HINT:#
The question asks us to determine the shortest distance from a black hole that energy could be emitted so that the radiation could escape the powerful gravitational influence of the black hole, and be detectable by the rest of the world.
HINT:#
The problem tells us that the mass of the black hole is equal to 109 solar masses, and that the size of Rs can be determined using an energy consideration.
HINT:#
If we consider a classical particle of mass m near a gravitational source the total energy of the particle is be the sum of its kinetic energy and potential energy If we consider a classical energy source emitting quantized bits of energy a finite distance from a black hole, it will emit photons of mass m2, in all directions. The total energy of each photon will be a sum of its kinetic energy and its potential energy.
ET = EK + EP
The kinetic energy of the particle will be given by: EK = 1/2m2v2
The potential energy of the particle will be given by: EP = -Gm1m2/R
where G is the universal gravitational constant.

Now, let's consider a particle emitted in the direction of an outsider observer. In other words, the particle is being emitted from a point in between the observer and the black hole, in the direction of the observer. Since the gravitational field is pointing away from the observer, and the particle is emitted toward the observer, the two energy terms have opposite signs. So,
ET = 1/2m2v2 - Gm1m2/R
Since we are looking for distance from the black hole at which the object is unable to escape from the the black hole, the sum of the two energies at Rs distance will be equal to zero. 0 = 1/2m2v2 - Gm1m2/R
Although it is strictly not correct to let the velocity, v equal to c, the speed of light, the answer agrees with that we would calculate if we took the effects of special relativity into account. 1/2m2c2 = Gm1m2/Rs
Solving for RS = 2Gm1/c2 By substituting the speed of light for v we can produce an accurate answer. Although, strictly speaking, this is not correct, as photons have no rest mass, but still potential energy.
HINT:#
G = 6.7 x 10-8 cm3/g*s2
m1 = mblack hole = 109 solar masses
c = 3.01 x 1010 cm/s
HINT:#
Let's use unit analysis:
RS = 2Gm1/c2
Since we have the mass of the black hole in terms of Solar Masses, and we are working with CGS (centimeter, grams, seconds) units we will have to convert the masses into grams.
m1 = mblack hole = 109 solar masses * 2 x 1033 grams/solar mass = 2 x 1042 g
HINT:#
RS = 2Gm1/c2
= (2)(6.7 x 10-8 cm3/g*s2)(2 x 1042 g)/(3.01 x 1010 cm/s)2 = 3 x 1014 cm
This is approximately the size of our solar system!

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