(a) We know that potential energy E = GMm/r, and "M" is the mass of
the stellar core, 1.4 solar masses.  Since it's the core itself that's
collapsing, we can say that "m" is also 1.4 solar masses.  Thus, if the
starting average radius is 10000 km (10^7 m) and the ending average
radius is 10 km (10^4 m), then the amount of energy released is the
difference between the two:

start E = (6.67 * 10^-11) * (1.4 * 2 * 10^30)^2 / 10^7 = 5.2 * 10^43 J

end E = (6.67 * 10^-11) * (1.4 * 2 * 10^30)^2 / 10^4 = 5.2 * 10^46 J

released E = (5.2 * 10^46 J) - (0.0052 * 10^46 J) = 5.2 * 10^46 J


(b) The supernova has a luminosity of 

L = (5.2 * 10^46 J) / (10 s) = 5.2 * 10^45 Watts.  This is

  (5.2 * 10^45 W)
  --------------- = 1.3 * 10^19 times (thirteen billion billion) times
  (3.9 * 10^26 W)

more luminous than the Sun!


(c) If one percent of the energy from 3(a) comes out as kinetic energy,
that's (0.01 * 5.2 * 10^46 J) = 5.2 * 10^44 J.  The stellar matter is
ten solar masses, or (10 * 2 * 10^30 kg) = 2 * 10^31 kg.  The kinetic
energy of the system is thus

  E = (1/2)mv^2 = 5.2 * 10^44 J = 0.5 * 2 * 10^31 kg * v^2

  v = sqrt(5.2 * 10^44 / 10^31) = 7.2 * 10^6 m/s

This matter would travel 1 AU (the distance between Earth and the Sun) in

  (1.5 * 10^11 m) / (7.2 * 10^6 m/s) = 21000 s or about 6 hours.

Don't worry, the Sun won't go supernova; but if it did, at least we'd have
a little time to prepare.