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

1836214271.77724 kg/m³

3268476.80175 m/s²

`-8.14198* 10^(-24)\ J`

Step-by-step explanation:

m = Mass of Sun =
`1.989* 10^(30)\ kg`

r = Radius of Earth = 6371000 m

Volume of Earth

`V=(4)/(3)\pi r^3\\\Rightarrow V=(4)/(3)\pi 6371000^3`

Density is given by

`\rho=(M)/(V)\\\Rightarrow \rho=(1.989* 10^(30))/((4)/(3)\pi 6371000^3)\\\Rightarrow \rho=1836214271.77724\ kg/m^3`

Density of the Sun would be 1836214271.77724 kg/m³

Acceleration due to gravity is given by

`g=(GM)/(r^2)\\\Rightarrow g=(6.67* 10^(-11)* 1.989* 10^(30))/(6371000^2)\\\Rightarrow g=3268476.80175\ m/s^2`

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

Potential energy is given by

`U=-(GMm)/(r)\\\Rightarrow U=-(6.67* 1.989* 10^(30)* 3.91)/(6371000)\\\Rightarrow U=-8.14198* 10^(-24)\ J`

The gravitational potential energy associated with the object is
`-8.14198* 10^(-24)\ J`