Chapter 7, Problem 36P

(a)

To determine

The average density of the white dwarf.

Answer to Problem 36P

The average density of the white dwarf is $1.83\times {10}^9{\text{kg/m}}^{\text{3}}$.

Explanation of Solution

Given info: The mass of the dwarf is $1.991\times {10}^{\text{30}}\text{kg}$ and the radius of the dwarf is $6.38\times {10}^6\text{m}$.

Explanation: The average density of the dwarf is defined as $\rho =M_{\text{Sun}}/V_{\text{Earth}}=M_{\text{Sun}}/\left[\left(4/3\right)\pi R_{\text{Earth}}^3\right]$ but in this case the mass of the sun remains constant after the collapse to a white dwarf state and its volume is reduced to the volume of the earth.

The formula for the average density of the white dwarf is,

$\rho =\frac{3M_{\text{dwarf}}}{4\pi R_{\text{dwarf}}^3}$

• $M_{\text{dwarf}}$ is mass of the dwarf.
• $R_{\text{dwarf}}$ is radius of the dwarf.

Substitute $1.991\times {10}^{\text{30}}\text{kg}$ for $M_{\text{dwarf}}$ and $6.38\times {10}^6\text{m}$ for $R_{\text{dwarf}}$ to find $\rho$.

$\begin{array}{c}\rho =\frac{3\left(1.991\times {10}^{30}\text{kg}\right)}{4\left(3.14\right){\left(6.38\times {10}^6\text{m}\right)}^3}\\ =1.83\times {10}^9{\text{kg/m}}^{\text{3}}\end{array}$

Thus, the average density of the dwarf is $1.83\times {10}^9{\text{kg/m}}^{\text{3}}$.

Conclusion:

Therefore, the average density of the dwarf is $1.83\times {10}^9{\text{kg/m}}^{\text{3}}$.

(b)

To determine

The surface free acceleration of the white dwarf.

Answer to Problem 36P

The surface free acceleration of the white dwarf is $3.26\times {10}^6\text{m/s}$.

Explanation of Solution

Given info: The gravitational constant is $6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$, the mass of the dwarf is $1.991\times {10}^{\text{30}}\text{kg}$, and the radius of the dwarf is $6.38\times {10}^6\text{m}$.

Explanation:

The formula for the surface free fall acceleration of the dwarf is,

$g=G\frac{M_{\text{dwarf}}}{R_{\text{dwarf}}^2}$

• $G$ is gravitational constant.
• $M_{\text{dwarf}}$ is mass of the dwarf.
• $R_{\text{dwarf}}$ is radius of the dwarf.

Substitute $6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $1.991\times {10}^{\text{30}}\text{kg}$ for $M_{\text{dwarf}}$, and $6.38\times {10}^6\text{m}$ for $R_{\text{dwarf}}$ to find $g$.

$\begin{array}{c}g=\left(6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\frac{\left(1.991\times {10}^{\text{30}}\text{kg}\right)}{{\left(6.38\times {10}^6\text{m}\right)}^2}\\ =3.26\times {10}^6{\text{m/s}}^{\text{2}}\end{array}$

Thus, the surface free fall acceleration of the dwarf is $3.26\times {10}^6{\text{m/s}}^{\text{2}}$.

Conclusion:

Therefore, the surface free fall acceleration of the dwarf is $3.26\times {10}^6{\text{m/s}}^{\text{2}}$.

(c)

To determine

The gravitational potential energy associated with $1.00\text{kg}$ object on the surface of the white dwarf.

Answer to Problem 36P

The gravitational potential energy associated with $1.00\text{kg}$ object on the surface of the white dwarf is $-2.08\times {10}^{13}\text{J}$.

Explanation of Solution

Given info The gravitational constant is $6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$, the mass of the dwarf is $1.991\times {10}^{\text{30}}\text{kg}$, the mass of the object is $1.00\text{kg}$, and the radius of the dwarf is $6.38\times {10}^6\text{m}$.

Explanation:

The formula for the gravitational potential energy of an object on the surface of the dwarf is,

$PE=-\frac{G{M}_{\text{dwarf}}m}{R_{\text{dwarf}}}$

• $G$ is gravitational constant.
• $m$ is the mass of the object on the surface of the dwarf.
• $M_{\text{dwarf}}$ is mass of the dwarf.
• $R_{\text{dwarf}}$ is radius of the dwarf.

Substitute $6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $1.991\times {10}^{\text{30}}\text{kg}$ for $M_{\text{dwarf}}$, $1.00\text{kg}$ for $m$, and $6.38\times {10}^6\text{m}$ for $R_{\text{dwarf}}$ to find $PE$.

$\begin{array}{c}PE=-\left(6.673\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\frac{\left(1.991\times {10}^{\text{30}}\text{kg}\right)\left(1.00\text{kg}\right)}{6.38\times {10}^6\text{m}}\\ =-2.08\times {10}^{13}\text{J}\end{array}$

Thus, the gravitational potential energy of the object on the surface of the dwarf is $-2.08\times {10}^{13}\text{J}$.

Conclusion:

Therefore, the gravitational potential energy of the object on the surface of the dwarf is $-2.08\times {10}^{13}\text{J}$.