Answer:
The escape velocity of photon from its surface=445625.7m/s
Explanation:
We are given that
Radius, R=240 solar radii=[tex]1.67\times 10^{11} m[/tex]
1 solar radii=[tex]6.957\times 10^8 m[/tex]
Mass, m=125 solar mass=[tex]2.486\times 10^{32}kg[/tex]
1 solar mass=[tex]1.989\times 10^{30}kg[/tex]
We have to find the escape velocity of photon from its surface.
We know that escape velocity
[tex]v={\sqrt{\frac{2GM}{R}}[/tex]
Using the formula
[tex]v=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 2.486\times 10^{32}}{1.67\times 10^{11}}}[/tex]
[tex]v=445625.7m/s[/tex]
Hence ,the escape velocity of photon from its surface=445625.7m/s
