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raspberriesareyummy , 11 months ago (edited 11 months ago) Accretion discs can be large enough that I am pretty sure a human body wouldn’t be torn apart at that distance (at least the outer bits) by the difference in gravity across it’s length. In the linked article about the supermassive black hole at the center of the Milky Way, we’re talking 1000 astronomic units, so 1.5 * 10^14 meters. The current value of this black hole’s mass is estimated at ca. 4.154±0.014 million solar masses. So let’s calculate the equivalent distance from the sun in terms of gravitational force on an object at the outer edge of the accretion disk: F_sun = C * (R_equivalent)^-2 * m_object F_black_hole = C * 4.15*10^6 * (R_accretion_disk)^-2 * m_object where C equals the gravity constant times the mass of our sun. ==> C * (R_equivalent)^-2 * m_object = C * 4.15*10^6 * (R_accretion_disk)^-2 * m_object divide by C and m_object: <=> (R_equivalent)^-2 = 4.15*10^6 * (R_accretion_disk)^-2 invert: <=> R_equivalent^2 = (1/4.15) * 10^-6 * (R_accretion_disk)^2 ==> R_equivalent^2 ~= 0.241 * 10^-6 * (R_accretion_disk)^2 square root (only the positive solution makes sense here): ==> R_equivalent ~= 0.491 * 10^-3 * R_accretion_disk with R_accretion_disk = 1000 astronomic units = 10^3 AU <=> R_equivalent ~= 0.491 * 10^-3 * 10^3 AU <=> R_equivalent ~= 0.491 AU Unless I have a mistake in my math, I sincerely hope you will agree that the gravitational field (tidal forces) of the sun is very much survivable at a distance of 0.491 astronomical units - especially since the planet Mercury approaches the sun to about 0.307 AUs in its perihelion.
Accretion discs can be large enough that I am pretty sure a human body wouldn’t be torn apart at that distance (at least the outer bits) by the difference in gravity across it’s length. In the linked article about the supermassive black hole at the center of the Milky Way, we’re talking 1000 astronomic units, so 1.5 * 10^14 meters.
The current value of this black hole’s mass is estimated at ca. 4.154±0.014 million solar masses.
So let’s calculate the equivalent distance from the sun in terms of gravitational force on an object at the outer edge of the accretion disk:
F_sun = C * (R_equivalent)^-2 * m_object
F_black_hole = C * 4.15*10^6 * (R_accretion_disk)^-2 * m_object
where C equals the gravity constant times the mass of our sun.
==> C * (R_equivalent)^-2 * m_object = C * 4.15*10^6 * (R_accretion_disk)^-2 * m_object
divide by C and m_object:
<=> (R_equivalent)^-2 = 4.15*10^6 * (R_accretion_disk)^-2
invert:
<=> R_equivalent^2 = (1/4.15) * 10^-6 * (R_accretion_disk)^2
==> R_equivalent^2 ~= 0.241 * 10^-6 * (R_accretion_disk)^2
square root (only the positive solution makes sense here):
==> R_equivalent ~= 0.491 * 10^-3 * R_accretion_disk
with R_accretion_disk = 1000 astronomic units = 10^3 AU
<=> R_equivalent ~= 0.491 * 10^-3 * 10^3 AU
<=> R_equivalent ~= 0.491 AU
Unless I have a mistake in my math, I sincerely hope you will agree that the gravitational field (tidal forces) of the sun is very much survivable at a distance of 0.491 astronomical units - especially since the planet Mercury approaches the sun to about 0.307 AUs in its perihelion.