Answer:
Explanation:
Given
Halley's closest distance from sun is [tex]r_1=0.59\ A.U.[/tex]
Greatest distance is [tex]r_2=35\ A.U.[/tex]
Comet's speed at closest approach is [tex]v_1=47\ km/s[/tex]
As there is no external torque so angular momentum of comet about the sun is conserved
[tex]L_1=L_2[/tex]
[tex]mr_1^2\times \omega _1=mr_2^2\times \omega _2[/tex]
where [tex]\omega =angular\ velocity [/tex]
This can be written as [tex]\omega =\frac{v}{r}[/tex]
Therefore
[tex]mr_1^2\times \frac{v_1}{r_1}=mr_2^2\times \frac{v_2}{r_2}[/tex]
[tex]r_1\times v_1=r_2\times v_2[/tex]
[tex]0.59\times 47=35\times v_2[/tex]
[tex]v_2=0.79\ km/s[/tex]
