Answered

As Halley's comet orbits the sun, its distance from the sun changes dramatically from 8.8 x 1010 m to 5.2 x 1012 m. if the comet's speed at closest approach is 5.4 x 104 m/s, what is its linear speed when it is farthest from the sun, if angular momentum is conserved

Answer :

Answer:

The linear speed when it is farthest point is 913.85 m/s

Explanation:

Given:

rp = radius at nearest point = 8.8x10¹⁰ m

ra = radius at farthest point = 5.2x10¹² m

vp = velocity nearest point = 5.4x10⁴m/s

The angular momentum is conserved, thus:

La = Lp

[tex]I_{a} w_{a} =I_{p} w_{p}\\m_{a}r_{a}^{2} w_{a} =m_{p}r_{p}^{2} w_{p}\\w_{a} =\frac{r_{a} }{v_{a} } \\w_{p} =\frac{r_{p} }{v_{p} } \\Replacing\\\frac{r_{a}^{2}v_{a} }{r_{a} } =\frac{r_{p}^{2}v_{p} }{r_{p} }\\v_{a} =\frac{r_{o}v_{p}}{r_{a} } =\frac{8.8x10^{10}*5.4x10^{4} }{5.2x10^{12} } =913.85m/s[/tex]

Other Questions