces centrifugal acceleration and Coriolis acceleration, which must be included separately.

The acceleration tidal tensor is $T_{jk}={\partial }_k{g}_j=-{\partial }_j{\partial }_k\Phi$ for gravitational acceleration $\mathbf{g}=-\nabla \Phi$. It describes differential acceleration $\Delta \mathbf{g}\simeq T\Delta \mathbf{r}$. Some authors instead call the positive Hessian matrix ${\partial }_j{\partial }_k\Phi$ the tidal tensor, reversing this sign convention. For a point mass and the positive potential convention $\mathbf{g}=\nabla \psi$, the tensor is $Gm(3\mathbf{R}{\mathbf{R}}^{\mathsf{T}}/R^5-1/R^3)$.

The tidal tensor is the Jacobian matrix of the Newtonian gravitational field, rather than the gravitational acceleration itself. Its entries have units ${\mathrm{s}}^{-2}$: multiplying by a displacement produces an acceleration. For a smooth Newtonian gravitational potential, the tidal tensor is a symmetric matrix, since mixed partial derivatives commute. A spatially uniform gravitational acceleration has zero tidal tensor, however large that acceleration is.

Remove a spherical void centered at $c$ from a uniform sphere centered at the origin. At points inside the void, superposition combines the sphere's interior field $-4\pi G\rho x/3$ with the removed sphere's field $+4\pi G\rho (x-c)/3$. Their sum is the displayed constant Newtonian gravitational field. A second removed mass outside this void adds a nonuniform contribution.

For a steady axisymmetric stellar population with no mean vertical motion, the vertical Jeans equation balances the gradient of the vertical random-motion stress and the radial gradient of the tilted velocity ellipsoid against the vertical Newtonian gravitational field. Neglecting the mixed moment gives ${\partial }_z(n{\sigma }_z^2)=-n{\partial }_z\Phi$. A tracer density $n$ need not equal the density sourcing the gravitational potential.

Weightlessness is the absence of supporting or constraining forces that would give a body an apparent weight. It does not require zero gravitational acceleration. A freely falling cabin and a small freely falling object inside share the same acceleration in a uniform field, so neither needs to support the other.

The circular orbit as continuous free fall gives a sustained example. Real spacecraft have small residual relative accelerations from tidal forces, atmospheric drag force and onboard disturbances; the ideal common-fall explanation does not assert that every point in an extended spacecraft experiences exactly the same Newtonian gravitational field. ^[ref]