Published equation contexts

$M\approx4\times10^6\,M_\odot$

Why this formula appears here

To make this concrete rather than asymptotic, take the mass of Sagittarius A* as measured by the Event Horizon Telescope, M $\approx$ 4 $\times$ 10^6\, $M_\odot$ , giving a Schwarzschild radius $r_s$ $\approx$ 1.18 $\times$ 10^{10}\, $\mathrm m$ and a horizon-scale acceleration factor GM/ $r_s^2$ $\approx$ 3.80 $\times$ 10^{6}\, $\mathrm{m\,s^{-2}}$ [ 14 ] . At a clearance of just one kilometer above the horizon — a distance smaller than a millionth of the horizon’s own radius — f $\approx$ 8.46 $\times$ 10^{-8} and the required proper acceleration is already a $\approx$ 1.3 $\times$ 10^{10}\, $\mathrm{m\,s^{-2}}$ , more than a billion Earth gravities. Even at a clearance of 1.8 $\times$ 10^6\, $\mathrm{km}$ — about fifteen percent of the horizon’s own…

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M

Symbol M

M is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

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M_\odot

Symbol M_odot

$M_o$ dot is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

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How to interpret it

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Research cited beside this formula

[14] First Sagittarius A* Event Horizon Telescope Results. I. The Shadow of the Supermassive Black Hole in the Center of the Milky Way ↗

Published contexts (1)

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M\approx4\times10^6\,M_\odot

Equation 64 · Evolutionary Physics

A Horizon Is a Toll Booth, Not a Loophole

This mathematical expression combines the displayed quantities; its precise role follows from the surrounding article text.

To make this concrete rather than asymptotic, take the mass of Sagittarius A* as measured by the Event Horizon Telescope, M $\approx$ 4 $\times$ 10^6\, $M_\odot$ , giving a Schwarzschild radius $r_s$ $\approx$ 1.18 $\times$ 10^{10}\, $\mathrm m$ and a horizon-scale acceleration factor GM/ $r_s^2$ $\approx$ 3.80 $\times$ 10^{6}\, $\mathrm{m\,s^{-2}}$ [ 14 ] . At a clearance of just one kilometer above the horizon — a distance smaller than a millionth of the horizon’s own radius — f $\approx$ 8.46 $\times$ 10^{-8} and the required proper acceleration is already a $\approx$ 1.3 $\times$ 10^{10}\, $\mathrm{m\,s^{-2}}$ , more than a billion Earth gravities. Even at a clearance of 1.8 $\times$ 10^6\, $\mathrm{km}$ — about fifteen percent of the horizon’s own…

Equation guide → · Article →

M≈4×106 M⊙M\approx4\times10^6\,M_\odotM≈4×106M⊙​