Given Data:
At distance d₁ = 4.5×10¹⁰ m, speed of comet is [tex]v_{i}[/tex] = 9.2×10⁴ m/s.
At distance d₂ = 6×10¹² m, speed of comet will be [tex]v_{f}[/tex] = ?
As there is no work done from the outside, so the initial energy of the system will be equal to the final state of energy.
[tex]E_{f} = E_{i}+W\\ K_{f} +U_{f} =K_{i} +U_{i} +0[/tex]
[tex]\frac{1}{2}mv^{2} _{f} + \frac{-GMm}{d_{2} } = \frac{1}{2}mv^{2} _{i} + \frac{-GMm}{d_{1} }[/tex]
[tex]v_{f} = \sqrt{v^{2} _{i}+2GM[\frac{1}{d_{2}} -\frac{1}{d_{1} } ] }[/tex]
Now by putting the values in the above equation we can find out [tex]v_{f}[/tex]. Here M is the mass of sun.
[tex]v_{f} = \sqrt{(9.2*10^{4})^{2} + 2(6.7*10^{-11})(1.98*10^{30})[\frac{1}{6*10^{12}}-\frac{1}{4.5*10^{10} } } } ][/tex]
= 51109.88 m/s
