Question

Black holes have a special boundary called the event horizon, beyond which not even something traveling at the speed of light can escape. Use what you know about escape velocity to find the event horizon distance for a 5 solar mass (1 Msolar = 2x1030 kg) black hole. Your answer will be in meters.

Answer 1

30 Kg Kg Kg = r mass of black hole =m=s Msolar where 30 1 M solar = 2 x 10² = m=5 X 2 X 100 m= 10 X 1030 kg = m= I x 703) Let event horizon distance Event horizon distance is distance between center of black hole and even horizon. Now, escape relocity due to black hole at event horizon by 26 m VE = 0 은 is giren こ - 1 3 where his Universal gravitational Constant and G 6. 674 X10-1) т kg s-2 Let speed of light =C c = 3 x 108 mi) Now at event horizon escape relocity speed of light VE=C 2om =C - > r 2 cm 2 Я. 31 = cz 20 2 m3 s =c2 ( By 3 yaring both sides) -11 2 Gm = e = QX6-674X10 m3 Ky '5² x 1x10 kg r 3 8/08 m š!)? 13.348 x 10 r > x= 1.4831X10, 9 x 1016 m2 s 2 e ñ 1.5 X loum. Hence, event horizon distance for black hole is 1.5 Xlum > 4 = -2 =)