Answer to Question #102695 in Astronomy | Astrophysics for sakshi

Question

A comet has an aphelion distance of 6.1 × 10¹¹m and perihelion distance of 6.1 × 10¹¹m. The mass of the sun is 2.0 × 10^30 .

Answer 1

Solution. Find semi-major axis a  

$$a=\frac{r_a+r_p}{2}=\frac{6.1\times10^{11}+6.1\times10^{11}}{2}=6.1\times10^{11}m$$

where r_a is aphelion distance; r_p is perihelion distance.

Find the orbital period T of a comet

$$\frac{a^3}{T^2}=\frac{G(M+m)}{4\pi^2}$$

where M=2.0 × 10^30kg is the mass of the sun; m is a comet mass; (M>>m). Therefore

$$T=\frac {2\pi \sqrt{a^3}}{GM}=\frac {2\pi \sqrt{(6.1\times10^{11})^3}}{\sqrt{6.67\times10^{-11} \times 2\times 10^{30}}}=2.59 \times 10^{8}s\approx 8.22 years$$

Average orbit speed

$$v=\frac{2\pi a}{T}=\frac{2\pi\times 6.1\times10^{11}}{2.59 \times 10^{8}}\approx 14/55 \frac{km}{s}$$