PartA Large stars can explode as they finish burning their nuclear fuel, causing
ID: 2032590 • Letter: P
Question
PartA Large stars can explode as they finish burning their nuclear fuel, causing a supenova. The explosion blows away the outer layers of the star. According to Newton's third law, the forces that push the outer layers away have reaction forces that are inwardly directed on the core of the star. These forces compress the core and can cause the core to undergo a gravitational collapse. The gravitational forces keep pulling all the matter together tighter and tighter, crushing atoms out of existence. Under these extreme conditions, a proton and an electron can be squeezed together to form a neutron. If the collapse is halted when the neutrons all come into contact with each other, the result is an object called a neutron star an entire star consisting of solid nuclear matter. Many neutron stars rotate about their axis with a period of1 s and, as they do so, send out a pulse of electromagnetic waves once a second. These stars were discovered in the 1960s and are called pulsars. Consider a neutron star with a mass equal to the sun, a radius of 19 km, and a rotation period of 1.0 s. What is the speed of a point on the equator of the star? Express your answer to two significant figures and include the appropriate units. ratValue Units Urot Submit Part B What is gat the surface of this neutron star? Express your answer to three significant figures and include the appropriate units. gValue UnitsExplanation / Answer
(a)
speed = circumference / period
v = 2?r / T
v = 2?( 19000 m ) / ( 1 sec )
v = 119380.5 m/s
(b)
a = GM / r²
a = ( 6.673 × 10^-11 N(m/kg)² )( 1.9891 × 10^30 kg ) / ( 19000 m )²
a = 3.68 × 10^11 m/s²
(c)
f = ma
f = ( 1 kg )( 3.68 × 10^11 m/s² )
f = 3.68 × 10^11 N
(d)
centripetal acceleration = gravitational acceleration
v²/r = GM / r²
v = ?( GM / r )
frequency = speed / circumference
= v / 2?r
= ?( GM / r ) / 2?r
= ?[ ( 6.673 × 10^-11 N(m/kg)² )( 1.9891 × 10^30 kg ) / ( 20000 m ) ] / 2?( 20000 m )
= 648 rev/sec
(e)
period = 1 / frequency = 1sec
2?r / ?( GM / r ) = 1 sec
2?r = ( 1 sec )?( GM / r )
4?²r² = ( 1 s² )( GM / r )
r^3 = ( 1 s² )( GM / 4?² )
r^3 = ( 1 s² )[ ( 6.673 × 10^-11 N(m/kg)² )( 1.9891 × 10^30 kg ) / 4?² ]
r^3 = 3.362 × 10^18 m^3
r = 1.498 × 10^6 m