Question

After the Sun exhausts its nuclear fuel, its ultimate fate will be to collapse to a white dwarf state. In this state, it would have approximately the same mass as it has now, but its radius would be equal to the radius of the Earth.
(a) Calculate the average density of the white dwarf.
kg/m3

(b) Calculate the surface free-fall acceleration.
m/s2

(c) Calculate the gravitational potential energy associated with a 3.91-kg object at the surface of the white dwarf.
J

Answer 1

Answer:

(a) $\rho= 1.84 \times 10^9\, kg.m^{-3}$

(b) $g=3.27\times 10^6 m.s^{-2}$

(c) $U=-8.14\times 10^{13} J$

Explanation:

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

$M_s = 1.989\times 10^{30}\,kg$

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

$r = 6.37\times 10^6m$

(a)

∴Density

$\rho= \frac{M_s}{\frac{4\pi .r^3}{3} }$

$\rho= \frac{1.989\times 10^{30}}{\frac{4\pi \times (6.37\times 10^6)^3}{3} }$

$\rho= 1.84 \times 10^9\, kg.m^{-3}$

(b)

The free fall acceleration is:

$g=G.\frac{M_s}{r^2}$

$g=6.67\times 10^{-11}\times \frac{1.989\times 10^{30}}{(6.37\times 10^6)^2}$

$g=3.27\times 10^6 m.s^{-2}$

(c)

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

$U=-G.\frac{M_s.M}{r}$

$U=-6.67\times 10^{-11}\times \frac{1.989\times 10^{30}\times 3.91}{6.37\times 10^6}$

$U=-8.14\times 10^{13} J$