Question

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.)

Short answer

The speed is $81396.88\mathrm{m}/\mathrm{s}$

Step-by-step solution

Identification of given data

- 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 $v$

Concept of Principle of Energy Conservation

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, $P_i+K_i=P_f+K_f\cdots \cdots \left(1\right)$

Determination of the speed of a comet

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$v=81396.88\mathrm{m}/\mathrm{s}$

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