Black Hole War

The mass of the sun after its density is increased to that of a black hole with Schwarzschild radius the same as the solar radius is given by $R_S=\frac{2MG}{c^2}$ , $M=\frac{R_{S}c^2}{2G}$ . where $R_s$ is the solar radius.

Then the Newtonian energy needed is dominated entirely by gravitational potential of $\frac{GMm}{R_A}$ , where $R_A$ is the distance between the sun and the Earth, and m is mass of the Earth. So substituting in M found earlier, E= $\frac{1}{2}\frac{R_S}{R_{Au}}mc^2$ .

Check that $E<<mc^2$ to ensure relativistic physics does not have to be took into account. As $\frac{R_S}{R_{Au}}$ is very small, $E<<mc^2$

Then converting this energy to mass via $E=Mc^2$ , just divide by $c^2$ , to get a mass of $\frac{1}{2}\frac{R_S}{R_{Au}}m$ , which comes out as $13.8*10^{21}$ , or 13.8 exatonnes =14 exatonnes to the nearest whole number.

By definition, the Schwarzschild radius is:

$R_s = \frac{2GM}{c^2}$

Isolating $M$ , which is sun's new mass

$M = \frac{R_s \cdot c^2}{2G}$

Calculating the Escape velocity for sun's new mass we have:

$v_e = \sqrt{\frac{2G M}{R}}$

By substituting $M$ :

$v_e = c\cdot \sqrt{\frac{R_s}{R}}$

Where $c$ is the Speed of Light in vacuum , $R_s$ is sun's radius (now its Schwarzschild radius) and $R$ is Earth's distance to the sun.

By substituting their known values ( $R_s = 6.975\cdot 10^8$ m, $R = 1.5\cdot 10^{11}$ m and $c = 299792458$ m/s )

$v_e \approx 20395754.94$ m/s

Which leads to a kinetic energy of:

$E_e = \frac{M_{Earth}v_e^2}{2}$

$E_e \approx 1.25 \cdot 10^{39}$ J

Each antimatter bomb will have an energy of:

$E_b = m_{b} c^2$ ( $m_b = 10^{21}$ kg)

$E_b = 8.99 \cdot 10^{37}$ J

So, it will be needed:

$\frac{E_e}{E_b} \approx 13.88 \rightarrow \fbox{14 bombs}$