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GM⊕=3.986×1014 m3 s−2GM_\oplus = 3.986\times10^{14}\,\mathrm{m^3\,s^{-2}}

Why this formula appears here

For Earth modelled as a point mass, Φ(r)\Phi(r) = -GM/r gives a radial second derivative of magnitude 2GM/R3R^3 and, since Φ\Phi is harmonic in vacuum, transverse second derivatives of magnitude GM/R3R^3 with the opposite sign — a trace-free tensor in the exact ratio 2:{-1}:{-1} . With the standard gravitational parameter GM⊕M_\oplus = 3.986×\times10^{14}\,m3 s−2\mathrm{m^3\,s^{-2}} and mean radius R⊕R_\oplus = 6.371×\times10^{6}\,m\mathrm m , this gives GM⊕M_\oplus/R⊕3R_\oplus^3 ≈\approx 1.541×\times10^{-6}\,s−2\mathrm{s^{-2}} and a radial magnitude of 3.08×\times10^{-6}\,s−2\mathrm{s^{-2}} , matching the standard geodetic free-air gravity gradient of about 3.086×\times10^{-6}\,s−2\mathrm{s^{-2}} , or 0.3086 milligal per metre,…

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M⊕M_\oplus

Symbol M_oplus

MoM_oplus is part of the quantity the equation computes from the expression on the right.

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GM⊕=3.986×1014 m3 s−2GM_\oplus = 3.986\times10^{14}\,\mathrm{m^3\,s^{-2}}

Equation 129 · Evolutionary Physics

The Bend an Elevator Cannot Fake

This equation states an equality: the expressions on both sides have the same value under the article’s assumptions.

For Earth modelled as a point mass, Φ(r)\Phi(r) = -GM/r gives a radial second derivative of magnitude 2GM/R3R^3 and, since Φ\Phi is harmonic in vacuum, transverse second derivatives of magnitude GM/R3R^3 with the opposite sign — a trace-free tensor in the exact ratio 2:{-1}:{-1} . With the standard gravitational parameter GM⊕M_\oplus = 3.986×\times10^{14}\,m3 s−2\mathrm{m^3\,s^{-2}} and mean radius R⊕R_\oplus = 6.371×\times10^{6}\,m\mathrm m , this gives GM⊕M_\oplus/R⊕3R_\oplus^3 ≈\approx 1.541×\times10^{-6}\,s−2\mathrm{s^{-2}} and a radial magnitude of 3.08×\times10^{-6}\,s−2\mathrm{s^{-2}} , matching the standard geodetic free-air gravity gradient of about 3.086×\times10^{-6}\,s−2\mathrm{s^{-2}} , or 0.3086 milligal per metre,…

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