Question

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

Answer 1

Answers:

1) 2951.39 m

2) 0.000952 m

3) $8.765(10)^{16}kg$

Explanation:

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

$RBH=\frac{2GM}{c^{2}}$   (1)

Where:

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ 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:

$RBH_{Sun}=\frac{2GM_{Sun}}{c^{2}}$   (2)

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

$RBH_{Mars}=\frac{2GM_{Mars}}{c^{2}}$   (5)

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

$M=\frac{RBH c^{2}}{2G}$   (8)

Where $RBH=1.30(10)^{-10}m$

Solving (8) with the known values:

$M=\frac{(1.30(10)^{-10}m)(3(10)^{8}m/s)^{2}}{2(6.674(10)^{-11}m^{3}/kgs^{2}}$   (9)

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