Question

A comet is approaching Venus on a parabolic path with perigee distance of 18,200 km . Calculate the total time of travel (in hours) that the comet distance to the center of Venus is between 39,000 km and 24,500 km.

Answer 1

Answer:

The time traveled is 1.39 hrs

Explanation:

Equation of Trajectory of a comet is given as

$r=\frac{h^2}{\mu}\frac{1}{1+cos \theta}$

Here

• h is the specific angular momentum given as

                                           $h=v_p r_p$

• μ is gravitational parameter whose value is $3.24859 \times 10^{14} \, \, m^3/s^2$ for Venus
• r_p is the perigee distance of parabolic which is 18200 km
• As the path of comet is parabolic so energy is conserved i.e

                                 $\frac{1}{2}m_c v_p^2-\frac{\mu m_c}{r_p}=0\\v_p=\sqrt{\frac{2 \mu}{r_p}}$

So h is given as

                                    $h=v_pr_p\\\\h=\sqrt{2 \mu r_p}\\h=\sqrt{2 \times 3.24859 \times 10^{14} \times 18200 \times 10^3}\\h=1.087 \times 10^{11}$

So for point a where r=24500 km

                                $r_1=\frac{h^2}{\mu}\frac{1}{1+cos \theta_1}\\24500 \times 10^3=\frac{1.1824 \times 10^{22}}{3.24859 \times 10^{14}}\frac{1}{1+cos \theta_1}\\0.6731=\frac{1}{1+cos \theta_1}\\1+cos \theta_1=\frac{1}{0.6731}\\cos \theta_1=1.4857-1\\cos \theta_1=0.4857\\\theta_1=cos^{-1}0.4857\\\theta_1=1.0636 rad$

So for point a where r=39000 km

                              $r_2=\frac{h^2}{\mu}\frac{1}{1+cos \theta_2}\\39000 \times 10^3=\frac{1.1824 \times 10^{22}}{3.24859 \times 10^{14}}\frac{1}{1+cos \theta_2}\\1.0715=\frac{1}{1+cos \theta_2}\\1+cos \theta_2=\frac{1}{1.0715}\\cos \theta_2=0.9332-1\\cos \theta_2=-0.0667\\\theta_2=cos^{-1}(-0.0667)\\\theta_2=1.6375 rad$

So as per the Barkers equation

                      $t_1-T=\sqrt{\frac{2 r_p^3}{\mu}}(D_1+\frac{D_1^3}{3})$

where

                      $D_1=tan (\theta_1/2)\\D_1=tan(0.5318)\\D_1=0.5883$

                   $t_2-T=\sqrt{\frac{2 r_p^3}{\mu}}(D_2+\frac{D_2^3}{3})$

where

                    $D_2=tan (\theta_2/2)\\D_1=tan(0.8187)\\D_2=1.0690$

So

$t_2-t_1=\sqrt{\frac{2 r_p^3}{\mu}}(D_2+\frac{D_2^3}{3}-D_1+\frac{D_1^3}{3})\\t_2-t_1=\sqrt{\frac{2 (18200 \times 10^3)^3}{3.24859 \times 10^{14}}}((1.0690)+\frac{(1.0690)^3}{3}-(0.5883)-\frac{(0.5883)^3}{3})\\t_2-t_1=\sqrt{\frac{2 (18200 \times 10^3)^3}{3.24859 \times 10^{14}}}(0.8201)\\t_2-t_1=(6092)(0.8201)\\t_2-t_1=4996.21 s\\t_2-t_1=1.39 hrs$

So the time traveled is 1.39 hrs