Comets move around the sun in very elliptical orbits. At its closet approach, in 1986, Comet Halley was 8.79 x 10⁷ km from the sun and moving with a speed of 54.6 km/s. What was the comet’s speed when it crossed Neptune’s orbit in 2006?

The distance of the Comet Halley from the Sun was 8.79 x 10⁷ km
the speed of the Halley Comet was 54.6 km/s
The mean radius of the Neptune orbit is 2.21 x 10⁷ km
The mass of the Sun is 1.99x 10³⁰ kg

From law of conservation of energy

K₁ + U₁ = K₂ + U₂ ...........................................................(1)

Where

K₁ = kinetic energy of Halley Comet in 1986 .
U₁ = potential energy of Halley Comet in 1986 .
K₂ = kinetic energy of Halley Comet in 2006 .
U₂ = potential energy of Halley Comet in 2006 .

Calculate the kinetic energy of Halley Comet in 1986 using

$K_1=\frac{1}{2}{m}_{\mathrm{c}}{v}_1^2$

and calculate the potential energy

$U_1=\frac{-GM{m}_{\mathrm{c}}}{r_1}$

Similarly calculate potential energy and kinetic energy in 2006

$K_2=\frac{1}{2}{m}_{\mathrm{c}}{v}_2^2$

and

$U_2=\frac{-GM{m}_{\mathrm{c}}}{r_2}$

Now substitute these in equation(1)

$$\begin{gathered} \frac{1}{2}{m}_{\mathrm{c}}{v}_1^2+\frac{-GM{m}_{\mathrm{c}}}{r_1}=\frac{1}{2}{m}_{\mathrm{c}}{v}_2^2+\frac{-GM{m}_{\mathrm{c}}}{r_2} \\ \frac{v_1^2}{2}-\frac{GM}{r_1}=\frac{v_2^2}{2}-\frac{GM}{r_2} \\ \frac{v_2^2}{2}=\frac{v_1^2}{2}-\frac{GM}{r_1}+\frac{GM}{r_2} \\ {v}_2=\sqrt{2\left(\frac{{v_1}^2}{2}-GM\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\right)} \end{gathered}$$

Now substitute the values given and calculate

$v_2=\sqrt{2\left(\frac{{(54.6\times {10}^{3}m/s)}^2}{2}-(6.67\times {10}^{-11}N{m}^2/k{g}^2)(1.99x{10}^{30}kg)\left(\frac{1}{(8.79x{10}^{10}m)}-\frac{1}{(4.5\times {10}^{12}m)}\right)\right)}$

v₂= 4.5 km/s