How does this compare to the total binding energy, in ergs, of the original white draw? Does the white dwarf explode completely, or is some mass left over in the form of a highly concentrated remnant?
Let's begin.
First let's find the energy of the explosion. It's time for a classic equation:
\[E=Mc^2\]
We are told that only 0.1% of the mass is converted to energy, so:
\[E=0.001Mc^2\]
Let's plug and chug.
\[E=0.001(1.4M_{\odot})(3\times 10^{10}cm/s)^2\]
\[E=0.001(2.8\times 10^{33}g)(3\times 10^{10}cm/s)^2\]
\[\boxed{E=2.52\times 10^{51} ergs}\]
Now lets find the potential energy of the original system.
\[U=\frac{3GM^2}{5R}\]
\[U=\frac{3(6.7\times 10^{-8}cm^3 g^{-1}s^{-2})(2.8\times 10^{33}g)^2}{5(12\times 10^8 cm)}\]
\[\boxed{E=2.52\times 10^{51} ergs}\]
Now lets find the potential energy of the original system.
\[U=\frac{3GM^2}{5R}\]
\[U=\frac{3(6.7\times 10^{-8}cm^3 g^{-1}s^{-2})(2.8\times 10^{33}g)^2}{5(12\times 10^8 cm)}\]
\[\boxed{U=2.6\times 10^{50} ergs}\]
Thanks to approximations and estimations of mass consumption etc, these values are not equal, however, in the grand scheme of things, they are the same. This implies that Type Ia SNe leave nothing behind and the white draw is completely blown apart!
I worked on the problem with B. Brzycki, G. Grell, and N. James.
nice job.
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