Question

The Schwarzschild radius RBH for an object of mass M is defined as: Rbh = (2GM)/(c^2)

Where c is the speed of light and G is the universal gravitational constant. RBH gives the radius of the event horizon of a black hole with mass M (in other words, it gives the radius to which some amount of mass M would need to be compressed in order to form a black hole).

1) The mass of the Sun is about 1.99 × 1030 kg. What would be the radius of a black hole with this mass?

2)The mass of Mars is about 6.42 × 1023 kg. What would be the radius of a black hole with this mass?

3) Suppose you want to make a black hole that is roughly the size of an atom (take RBH = 1.30 × 10-10 m). What would be the mass M of such a black hole?

Answer 1

Answers:

1) 2951.39 m

2) 0.000952 m

3)
`8.765(10)^(16)kg`

Step-by-step explanation:

1) We know the Schwarzschild radius
`RBH` is given by the following equation:

`RBH=(2GM)/(c^(2))` (1)

Where:

`G=6.674(10)^(-11)(m^(3))/(kgs^(2))` is the Universal Gravitational Constant

`M` the mass of the black hole

`c=3(10)^(8)m/s` is the speed of light

Now, if we have a black hole with the mass of the Sun (
`M_(Sun)=1.99(10)^(30)kg`), its radius will be:

`RBH_(Sun)=(2GM_(Sun))/(c^(2))` (2)

`RBH_(Sun)=\frac{2(6.674(10)^(-11)(m^(3))/(kgs^(2)))(1.99(10)^(30)kg)}{{(3(10)^(8)m/s)}^(2)}` (3)

`RBH_(Sun)=2951.39m` (4) This is the radius of the black hole with the mass of the Sun

2) On the other hand, if the black hole has the mass of Mars (
`M_(Mars)=6.42(10)^(23)kg`), its radius will be:

`RBH_(Mars)=(2GM_(Mars))/(c^(2))` (5)

`RBH_(Mars)=\frac{2(6.674(10)^(-11)(m^(3))/(kgs^(2)))(6.42(10)^(23)kg)}{{(3(10)^(8)m/s)}^(2)}` (6)

`RBH_(Mars)=0.000952m` (7) This is the radius of the black hole with the mass of Mars

3) In this case, we have to isolate
`M` from (1):

`M=(RBH c^(2))/(2G)` (8)

Where
`RBH=1.30(10)^(-10)m`

Solving (8) with the known values:

`M=((1.30(10)^(-10)m)(3(10)^(8)m/s)^(2))/(2(6.674(10)^(-11)m^(3)/kgs^(2))` (9)

`M=8.765(10)^(16)kg` (10) This is the mass of the black hole