Respuesta :
Answer:
(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]
