A comet is in elliptical orbit around the Sun. Its closest approach to the Sun is a distance of $\mathbf{4}\mathbf{\times }{\mathbf{10}}^{\mathbf{10}}\mathbf{}\mathbf{m}$ (inside the orbit of Mercury), at which point its speed is $\mathbf{8}\mathbf{.}\mathbf{17}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is $\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{12}}\mathbf{}\mathbf{m}$ from the Sun? (This is the approximate distance of Pluto from the Sun.)

- The closest distance is $r=4\times {10}^{10}\mathrm{m}$

- The first speed is $u=8.17\times {10}^4\mathrm{m}/\mathrm{s}$

- The final distance is $R=6\times {10}^{12}\mathrm{m}$

- The mass of Sun is $M=2\times {10}^{30}\mathrm{kg}$

- The Gravitational Constant is $G=6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2$

- The mass of a comet is $m$

- Final Speed is $v$

The principle of energy conservation states that, the addition of initial kinetic and potential energy is equal to the addition of final potential and kinetic energy. The expression will be, $P_i+K_i=P_f+K_f\cdots \cdots \left(1\right)$

The potential energy and kinetic energy is conserved for the system, From Equation (1), we can get the expression for speed.

$-\frac{GMm}{r}+\frac{1}{2}m{u}^2=-\frac{GMm}{R}+\frac{1}{2}m{v}^2$

$-\frac{GM}{r}+\frac{1}{2}{u}^2=-\frac{GM}{R}+\frac{1}{2}{v}^2$

$GM\left(-\frac{1}{r}+\frac{1}{R}\right)+0.5u^2=0.5v^2$

$0.5(v{)}^2=\left(\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\times \left(2\times {10}^{30}\mathrm{kg}\right)\right)·\left(-\frac{1}{4\times {10}^{10}\mathrm{m}}+\frac{1}{6\times {10}^2\mathrm{m}}\right)$

$0.5\times {\left(8.17\times {10}^4\mathrm{m}/\mathrm{s}\right)}^2$

$0.5(v{)}^2=\left(\left(6.67\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\times \left(2\times {10}^{30}\mathrm{kg}\right)\right)·\left(\frac{1}{4\times {10}^{\text{'}0}\mathrm{m}}+\frac{1}{6\times {10}^{\text{'}2}\mathrm{m}}\right)$

$+0.5\times {\left(8.17\times {10}^4\mathrm{m}/\mathrm{s}\right)}^2$

$=\left[\left(6.67\times {10}^{-11}\right)\times \left(2\times {10}^{\mathrm{sp}}\right)\times \left(-\frac{1}{4\times {10}^{10}}+\frac{1}{6\times {10}^{-2}}\right)·\left(\frac{1\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\times 1\mathrm{kg}}{1\mathrm{m}}\right)\right.$

$+0.5\times {\left(8.17\times {10}^2\mathrm{m}/\mathrm{s}\right)}^2$

$=\left[\left(6.67\times {10}^{11}\right)\times \left(2\times {10}^{30}\right)\times \left(-\frac{1}{4\times {10}^{10}}+\frac{1}{6\times {10}^{22}}\right)·\left(\frac{1\mathrm{N}·{\mathrm{m}}^2·1\mathrm{kg}}{1{\mathrm{kg}}^2·1\mathrm{m}}\right)\right.$

$+0.5\times {\left(8.17\times {10}^2\right)}^2\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}$

$=\left[\left(6.67\times {10}^{-11}\right)\times \left(2\times {10}^{30}\right)\times \left(-\frac{1}{4\times {10}^{10}}-\frac{1}{6\times {10}^2}\right)\right]·\left(\frac{1\mathrm{N}·\mathrm{m}}{1\mathrm{kg}}\right)$

$+0.5\times {\left(8.17\times {10}^4\right)}^2\frac{{\mathrm{m}}^2}{{\mathrm{s}}^2}$

$v^2=6625452074{\mathrm{m}}^2/{\mathrm{s}}^2$

Hence, the speed of a comet in a circular orbit near the Earth is $81396.88\mathrm{m}/\mathrm{s}$