Question

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 1

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

Step-by-step 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

`I_(a) w_(a) =I_(p) w_(p)\\m_(a)r_(a)^(2) w_(a) =m_(p)r_(p)^(2) w_(p)\\w_(a) =(r_(a) )/(v_(a) ) \\w_(p) =(r_(p) )/(v_(p) ) \\Replacing\\(r_(a)^(2)v_(a) )/(r_(a) ) =(r_(p)^(2)v_(p) )/(r_(p) )\\v_(a) =(r_(o)v_(p))/(r_(a) ) =(8.8x10^(10)*5.4x10^(4) )/(5.2x10^(12) ) =913.85m/s`