Answer:
Its speed is [tex]v_{b} = 1185m/s[/tex].
Explanation:
The angular momentum is defined as:
[tex]L = mrv[/tex] (1)
Since there is no torque acting on the system, it can be express in the following way:
[tex]t = \frac{\Delta L}{\Delta t}[/tex]
[tex]t \Delta t = \Delta L[/tex]
[tex]\Delta L = 0[/tex]
[tex]L_{a} - L_{b} = 0[/tex]
[tex]L_{a} = L_{b}[/tex] (2)
Replacing equation 1 in equation 2 it is gotten:
[tex]mr_{a}v_{a} =mr_{b}v_{b}[/tex] (3)
As can be seen in equation 3 the angular momentum is conserved.
Where m is the mass of the comet, [tex]r_{a}[/tex] is the orbital radius when it is farther form the Sun, [tex]v_{a}[/tex] is the speed when is farther from the Sun, [tex]r_{b}[/tex] is the orbital radius when is closer to the Sun, [tex]v_{b}[/tex] is the speed when it is closer to the Sun.
From equation 3 [tex]v_{b}[/tex] will be isolated:
[tex]v_{b} = \frac{mr_{a}v_{a}}{mr_{b}}[/tex]
[tex]v_{b} = \frac{r_{a}v_{a}}{r_{b}}[/tex] (4)
[tex]v_{b} = \frac{(9.3x10^{10}m)(5.1x10^{4}m/s)}{(4.0x10^{12}m)}[/tex]
[tex]v_{b} = 1185m/s[/tex]
Hence, its speed is [tex]v_{b} = 1185m/s[/tex].
