# After the sun exhausts its nuclear fuel, its ultimate fate may be to collapse to a white dwarf state.

###### Question:

## Answers

To calculate the density of the white dwarf we need the mass of the Sun and the radius of the Earth:

- Sun mass: [tex]M=1.99 \cdot 10^{30}kg[/tex]

- Earth radius: [tex]r=6.37 \cdot 10^6 m[/tex]

Assuuming the dwarf to be a perfect sphere, its volume is [tex]V= \frac{4}{3} \pi r^3[/tex].

The average density is given by

[tex]d= \frac{M}{V}[/tex]

so, substituting we find

[tex]d= \frac{M}{ \frac{4}{3}\pi r^3 }=1.83 \cdot 10^9 kg/m^3[/tex]

(a) [tex]\rho= 1.84 \times 10^9\, kg.m^{-3}[/tex]

(b) [tex]g=3.27\times 10^6 m.s^{-2}[/tex]

(c) [tex]U=-8.14\times 10^{13} J[/tex]

Explanation:

The white dwarf will have a mass almost the same mass as the sun:

[tex]M_s = 1.989\times 10^{30}\,kg[/tex]

but will have a smaller radius comparable to earth’s :

[tex]r = 6.37\times 10^6m[/tex]

(a)

∴Density

[tex]\rho= \frac{M_s}{\frac{4\pi .r^3}{3} }[/tex]

[tex]\rho= \frac{1.989\times 10^{30}}{\frac{4\pi \times (6.37\times 10^6)^3}{3} }[/tex]

[tex]\rho= 1.84 \times 10^9\, kg.m^{-3}[/tex]

(b)

The free fall acceleration is:

[tex]g=G.\frac{M_s}{r^2}[/tex]

[tex]g=6.67\times 10^{-11}\times \frac{1.989\times 10^{30}}{(6.37\times 10^6)^2}[/tex]

[tex]g=3.27\times 10^6 m.s^{-2}[/tex]

(c)

Potential energy of the 3.91 kg object at the surface of the white dwarf is :

[tex]U=-G.\frac{M_s.M}{r}[/tex]

[tex]U=-6.67\times 10^{-11}\times \frac{1.989\times 10^{30}\times 3.91}{6.37\times 10^6}[/tex]

[tex]U=-8.14\times 10^{13} J[/tex]

1836214271.77724 kg/m³

3268476.80175 m/s²

[tex]-8.14198\times 10^{-24}\ J[/tex]

Explanation:

m = Mass of Sun = [tex]1.989\times 10^{30}\ kg[/tex]

r = Radius of Earth = 6371000 m

Volume of Earth

[tex]V=\frac{4}{3}\pi r^3\\\Rightarrow V=\frac{4}{3}\pi 6371000^3[/tex]

Density is given by

[tex]\rho=\frac{M}{V}\\\Rightarrow \rho=\frac{1.989\times 10^{30}}{\frac{4}{3}\pi 6371000^3}\\\Rightarrow \rho=1836214271.77724\ kg/m^3[/tex]

Density of the Sun would be 1836214271.77724 kg/m³

Acceleration due to gravity is given by

[tex]g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{6371000^2}\\\Rightarrow g=3268476.80175\ m/s^2[/tex]

Acceleration due to gravity on the Sun would be 3268476.80175 m/s²

Potential energy is given by

[tex]U=-\frac{GMm}{r}\\\Rightarrow U=-\frac{6.67\times 1.989\times 10^{30}\times 3.91}{6371000}\\\Rightarrow U=-8.14198\times 10^{-24}\ J[/tex]

The gravitational potential energy associated with the object is [tex]-8.14198\times 10^{-24}\ J[/tex]