Chapter 13, Problem 61AP

(a)

To determine

The magnitude of the speed of each planets and relative speed.

Answer to Problem 61AP

The magnitude of speed of each planet respectively $m_2\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}$, $m_1\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}$ and their relative velocity is $\sqrt{\frac{2G\left(m_1+m_2\right)}{d}}$.

Explanation of Solution

Initially both planets will have zero potential and kinetic energy.

Write expression for the conservation of energy.

  $0=\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{v}_2{}^2-\frac{G{m}_1{m}_2}{d}$                                                                                  (I)

Here, $v_1$ is the speed of the planet with mass $m_1$, $v_2$ is the speed of the planet with mass $m_2$.

The initial momentum of the both planet is zero.

Write the expression for the conservation of momentum.

  $0=m_1{v}_1-m_2{v}_2$

Rewrite the above equation in terms of $v_2$.

  $\begin{array}{c}{m}_2{v}_2=m_1{v}_1\\ v_2=\frac{m_1{v}_1}{m_2}\end{array}$                                                                                                              (II)

Write the expression to calculate the relative velocity of the two planets.

  $V=v_1-\left(-v_2\right)$                                                                                                         (III)

Here, V is the relative velocity of the planets.

Conclusion:

Substitute the equation (II) in (I) to rewrite in terms of $v_1$.

  $\begin{array}{c}\frac{1}{2}{m}_1{v}_1{}^2+\frac{1}{2}{m}_2{\left(\frac{m_1{v}_1}{m_2}\right)}^2=\frac{G{m}_1{m}_2}{d}\\ v_1{}^2+\left(\frac{m_1}{m_2}\right)v_1{}^2=\frac{2G{m}_2}{d}\\ v_1{}^2\left(1+\frac{m_1}{m_2}\right)=\frac{2G{m}_2}{d}\\ v_1=\sqrt{\frac{2G{m}_2{}^2}{d\left(m_1+m_2\right)}}\\ v_1=m_2\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}\end{array}$

Use the above expression in the equation (II) to rewrite in terms of

  $\begin{array}{c}{v}_2=\frac{m_1\left(m_2\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}\right)}{m_2}\\ =\left(m_1\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}\right)\end{array}$

Substitute the expression for $v_1$ and $v_2$ in the equation (III) to calculate V.

  $\begin{array}{c}V=\left(m_2\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}\right)-\left(-\left(m_1\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}\right)\right)\\ =\left(m_1+m_2\right)\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}\\ =\sqrt{\frac{2G\left(m_1+m_2\right)}{d}}\end{array}$

Therefore, the magnitude of speed of each planet respectively $m_2\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}$, $m_1\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}$ and their relative velocity is $\sqrt{\frac{2G\left(m_1+m_2\right)}{d}}$.

(b)

To determine

The kinetic energy of each planet.

Answer to Problem 61AP

The kinetic energy of each planet is respectively $1.06\times {10}^{32}\text{J}$ and $2.66\times {10}^{31}\text{J}$.

Explanation of Solution

Write the expression to calculate relative distance between the two planets before collision.

  $d=r_1+r_2$                                                                                                                (IV)

Write the expression to calculate the velocity of the planet of mass $m_1$.

  $v_1=m_2\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}$                                                                                             (V)

Write the expression to calculate the velocity of the planet of mass $m_2$.

  $v_2=m_1\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}$                                                                                              (VI)

Write the expression to calculate the kinetic energy of the planet of mass $m_1$.

  $K_1=\frac{1}{2}{m}_1{v}_1{}^2$                                                                                                          (VII)

Here, $K_{1`}$ is the kinetic energy of the planet with mass $m_1$.

Write the expression to calculate the kinetic energy of the planet with mass $m_2$.

  $K_2=\frac{1}{2}{m}_2{v}_2{}^2$                                                                                                  (VIII)

Here, $K_2$ is the kinetic energy of the planet with mass $m_2$.

Conclusion:

Substitute $3.00\times {10}^6\text{m}$ for $r_1$ and $5.00\times {10}^6\text{m}$ for $r_2$ in the equation (IV) to calculate d.

  $\begin{array}{c}d=3.00\times {10}^6\text{m}+5.00\times {10}^6\text{m}\\ =8.00\times {10}^6\text{m}\end{array}$

Substitute $8.00\times {10}^6\text{m}$ for d, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for G, $2.00\times {10}^{24}\text{kg}$ for $m_1$ and $8.00\times {10}^{24}\text{kg}$ for $m_2$ in the above equation to calculate $v_1$.

  $\begin{array}{c}{v}_1=8.00\times {10}^{24}\text{kg}\sqrt{\frac{2\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)}{8.00\times {10}^6\text{m}\left(2.00\times {10}^{24}\text{kg}+8.00\times {10}^{24}\text{kg}\right)}}\\ =1.03\times {10}^4\text{m}/\text{s}\end{array}$

Substitute $8.00\times {10}^6\text{m}$ for d, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for G, $2.00\times {10}^{24}\text{kg}$ for $m_1$ and $8.00\times {10}^{24}\text{kg}$ for $m_2$ in the above equation to calculate $v_2$.

  $\begin{array}{c}{v}_2=2.00\times {10}^{24}\text{kg}\sqrt{\frac{2\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)}{8.00\times {10}^6\text{m}\left(2.00\times {10}^{24}\text{kg}+8.00\times {10}^{24}\text{kg}\right)}}\\ =2.58\times {10}^3\text{m}/\text{s}\end{array}$

Substitute $2.00\times {10}^{24}\text{kg}$ for $m_1$ and $1.03\times {10}^4\text{m}/\text{s}$ for $v_1$ in the equation (VII) to calculate $K_1$.

  $\begin{array}{c}{K}_1=\frac{1}{2}\left(2.00\times {10}^{24}\text{kg}\right){\left(1.03\times {10}^4\text{m}/\text{s}\right)}^2\\ =1.06\times {10}^{32}\text{J}\end{array}$

Substitute $8.00\times {10}^{24}\text{kg}$ for $m_2$ and $2.58\times {10}^3\text{m}/\text{s}$ for $v_2$ in the equation (VIII) to calculate $K_2$.

  $\begin{array}{c}{K}_2=\frac{1}{2}\left(8.00\times {10}^{24}\text{kg}\right){\left(2.58\times {10}^3\text{m}/\text{s}\right)}^2\\ =2.66\times {10}^{31}\text{J}\end{array}$

Therefore, the kinetic energy of each planet is respectively $1.06\times {10}^{32}\text{J}$ and $2.66\times {10}^{31}\text{J}$.