Chapter 7, Problem 23Q

To determine

(a)

The composition of a transneptunian object of same mass as earth.

Answer to Problem 23Q

TNO at the distance of $50\text{AU}$ would be made up of ice and light elements of about $45-50\%$ from its interior.

Explanation of Solution

The transneptunian objects or TNOs, first found about in February, $1930$ Pluto was also discovered during the twentieth century. Pluto has five moons and another transneptunian named Eris, the most massive TNO was discovered in $2005$ is about similar in diameter as Pluto and has one moon.

The largest transneptunian are also named as dwarf planets.A trans-Neptunian object is a kind of a planet or a minor planet among other celestial objects in the solar system that orbits the sun at greater orbital radii than Neptune and also a larger orbital speed.

As first TNO was Pluto that brought a new revolution in astronomy and after whose discovery, many astronomers wanted more TNOs like that. One of the astronomers, an American astronomer Clyde Tombaugh started researching for more such objects for a couple of years, but he found nothing.

After that for a long time, no astronomer had the guts to go for research for such objects as sake of saving time and spending it in some productive work and also as they believed that there would be no more of such objects like transneptunian.

Average distance between the sun and the Neptune is $30.1\text{AU}$. Therefore, a transneptunian at the distance of $50\text{AU}$ would be far away from the Neptune. All Jovian planets and their moons are all made up of either ice or the lighter elements.

Conclusion:

Thus, TNO at the distance of $50\text{AU}$ would be made up of ice and light elements of about $45-50\%$ from its interior.

To determine

(b)

The diameter of a transneptunian and the comparison of it with that of the Earth.

Answer to Problem 23Q

The diameter of a transneptunian is $1.654\times {10}^7\text{m}$ and it is $1.298$ times that of earth.

Explanation of Solution

Concept used:

The mass of the transneptunian is equal to the mass of the earth, $5.97\times {10}^{24}\text{kg}$ .As the transneptunian are less dense because of presence of ice and lighter elements. Therefore, assuming it as have a density equal to the largest TNO, Eris which is given by $2.52{\text{g/cm}}^{\text{3}}$.

Write the expression for the density of the transneptunian.

$$

$\rho =\frac{m}{V}$

Here, $m$ is the mass, $V$ is the volume and $\rho$ is the density.

Rearrange the above equation for $V$.

$V=\frac{m}{\rho }$ …… (1)

Write the expression for the volume.

$V=\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3$

Here, $D$ is the diameter of the transneptunian.

Substitute $\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3$ for $V$ in equation (1).

$\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3=\frac{m}{\rho }$

Rearrange the above expression for $D$.

$D=\sqrt[3]{\frac{6m}{\pi \rho }}$ …… (2)

Write the expression for the ratio of the diameters of the above TNO to that of earth.

$R=\frac{D}{D_{earth}}$ …… (3) Here, $R$ is the ratio of the diameters of the above TNO to that of earth and $D_{earth}$ is the diameter of the earth.

Calculation:

Substitute $5.97\times {10}^{24}\text{kg}$ for $m$ and $2.52{\text{g/cm}}^{\text{3}}$ for $\rho$ in equation (2).

$\begin{array}{c}D=\sqrt[3]{\frac{6\times 5.97\times {10}^{24}\text{kg}}{3.14\times 2.52{\text{g/cm}}^{\text{3}}}}\\ =\sqrt[3]{\frac{35.82\times {10}^{24}\text{kg}}{3.14\times 2.52\frac{\text{g}}{{\text{cm}}^{\text{3}}}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\times \frac{\text{1}{\text{cm}}^{\text{3}}}{{10}^{-6}{\text{m}}^3}\right)}}\\ =\sqrt[3]{\frac{35.82\times {10}^{24}\text{kg}}{3.14\times 2520{\text{kg/m}}^{\text{3}}}}\\ =1.654\times {10}^7\text{m}\end{array}$

The diameter of the TNO is $1.654\times {10}^7\text{m}$.

Substitute $1.654\times {10}^7\text{m}$ for $D$ and $1.274\times {10}^7 \text{m}$ for $D_{earth}$ in equation (3).

$\begin{array}{c}R=\frac{1.654\times {10}^7\text{m}}{1.274\times {10}^7 \text{m}}\\ =1.298\end{array}$

Conclusion:

Thus, the diameter of a transneptunian is $1.654\times {10}^7\text{m}$ and it is $1.298$ times that of earth.