Answer to Question #247867 in Molecular Physics | Thermodynamics for Kelani

Question

A comet has a period of 80.8 years and moves in an elliptical orbit in which its perihelion (closest approach to the Sun) is 0.670 AU. Find the semimajor axis of the comet and an estimate of the comet's maximum distance from the Sun, both in astronomical units.

(a)

the semimajor axis of the comet (in AU)

 __AU

(b)

an estimate of the comet's maximum distance (in AU) from the Sun

 __AU

Answer 1

(a) For the third Kepler's law, we have the relation between the period of the elliptic motion and the length of the semi-major axis a:

$T=\cfrac{2\pi a^{3/2}}{\sqrt{Gm_S}}=2\pi\sqrt{\cfrac{ a^{3}}{Gm_S}} \\ \text{From that definition, we can find a:} \\ a= \Bigg[ \bigg( \cfrac{T}{2\pi} \bigg)^2Gm_S \Bigg]^{1/3}$

Now, we have the mass of the Sun $m_S=1.99\times10^{30}\,{kg}$, the period of the movement of the

comet as $T=80.8\text{ years }(3.156\times10^7\,{s}/{year})=2.550048\times10^9\,{s}$, and we proceed to substitute to find the length of the semi-major axis:

$\\ a= \Bigg[ \bigg( \cfrac{2.550048\times10^9\,{s}}{2\pi} \bigg)^2(6.67428\times10^{-11}\frac{N\cdot m^2}{{kg}^2})(1.99\times10^{30}\,{kg}) \Bigg]^{1/3} \\ a=2.7968\times10^{12}\,{m} (\frac{1\,AU}{1.49598\times 10^{11}\,m})=18.695\,AU$

Now, for the elliptic motion, we know that the semimajor axis of the comet is related to the minimum length between the comet and the Sun (perihelion) and the maximum length or aphelion as:

$2a\approxeq a_{perihelion}+a_{aphelion} \\ \implies a_{aphelion}=2a-a_{perihelion} \\ \therefore a_{aphelion} =[2(18.695)-0.670)]\,AU=36.720\,AU$

In conclusion, we were able to find that (a) the semimajor axis of the comet is a = 18.695 AU and (b) an estimate of the comet's maximum distance from the Sun (or aphelion) is a_aphelion = 36.720 AU.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.