Understand the relationship
The reasoning behind the result
The two-fifths power is a perturbation scale
rL/a=q^(2/5), where q=m/M
The conventional Laplace sphere-of-influence scale compares the relative importance of perturbations when describing a small test object's motion with different central bodies. Here M is the large primary mass, m is the smaller secondary mass and a is the assumed circular outer-orbit radius of the secondary around the primary.
This is not the point where two direct gravitational forces become equal, and gravity does not stop at the displayed radius. The sphere is an approximate scale used in simplified hierarchical orbital descriptions, not a sharp material boundary or a proof of a test object's actual trajectory.
The inverse changes which assumption is unknown
q=(rL/a)^(5/2); a=rL/q^(2/5)
If the outer orbit and desired Laplace scale are specified, their dimensionless ratio determines the mass ratio required by this formula. If mass ratio and Laplace scale are specified, dividing by q to the two-fifths power gives the required outer-orbit radius.
A desired Laplace radius at least as large as the outer orbit would imply q at least one and conflicts with the smaller-secondary interpretation. It is rejected. The inverse does not create an observed body mass or an actual orbital configuration; it states what the chosen approximate scale relationship requires.
Mass and distance scale differently
At fixed mass ratio, doubling the outer-orbit radius doubles the Laplace radius. At fixed outer-orbit radius, multiplying the mass ratio by ten multiplies the scale by 10^0.4, approximately 2.51189. A tenfold change in radius fraction would require a 10^2.5, approximately 316.228-fold change in mass ratio.
The graph places log10(q) on the horizontal axis and the radius fraction on the vertical axis. A horizontal increase of one therefore means a tenfold mass-ratio increase, not adding one to the ratio. Actual scenario markers and full dimensional radii remain available in the result table.
Laplace and Hill radii answer different model questions
A Hill approximation concerns a different three-body stability and tidal geometry and uses a cube-root dependence. It must not be substituted for this Laplace two-fifths scale or described as numerically identical. This workflow intentionally keeps the two models separate.
Real retention and stability depend on orbital orientation, eccentricity, phase, other bodies and the trajectory itself. This calculation assumes a circular outer scale and supplies no eccentric-orbit correction, capture condition, escape boundary or maximum stable satellite distance.
Follow the numbers
A one-percent Laplace radius fraction
- Let the outer circular-orbit radius be a = 1,000,000 km and the smaller-to-larger mass ratio be q = 10⁻⁵.
- The radius fraction is q^(2/5) = (10⁻⁵)^0.4 = 10⁻² = 0.01. The Laplace scale is 1,000,000 × 0.01 = 10,000 km.
- In reverse, (10,000 / 1,000,000)^(5/2) = 0.01^2.5 = 10⁻⁵, recovering the entered ratio. If M were independently known to be 10³⁰ kg, the corresponding m would be 10²⁵ kg.
The scale arithmetic is internally reversible. It does not establish a 10,000 km stable-satellite boundary.
Quick guide
How to use this calculator
- Choose known masses, a known mass ratio or one of the inverse scale calculations.
- Use an assumed circular outer orbit for the smaller body's motion around the primary. All radius values are centre-to-centre scale quantities.
- Keep the secondary mass much smaller than the primary for the hierarchical interpretation. A third test object's mass is neglected.
- Read the Laplace radius alongside its fraction of the outer orbit and the entered or inferred mass ratio. No radius here establishes that a satellite is stable or retained.
Calculation method
Calculation and interpretation
Make the assumptions, radius scale and inverse relationships of the Laplace approximation inspectable without treating them as satellite-stability boundaries.
rL=a(m/M)^(2/5); q=m/M; q=(rL/a)^(5/2); a=rL/q^(2/5).
Worked example
A one-percent Laplace radius fraction
The scale arithmetic is internally reversible. It does not establish a 10,000 km stable-satellite boundary.
rL=a(m/M)^(2/5); q=m/M; q=(rL/a)^(5/2); a=rL/q^(2/5).
Supported inputs
Precision and limits
Hierarchical circular approximation
The physical interpretation requires m much smaller than M and a negligible third test-object mass. The arithmetic accepts 0<m/M<1 without assigning an accuracy to that approximation. Comparable masses do not satisfy the hierarchy.
No actual dynamical boundary
The scale is not a direct-force equality, gravity cutoff, stable-orbit guarantee, capture rule or escape decision. Actual three-body trajectories require more information.
No eccentric outer-orbit or Hill substitution
The entered a is an assumed circular outer-orbit radius. Eccentric-orbit variation and Hill-sphere arithmetic are separate models and are not inferred here.
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