## Tuesday, 15 January 2013

### Rapid rotation o neutron star

Neutron stars spin very fast, lets fid in what time it completes on revolution.

$Formula P_{ns} = P_c(\frac{R_{ns}}{R_c})^2$

$P_{ns} = period of neutron star$

$P_c = period of core$

$R_{ns} = radius of neutron star$

$R_c = Radius of core$

Ans::

Cross-section of neutron star. Densities are in terms of ρ0 the saturation nuclear matter density, where nucleons begin to touch.

So the radius of the core will be taken as 3Km and radius of neutron star is 17Km, period of core is say 0.000007seconds

$P_{ns} = 0.0007(\frac{17}{3})^2 = 0.000335sec$

now
$R_c = 0.7Km, R_{ns} = 20Km, P_c=3.6e-6sec$

so

$P_{ns} = 3.6e-6(\frac{20}{0.7})^2 = 0.0029sec$

Now I exagerate:

$R_{ns} = 15.5Km, R_c = 0.06Km, P_c = 0.000000006sec$

so:

$P_{ns} = 0.000000006(\frac{15.5}{0.06})^2 = 0.0004sec$

## Friday, 11 January 2013

### sun

The Sun, in its T-Tauri phase, may have been losing mass at a rate of 108M/yr for up to ten million years, ending with a mass M. As a main sequence star, it loses mass at a rate of about 2×1014M/yr for ten billion years. As a red giant, it may lose up to 28% of the remaining mass. Estimate, in terms of M, the mass at the start of the T-Tauri phase, the mass of the remaining star at the end of the red giant phase. Round to two significant figures.

Yayy!! Done

First I found Mass at the

T-Tauri stage by

$M_{T-Tauri}=((M_{lost 1}*T_{1})*M_{\odot})+M_{\odot}$

where

$M_{lost 1} = 1*10^-8M_{\odot}/yr$

and

$T_{1}=1*10^{10}years$

and

$M_{\odot} = 1.989*10^{30}kg$

Then I found the mass when sun is a main sequence by

$M_{main.sequence} = ((M_{lost2}*T_{2})*M_{\odot})-M_{\odot}))$

where

$M_{lost2} = 2*10^{-14}M_{\odot}/yr$

and

$T_2=1*10^{10}years$

Finally I find the mass when sun is red giant by

$M_{red.giant} =(28/100*M_{main.sequence})-M_{main.sequence}$

then I finnally divide the

$\frac{M_{red.giant}}{T-Tauri}$