Question

The perihelion of the comet TOTAS is 1.69 AU and the aphelion is 4.40 AU. Given that its speed at perihelion is 28 km/s, what is the speed (in km/s) at aphelion (1 AU = 1.496 ✕ 1011 m)? (Hint: You may use either conservation of energy or angular momentum, but the latter is much easier.)

Answer 1

The speed at the aphelion is 10.75 km/s.

Step-by-step explanation:

The angular momentum is defined as:

`L = mrv` (1)

Since there is no torque acting on the system, it can be expressed in the following way:

`t = (\Delta L)/(\Delta t)`

`t \Delta t = \Delta L`

`\Delta L = 0`

`L_(a) - L_(p) = 0`

`L_(a) = L_(p)` (2)

Replacing equation 1 in equation 2 it is gotten:

`mr_(a)v_(a) =mr_(p)v_(p)` (3)

Where m is the mass of the comet,
`r_(a)` is the orbital radius at the aphelion,
`v_(a)` is the speed at the aphelion,
`r_(p)` is the orbital radius at the perihelion and
`v_(p)` is the speed at the perihelion.

From equation 3 v_{a} will be isolated:

`v_(a) = (mr_(p)v_(p))/(mr_(a))`

`v_(a) = (r_(p)v_(p))/(r_(a))` (4)

Before replacing all the values in equation 4 it is necessary to express the orbital radius for the perihelion and the aphelion from AU (astronomical units) to meters, and then from meters to kilometers:

`r_(p) = 1.69 AU x (1.496x10^(11) m)/(1 AU)` ⇒
`2.528x10^(11) m`

`r_(p) = 2.528x10^(11) m x (1km)/(1000m)` ⇒
`252800000 km`

`r_(a) = 4.40 AU x (1.496x10^(11) m)/(1 AU)` ⇒
`6.582x10^(11) m`

`r_(p) = 6.582x10^(11) m x (1km)/(1000m)` ⇒
`658200000 km`

Then, finally equation 4 can be used:

`v_(a) = ((252800000 km)(28 km/s))/((658200000 km))`

`v_(a) = 10.75 km/s`

Hence, the speed at the aphelion is 10.75 km/s.