Chapter 6, Problem 55P

To determine

The comet’s speed at its perihelion.

Answer to Problem 55P

The comet’s speed at its perihelion is $55\text{ km/s}$.

Explanation of Solution

The total mechanical energy of the comet will be constant.

Write the equation for the conservation of energy of the comet.

  $K_p+U_p=K_a+U_a$        (I)

Here, $K_p$ is the kinetic energy of the comet at the perihelion, $U_p$ is the potential energy of the comet at the perihelion, $K_a$ is the kinetic energy of the comet at the aphelion and $U_a$ is the potential energy of the comet at the aphelion.

Write the equation for $K_p$.

  $K_p=\frac{1}{2}m{v}_p^2$        (II)

Here, $m$ is the mass of the comet and $v_p$ is the speed of the comet at the perihelion.

Write the equation for $U_p$.

  $U_p=-\frac{G{M}_{S}m}{r_p}$        (III)

Here, $G$ is the universal gravitation constant, $M_S$ is the mass of the sun and $r_p$ is the distance from the perihelion to the sun.

Write the equation for $K_a$.

  $K_a=\frac{1}{2}m{v}_a^2$        (IV)

Here, $v_a$ is the speed of the comet at the aphelion.

Write the equation for $U_a$.

  $U_a=-\frac{G{M}_{S}m}{r_a}$        (V)

Here, $r_a$ is the distance from the aphelion to the sun.

Put equations (II) to (V) in equation (I).

  $\begin{array}{c}\frac{1}{2}m{v}_p^2+\left(-\frac{G{M}_{S}m}{r_p}\right)=\frac{1}{2}m{v}_a^2+\left(-\frac{G{M}_{S}m}{r_a}\right)\\ \frac{1}{2}{v}_p^2-\frac{G{M}_S}{r_p}=\frac{1}{2}{v}_a^2-\frac{G{M}_S}{r_a}\\ \frac{1}{2}{v}_p^2=\frac{1}{2}{v}_a^2+G{M}_S\left(\frac{1}{r_p}-\frac{1}{r_a}\right)\\ v_p=\sqrt{v_a^2+2G{M}_S\left(\frac{1}{r_p}-\frac{1}{r_a}\right)}\end{array}$        (VI)

Conclusion:

The value of $G$ is $6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2$ and the mass of the sun is $1.989\times {10}^{30}\text{ kg}$.

Substitute $10.0\text{ km/s}$ for $v_a$ , $6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2$ for $G$ , $1.989\times {10}^{30}\text{ kg}$ for $M_S$ , $5.3\times {10}^{12}\text{ m}$ for $r_a$ and $8.9\times {10}^{10}\text{ m}$ for $r_p$ in equation (VI) to find $v_p$.

  $\begin{array}{c}{v}_p=\sqrt{{\left(10.0\text{ km/s}\frac{1000\text{ m}}{1\text{ km}}\right)}^2+2\left(6.67\times {10}^{-11}{\text{ m}}^3\text{/kg}\cdot {\text{s}}^2\right)\left(1.989\times {10}^{30}\text{ kg}\right)\left(\frac{1}{8.9\times {10}^{10}\text{ m}}-\frac{1}{5.3\times {10}^{12}\text{ m}}\right)}\\ =55.0\times {10}^3\text{ m/s}\\ =55.0\times {10}^3\text{ m/s}\frac{1\text{ km}}{1000\text{ m}}\\ =55\text{ km/s}\end{array}$

Therefore, the comet’s speed at its perihelion is $55\text{ km/s}$.