Respuesta :
Answer:
(I). The Schwarzschild radius is [tex]2.94\times10^{8}\ km[/tex]
(II). The Schwarzschild radius is 17.7 km.
(III). The Schwarzschild radius is [tex]1.1\times10^{-7}\ km[/tex]
(IV). The Schwarzschild radius is [tex]7.4\times10^{-29}\ km[/tex]
Explanation:
Given that,
Mass of black hole [tex]m= 1\times10^{8} M_{sun}[/tex]
(I). We need to calculate the Schwarzschild radius
Using formula of radius
[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]
Where, G = gravitational constant
M = mass
c = speed of light
Put the value into the formula
[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}[/tex]
[tex]R_{g}=2.94\times10^{8}\ km[/tex]
(II). Mass of block hole [tex]m= 6 M_{sun}[/tex]
We need to calculate the Schwarzschild radius
Using formula of radius
[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]
Put the value into the formula
[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}[/tex]
[tex]R_{g}=17.7\ km[/tex]
(III). Mass of block hole m= mass of moon
We need to calculate the Schwarzschild radius
Using formula of radius
[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]
Put the value into the formula
[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}[/tex]
[tex]R_{g}=1.1\times10^{-7}\ km[/tex]
(IV). Mass = 50 kg
We need to calculate the Schwarzschild radius
Using formula of radius
[tex]R_{g}=\dfrac{2MG}{c^2}[/tex]
Put the value into the formula
[tex]R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}[/tex]
[tex]R_{g}=7.4\times10^{-29}\ km[/tex]
Hence, (I). The Schwarzschild radius is [tex]2.94\times10^{8}\ km[/tex]
(II). The Schwarzschild radius is 17.7 km.
(III). The Schwarzschild radius is [tex]1.1\times10^{-7}\ km[/tex]
(IV). The Schwarzschild radius is [tex]7.4\times10^{-29}\ km[/tex]
The energy conservation allows to find the Schwarschild radius for several bodies of different masses are:
1) Black hole quasar is: r = 2.9 10⁸ km
2) Blsck hole supernove is: r = 17.7 km
3) Mini black hole is: r = 1.1 10⁻⁷ km
4) Human body is: r= 7 10⁻²⁹ km
The schwarschild radius is defined as the distance from a black hole center at radius which the escape velocity is equal to the light speed, in some cases it is also called the event horizon.
Let's use Newton's second law where force is the universal law of attraction and acceleration is centripetal.
F = ma
F = [tex]G \frac{Mm}{r^2}[/tex]
Where F is the force, M the mass of the black hole, m the handle of the body, r the radius and v the speed of the body.
The energy of the gravitational field is
F = [tex]- \frac{dU}{dr }[/tex]
U = [tex]-G \frac{Mm}{r}[/tex]
Let's use conservation of energy
Em₀ = K + U = ½ m v² - [tex]G \frac{Mm}{r}[/tex]
In infinity the energy
Em_f = 0
energy is conserved
Em₀ = Em_f
½ m v² - [tex]G \frac{Mm }{r}[/tex] = 0
r = [tex]\frac{2GM}{v^2}[/tex]
From the definition of the Schwarschild radius this speed is equal to the light speed
v = c
r = [tex]\frac{2GM}{c^2 }[/tex]
They ask to calculate the radius for several cases of different mass, claculate the constant value
V = [tex]\frac{2 \ 6.67 \ 10^{-11} }{(3 \ 10^8) ^2 }[/tex]
V = 1.482 10⁻²⁷
1) A black hole of mass M = 1 10⁸ [tex]M_{sum}[/tex]
The tabulated mass of the sun is [tex]M_{sum}[/tex] = 1.989 10³⁰ kg
Let's substitute
r = 1.482 10⁻²⁷ 1 10⁸ 1.989 10³⁰
r = 2.94 10⁸ km
With two significant figures
r = 2.9 10⁸ km
2) A black hole of mass M = 6 [tex]M_{sum}[/tex]
r = 1.482 10⁻²⁷ 6 1.989 10-30
r = 17.7 km
3) a mini black hole with the mass of the moon
Tabulated mass of the moon M = 7.35 10²² kg
r = 1.482 10⁻²⁷ 7.35 10²²
r = 1.1 10⁻⁷ km
4) A person of M = 50 kg
r = 1.482 10⁻²⁷ 50
r= 7 10-29 km
In conclusion using the conservation of energy we can find the Schwarschild radius for several bodies of different masses are:
1) Black hole quasar is: r = 2.9 10⁸ km
2) Blsck hole supernove is: r = 17.7 km
3) Mini black hole is: r = 1.1 10⁻⁷ km
4) Human body is: r= 7 10⁻²⁹ km
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