Question

As Halley’s comet orbits the sun, its distance from the sun changes dramatically. If the comet’s speed at a distance of 9.8 × 1010 m from the sun is 5.3 × 104 m/s and angular momentum is conserved, what is its speed when it is 3.6 × 1012 m from the sun? Assume the comet can be treated as a point mass. Ignore radial components of momentum

Answer 1

The speed of comet is 1442.77 m/s.

Step-by-step explanation:

Given that,

Speed of comet's
`v_(c)= 5.3*10^(4)\ m/s`

Distance from the sun
`r_(d)=9.8*10^(10)\ m`

Distance
`r_(far)=3.6*10^(12)\ m`

We need to calculate the speed of comet

Using conservation of angular momentum

`L_(f)=L_(i)`

`I\omega=I\omega`

Here.
`v = r\omega`

`\omega=(v)/(r)`

`mr_(far)^2*(v_(far))/(r_(far))=mr_(d)^2*(v_(d))/(r_(d))`

`r_(far)* v_(far)=r_(d)* v_(d)`

Put the value into the formula

`3.6*10^(12)* v_(far)=9.8*10^(10)*5.3*10^(4)`

`v_(far)=(9.8*10^(10)*5.3*10^(4))/(3.6*10^(12))`

`v_(far)=1442.77\ m/s`

Hence, The speed of comet is 1442.77 m/s.