Question

#2 please !

Ro le Astronomy 1F03 Problem Set 3 Thursday, Nov 10th, 2019 Write your answers on a separate sheet and show your work 1. The protesproton chain effectively c erta hyre en into hem according to 4 He. A Helium wus weighs 6.64518 x 10 IT kg and Hydrogen mucha proton) welches 1.67262 x 10 Workout the change in a when I kg of Hydrogen is converted completely into Helium 4. Convert the change in mass Per burning G inte energy created, Joules per kg of Hydrogen, using Finstein's formula E m e. (For comparison line releases only 5 x 10 Joules per kg.) ASIA berts worte b) The rate that the sun burns in Joules per second is given by the solar luminosity, 4 x 10-Watts. Express dhe burning rate in terms of kg of Hydrogen converted per second. The mass of the Sun is 2 x 10ks How long can the sun burn at this rate? fode to, che eigande y ESE of meaning S het GO BEDA S 2. Stars spend most of their lives on the main sequence and burn an appreciable fraction of their total hydrogen to hellum in this period. In the following assume that stars spend their whole lives on the main sequence burning at a fixed luminosity and that all star burn the same fixed fraction of their total mans doing so. The Sun's lifetime is about 10 billion years. Use this to estimate the lifetimes of the following stars. Show working. a) A 0.2 solar mas, 0.01 solar luminosity red dwarf star b) A larm , 200 solar luminosity blue giant star 3. Stars move up and down the Hertsprung-Russell diagram as they evolve. Consider a small, main-sequence buiza a) The star Ascends the red giant branch, during which time its surface temperature drops by a factor of 3 and its radius increases by a factor of 50. Determine how luminous the star is now compared to its original luminosity, that is, determine D GLASE b) The star puffs off its outer layers as a planetary nebula exposing the compact core of the old star as a hot, new white dwarf star. The white dwarf's surface temperature is 5 times the original star's surface temperature but its radius is 100 times smaller. Determine TE DWAL at this stage. (NB: This is relative to the original star, not the red giant). 4. When a massive star is at the end of its life, the inner core that is perhaps 2 solar masses shrinks in radius from a radius of around 0.01 solar radii to become a neutron star with a radius of just 15 km. a) Calculate the gravitational potential energy of the core EPOTENTIAL = -0.6 before and after the collapse in Joules. b) Energy is conserved. Whenever something collapses under gravity, the decrease in the potential energy releases other forms of energy. Calculate the amount of energy liberated due to the formation of the neutron star. c) The liberated energy goes into expanding the outer layers of the star at very high speed, creating neutrinos and producing light which we see as a Supernova. Assume that just 1/1000th (0.1 percent of this energy emerges as light over a period of 100 days. Determine the luminosity of the star over this period in Watts. d) Convert the luminosity of the Supernova into solar luminosities. (For comparison the entire Milky Way alaxy has a luminosity of around 2 x 1010 Solar Luminosities.)

Answer 1

Life time of sun

10 10 yr 9 10 billion years TO

2)

For red dwarf star $M=0.2\,M_\odot \,L=0.01\,L_\odot$

Life time of star $\tau\propto \frac{M}{L}$

( M is mass of star, L is luminosity of star)

For Sun, life time is $\tau_\odot \propto \frac{M_\odot }{L_\odot }$

($M_{\odot }$ is solar mass , $L_{\odot }$ is luminosity of sun)

$\frac{\tau}{\tau_\odot }=\frac{M/L}{M_\odot /L_\odot }$

$\frac{\tau}{\tau_\odot }=\frac{M}{L}\,\frac{L_\odot }{M_{\odot }}$

$\frac{\tau}{\tau_\odot }=\frac{0.2M_\odot }{0.01L_\odot }\,\frac{L_\odot }{M_{\odot }}=20$

$\tau=20\,\tau_\odot=20*(10*10^9)=200*10^9\,yr=200\,\text{billion\,years}$

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For blue giant star,

$M=4\,M_\odot \,L=200\,L_\odot$

Life time of star $\tau\propto \frac{M}{L}$

( M is mass of star, L is luminosity of star)

For Sun, life time is $\tau_\odot \propto \frac{M_\odot }{L_\odot }$

($M_{\odot }$ is solar mass , $L_{\odot }$ is luminosity of sun)

$\frac{\tau}{\tau_\odot }=\frac{M/L}{M_\odot /L_\odot }$

$\frac{\tau}{\tau_\odot }=\frac{M}{L}\,\frac{L_\odot }{M_{\odot }}$

$\frac{\tau}{\tau_\odot }=\frac{4M_\odot }{200L_\odot }\,\frac{L_\odot }{M_{\odot }}=0.02$

$\tau=0.02\,\tau_\odot=0.02*(10*10^9)=0.2*10^9\,yr=0.2\,\text{billion\,years}$

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