M080 · Differential gravity

Tidal Field Explorer

Move a probe over a three-dimensional globe. Subtract the acceleration of its center and see why gravity stretches both ends while squeezing the sides.

Radial differential acceleration $\text{Ready}$

Differential acceleration magnitude $\text{Ready}$

Unmagnified equilibrium height $\text{Ready}$

Linear-field relative error $\text{Ready}$

Center acceleration $\text{Ready}$

Surface display magnification $\text{Ready}$

GRAVITY / GEOMETRY / MOTION

Move the probe. Remove the common fall.

Companion distance / globe radius

d/R

4.50\,

Companion / central mass

M_2/M_1

2.00\,

Companion azimuth

\phi

0.00\,

Probe longitude

\lambda

0.00\,

Probe latitude

\beta

0.00\,

Surface display magnification

G_h

6.00\,

Globe radius, central gravitational parameter and surface gravity are one. Coordinates are right handed; the companion lies in the equatorial plane. Gold arrows show companion gravity, coral arrows show its difference from the center acceleration. Every arrow uses the same fixed display scale.

The blue surface uses the exact companion tidal potential divided by unit self-gravity, a first-order equilibrium response. Its maximum unmagnified height is below five percent of the radius; the selected gain changes only the drawn height. No continents, ocean dynamics, centrifugal potential, elastic response, self-gravity perturbation or Roche disruption.

The linear comparison is the first spatial derivative of the companion field. It stretches the radial direction and compresses two transverse directions. The far side is still attracted toward the companion; it falls less quickly than the center.

Exact companion point-mass field; linear equilibrium surface response shown with explicit magnification. A static gravity experiment, not an ocean forecast or disruption simulation.

Start with the essentials

Focus question: Why does the far side stretch away while gravity still pulls inward?
One-sentence intuition: Tides depend on differences in gravity, not on the common acceleration.

Core mathematical model

Exact differential acceleration

\Delta\mathbf a(\mathbf r)=\mu_2\left[\frac{d\hat{\mathbf e}-\mathbf r}{|d\hat{\mathbf e}-\mathbf r|^3}-\frac{\hat{\mathbf e}}{d^2}\right]

The second term is the center acceleration. Both absolute pulls still point toward the companion.

Far-field tensor

\Delta\mathbf a_{\rm lin}=\frac{\mu_2}{d^3}\left[3(\mathbf r\cdot\hat{\mathbf e})\hat{\mathbf e}-\mathbf r\right]

Its three eigenvalues are proportional to two, minus one and minus one. The trace vanishes outside the point source.

Equilibrium surface, with an explicit display gain

U_t=-\frac{\mu_2}{|d\hat{\mathbf e}-\mathbf r|}+\frac{\mu_2}{d}+\frac{\mu_2\mathbf r\cdot\hat{\mathbf e}}{d^2},\quad h=-\frac{U_t}{g_0},\quad r_{\rm drawn}=R+G_hh

The first-order surface response is evaluated on the undeformed unit sphere. Changing the gain does not change the physical height or force.

Common difficulties

Outward is relative

Typical misconceptionThe far-side gravity points away from the companion.

Better mental modelSwitch to absolute gravity. The far side is attracted toward the companion, more weakly than the center.

Run the experiment

01
Two outward sides
Compare Near side and Far side using Differential gravity.
What to observe: Both radial components are positive even though both absolute pulls point toward the companion.
02
Squeezing sideways
Choose Sideways, then drag the probe to another latitude.
What to observe: The differential field compresses transverse directions. The force diagram and probe readouts agree.
03
Test distance and display gain
Double the distance. Compare Linear approximation, then change only the surface magnification.
What to observe: The leading tide weakens eightfold. Linear-field error shrinks; display magnification leaves physical readouts unchanged.

Primary sources: NOAA · Tides