Respuesta :
Answer:
[tex]v^2_{esc} = \sqrt{ \frac{2GM}{R}}[/tex]
[tex]\Delta x =\sqrt{ \frac{Gh}{c^3}[/tex]
Explanation:
a) Let the average kinetic energy = gravitational potential energy
Then; by equation :
[tex]\frac{1}{2}mv^2_{esc} = \frac{GmM}{R}[/tex]
here;
m = mass of the confined photons
[tex]v_{esc}[/tex] = escape velocity
M = mass of the star
R = radius
[tex]\frac{1}{2}v^2_{esc} = \frac{GM}{R}[/tex]
[tex]v^2_{esc} = \frac{2GM}{R}[/tex]
[tex]v^2_{esc} = \sqrt{ \frac{2GM}{R}}[/tex]
b) Let [tex]v_{esc} = c[/tex] (where c = speed of light)
[tex]c = \sqrt{ \frac{2GM}{R}}[/tex]
[tex]c^2 ={ \frac{2GM}{R}}[/tex]
where
R =[tex]\Delta x[/tex]
[tex]c^2 ={ \frac{2GM}{\Delta x}}[/tex]
By using the expression ;
[tex]E = mc^2[/tex] and [tex]E = \rho \ c[/tex]
Then;
[tex]mc^2 = \rho c[/tex]
[tex]m = \frac{\rho}{c}[/tex]
[tex]m = \frac{h}{2 \Delta xc}[/tex] since ( [tex]\rho = \frac{h}{2 \Delta x}[/tex])
Replacing m for M in [tex]c^2 ={ \frac{2GM}{\Delta x}}[/tex]; we have:
[tex]c^2 = \frac{2 \ G}{\Delta x} \frac{h}{2 \Delta x}[/tex]
[tex]c^2 = \frac{Gh}{\Delta x^2 c}[/tex]
[tex]\Delta x^2 = \frac{Gh}{c^3}[/tex]
[tex]\Delta x =\sqrt{ \frac{Gh}{c^3}[/tex]