An astronaut of mass m is falling head first toward a black hole\'s event horizo

Question

An astronaut of mass m is falling head first toward a black hole's event horizon. The black hole has mass M; the astronaut's height is h (distance from helmet to boots). She is a distance r away from the black hole's center (r h). Using the approximation that half the astronaut's mass is located at her helmet and half at her boots, show that the difference in force between her helmet and boots is given by1 If the astronaut is, instead, a distance d from the event horizon, and the event horizon is the distance from the black hole's center at which v_esc = c, show that the tidal force is given by

Explanation / Answer

part a:

mass concentrated at head=mass concentrated at boots=m/2

distance of head from the blackhole=r

then force on head=G*M*0.5*m/r^2

distance of boot from blackhole=r+h

then force on boot=G*M*0.5*m/(r+h)^2

then difference between forces=

0.5*G*M*m*((1/r^2)-(1/(r+h)^2)

=0.5*G*M*m*((r+h)^2-r^2)/(r^2*(r+h)^2)

now, (r+h)^2-r^2=r^2+h^2+2*r*h-r^2=h^2+2*r*h

as r>>h,

h^2<<2*r*h

hence (r+h)^2-r^2=2*r*h

as r>>h, r+h=r

hence total expression for difference of forces is given by

T=0.5*G*M*m*2*r*h/r^4

=G*M*m*h/r^3

part B:

let radius of blackhole's event horizon(with blackhole's center as the center for event horizon circle too)

be r1.

as escape velocity=sqrt(2*G*M/r1)

==>c=sqrt(2*G*M/r1)

==>c^2=2*G*M/r1

==>r1=2*G*M/c^2

total distance of the astronaut from the blackhole's center=d+r1

=d+(2*G*M/c^2)

using this value as r in expression obtained from part A,

we get tidal force=G*M*m*h/(d+(2*G*M/c^2))^3

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