A comet that was seen in April 574 by Chinese astronomerson a day known by them as the Woo Woo day was spotted again inMay 1994. Assume the time between observations is the period of the Woo Woo day comet and take its eccentricity as $\mathbf{0}\mathbf{.11}$ .

What are(a) the semi major axis of the comet’s orbit and

(b) its greatest distance from the Sun in terms of the mean orbital radius ${\mathbf{R}}_P$ of Pluto?

$\begin{array}{c}\text{T}=1994\text{\hspace{0.17em}years}-574\text{\hspace{0.17em}years}\\ =1420\text{years}\end{array}$

Eccentricity $\left(\text{e}\right)=0.9932$

${\text{M}}_{\text{sun}}=1.99\times {10}^{30}\text{\hspace{0.17em}kg}$

$\text{G}=6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}$

Semi major axis of Pluto$=5.9\times {10}^{12}\text{\hspace{0.17em}m}$

As, the time period of the comet is given indirectly, using Kepler’s 3^rd law, find the semi major axis of the comet.

Once the semi major axis of the comet is found, and the semi major axis of Pluto is known, find the comet’s greatest distance from the sun in terms of the mean orbital radius ${\mathbf{R}}_P\mathbf{.}$According to Kepler’s third law, the squares of the orbital periods of the planets are directly proportional to the cubes of the semi major axes of their orbits.

Formula is as follow:

${\mathrm{T}}^2=\left(\frac{4{\mathrm{\pi }}^2}{\mathrm{GM}}\right){\mathrm{a}}^3$

where, T is time, G is gravitational constant, M is mass and a is semi major axis.

The time period is given in years. Therefore, convert them into seconds,

$\text{T}=1420\text{\hspace{0.17em}years}$

$1420\text{years}=1420\text{years}\times \left(\frac{\text{365days}\times \text{24hrs}\times \text{3600s}}{\text{1year}}\right)$

$1420\text{years}=4.48\times {10}^{10}\text{\hspace{0.17em}s}$

${\mathrm{T}}^2=\left(\frac{4{\mathrm{\pi }}^2}{\mathrm{GM}}\right){\mathrm{a}}^3$

$\begin{array}{c}(4.48\times {10}^{10}\mathrm{s})=\left(\frac{4{\mathrm{\pi }}^2}{\left(6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times (1.99\times {10}^{30}\mathrm{kg})}\right)\times {\mathrm{a}}^3\\ 4{\mathrm{\pi }}^2{\mathrm{a}}^3=\left(6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times (1.99\times {10}^{30}\mathrm{kg})\times (4.48\times {10}^{10}\mathrm{s})\\ \mathrm{a}={\left[\frac{\left(6.67\times {10}^{-11}\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times (1.99\times {10}^{30}\mathrm{kg})\times (4.48\times {10}^{10}\mathrm{s})}{4{\mathrm{\pi }}^2}\right]}^{1/3}\\ \mathrm{a}=1.9\times {10}^{13}\mathrm{m}\end{array}$

Hence, the semi-major axis of the comet’s orbit is ${\text{1.9×10}}^{\text{13}}\text{m}$.

As the eccentricity of the orbit of the comet is e, the distance from the focus to its center is ea,

$\begin{array}{c}{\mathrm{D}}_{\mathrm{comet}\mathrm{and}\mathrm{sun}}=\mathrm{ea}+\mathrm{a}\\ =\mathrm{a}(\mathrm{e}+1)\\ =1.9\times {10}^{13}\mathrm{m}(0.9932+1)\\ =1.9\times {10}^{13}\mathrm{m}\left(1.9932\right)\\ =3.78\times {10}^{13}\mathrm{m}\\ =3.78\times {10}^{13}\mathrm{km}\end{array}$

Mean Orbital radius of Pluto is${\mathrm{R}}_{\mathrm{P}},$

$\begin{array}{c}{\left({\mathrm{D}}_{\mathrm{comet}\mathrm{and}\mathrm{sun}}\right)}_{\mathrm{in}\mathrm{terms}\mathrm{of}{\mathrm{R}}_{\mathrm{P}}}={\mathrm{R}}_{\mathrm{P}}\times \left(\frac{{\mathrm{D}}_{\mathrm{comet}\mathrm{and}\mathrm{sun}}}{\text{semimajoraxisofpluto}}\right)\\ =\left(\frac{3.78\times {10}^{10}\mathrm{km}}{5.9\times {10}^{12}\mathrm{km}}\right)\times {\mathrm{R}}_{\mathrm{P}}\\ =6.4\times {\mathrm{R}}_{\mathrm{P}}\end{array}$.

Hence, the comet’s greatest distance from the sun in terms of the mean orbital radius ${\text{R}}_{\text{P}}$of Pluto is $6.4\times {\mathrm{R}}_{\mathrm{P}}$.

Therefore, using Kepler’s 3^rd law, the semi major axis of the comet can be found. Using this semi major axis and eccentricity, the distance of the comet from the sun can be found in terms of the orbital radius of the comet.