Chapter 7, Problem 24Q

To determine

(a)

The orbital period of a hypothetical transneptunian object located at $100\text{AU}$ from the sun.

Answer to Problem 24Q

The orbital period is $1006.35\text{years}$.

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 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.

The centripetal force is required for a body to be in a circular motion and that is provided by the gravitational force in case of celestial bodies.

$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$ ....... (1)

Here, $G$ is gravitational constant, $M$ is the mass of the sun, $m$ is the mass of the TNO, $v$ is the orbital velocity of the TNO and $r$ is the distance between the sun and the TNO.

Write the expression for the time period of an object in a circular motion.

$T=\frac{2\pi r}{v}$

Rearrange the above expression for $v$.

$v=\frac{2\pi r}{T}$

Here, $T$ is the orbital period of the TNO, $r$ is the distance between the sun and the TNO and $v$ is the orbital velocity of the TNO.

Substitute $\frac{2\pi r}{T}$ for $v$ in equation (1).

$\frac{GMm}{r^2}=\frac{m{\left(\frac{2\pi r}{T}\right)}^2}{r}$

Rearrange the above expression for $T$.

$T=\sqrt{\frac{4{\pi }^2{r}^3}{GM}}$ ....... (2)

Calculation:

Substitute $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$, $1.98\times {10}^{30}\text{kg}$ for $M$ and $100\text{AU}$ for $r$.

$\begin{array}{c}T=\sqrt{\frac{4{\pi }^2{\left(100\text{AU}\right)}^3}{\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(1.98\times {10}^{30}\text{kg}\right)}}\\ =\sqrt{\frac{4{\left(3.14\right)}^2{\left(100\text{AU}\left(\frac{1.5\times {10}^{11}\text{m}}{1\text{AU}}\right)\right)}^3}{\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(1.98\times {10}^{30}\text{kg}\right)}}\\ =3.17\times {10}^{10}\text{s}\left(\frac{1\text{year}}{31536000\text{s}}\right)\\ =1006.35\text{years}\end{array}$

Conclusion:

Thus, the orbital period is $1006.35\text{years}$.

To determine

(b)

The time taken by a transneptunian object to move $1\text{arcminute}$ across the sky.

Answer to Problem 24Q

The time taken by a transneptunian object to move $1\text{arcminute}$ is $17.0084\text{days}$

Explanation of Solution

Given:

There are $360°$ in a circle.

There are $60\text{arcminiute}$ in $1°$.

The orbital period is $1006.35\text{years}$.

Concept used:

The circle has a total arc argument of $2\pi$ in radians and $360°$ in degrees.

Write the expression of the relation between the degree and the arcminute.

$1°\text{=}\text{60}\text{arcminute}$ ....... (4)

Simplify above for $360°$.

$360°=21600\text{arcminute}$

The time taken to complete a circle is called as the orbital period or time taken to traverse $21600\text{arcminute}$ is $1006.35\text{years}$.

Write the expression for the time taken to traverse $\text{arcminutes}$.

$t=\frac{a}{A}T$ ....... (5)

Here, $t$ is the time taken to complete $a\text{arcminutes}$, $a$ is the magnitude of $\text{arcminutes}$ traversed in $t$ time, $T$ is the time taken to complete $A\text{arcminutes}$ and $A$ is the $\text{arcminutes}$ traversed in $T$ time.

Calculation:

Substitute $1\text{arcminute}$ for $a$, $21600\text{arcminutes}$ for $A$ and $1006.35\text{years}$ for $T$ in equation (5).

$\begin{array}{c}t=\frac{1\text{arcminutes}}{21600\text{arcminutes}}\left(1006.53\text{years}\right)\\ =4.6\times {10}^{-2}\text{year}\left(\frac{365\text{days}}{1\text{year}}\right)\\ =17.0084\text{days}\end{array}$

Conclusion:

Thus, the time taken by a transneptunian object to move $1\text{arcminute}$ across is $17.0084\text{days}$.

To determine

(c)

The reason that the astronomical discoveries require patience.

Answer to Problem 24Q

The discoveries related to transneptunian objects require patience because it asks for a huge amount of time too.

Explanation of Solution

Concept used:

As 1st 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, 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 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.

They thought that there would be no more such objects because the transneptunian objects or TNOs take a huge amount of time to complete the orbit around sun, they are so slow that approximately they take $1000\text{years}$.

This is because the discoveries related to transneptunian object take a lot amount of time and with that requires patience.

Conclusion:

Thus, the discoveries related to transneptunian object require patience because it asks for a huge amount of time too.

To determine

(d)

The reason that transneptunian objects requires large telescopes equipped with sensitive detectors.

Answer to Problem 24Q

The discovery of transneptunian objects requires large telescopes equipped with sensitive detectors because of larger distance and smaller orbital radii of them.

Explanation of Solution

Concept used:

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 with also a larger orbital speed.

The name transneptunian also suggests that these objects are far beyond the Neptune and therefore, this huge distance of about $30\text{AU}$ requires a lot to work on and the same asks for a huge amount of time.

This huge magnitude of distance requires the large telescopes with larger range and also the TNOs are very small in diameter, that’s because to make successful studies over there, the telescopes equipped with sensitive detectors.

Conclusion:

Thus, the discovery of transneptunian objects requires large telescopes equipped with sensitive detectors because of larger distance and smaller orbital radii of them.