91Ó°ÊÓ

The Schwarzschild radius is given by,

\({R_s} = \frac{{2GM}}{{{c^2}}}\)

Here\({R_s}\)is the Schwarzschild radius,\(G\)is the universal gravitational constant,\(M\)is the mass of the black hole,\(c\)is the speed of the light.

(a)

The Schwarzschild radius of a supermassive black hole is evaluated.

It has a mass,\({\rm{M = 1}}{{\rm{0}}^{\rm{9}}}{{\rm{M}}_{{\rm{sun}}}}\), then the radius is calculated by,

\(\begin{align}{R_s} &= \frac{{2GM}}{{{c^2}}}\\ &= {\rm{ }}\frac{{{\rm{2 \times 6}}{\rm{.67 \times 1}}{{\rm{0}}^{{\rm{ - 11}}}}\frac{{{\rm{N}}{{\rm{m}}^{\rm{2}}}}}{{{\rm{k}}{{\rm{g}}^{\rm{2}}}}}{\rm{ \times 1}}{{\rm{0}}^{\rm{9}}}{\rm{ \times 2 \times 1}}{{\rm{0}}^{{\rm{30}}}}{\rm{kg}}}}{{{{{\rm{(3 \times 1}}{{\rm{0}}^{\rm{8}}}\frac{{\rm{m}}}{{\rm{s}}}{\rm{)}}}^{\rm{2}}}}}{\rm{ }}\\ &= {\rm{ 2}}{\rm{.97 \times 1}}{{\rm{0}}^{{\rm{12}}}}{\rm{m}}\end{align}\)

Therefore, the radius of the object is\({\rm{2}}{\rm{.97 \times 1}}{{\rm{0}}^{{\rm{12}}}}{\rm{ m}}\).

(b)

The radius is obtained as:

\(\begin{align}{R_s} &= {\rm{ 2}}{\rm{.97 \times 1}}{{\rm{0}}^{{\rm{12}}}}{\rm{m}}\frac{{{\rm{1ly}}}}{{{\rm{9}}{\rm{.5 \times 1}}{{\rm{0}}^{{\rm{15}}}}{\rm{m}}}}{\rm{ }}\\ &= {\rm{ 3}}{\rm{.13 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ ly}}\end{align}\)

Therefore, the radius in light years is \({\rm{3}}{\rm{.13 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ ly}}\).