Chapter 11, Problem 57P

(a)

To determine

The expression for the speed of each planet and for their relative speed.

Answer to Problem 57P

The expression for the speed of planet $1$ is $\underset{\_}{m_2\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}}$, planet $2$ is $\underset{\_}{m_1\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}}$ and for their relative speed is $\underset{\_}{\sqrt{\frac{2G\left(m_1+m_2\right)}{d}}}$.

Explanation of Solution

The system of two planets are isolated, so the energy as well as the momentum is conserved.

Write the expression for the conservation of energy.

    $\begin{array}{c}\text{initial energy}=\text{final energy}\\ E_i=E_f\end{array}$        (I)

Here, $E_i$ is the initial energy, $E_f$ is the final energy.

Write the expression for $E_i$.

    $E_i=K_{1i}+K_{2i}+U_i$        (II)

Here, $K_{1i}$ is the initial kinetic energy of planet $1$, $K_{2i}$ is the initial kinetic energy of planet $2$, $U_i$ is the initial gravitational potential energy between the two planets

Write the expression for the $E_f$.

    $E_f=K_{1f}+K_{2f}+U_f$        (III)

Here, $K_{1f}$ is the final kinetic energy of planet $1$, $K_{2f}$ is the kinetic energy of planet $2$, , $U_f$ is the final gravitational potential energy between the planets.

Use equation (II) and (III) in (I) to solve for the conservation of energy.

    $K_{1i}+K_{2i}+U_i=K_{1f}+K_{2f}+U_f$        (IV)

In the initial condition, the two planets are nearly at rest and they are infinite distance apart.

    $\begin{array}{c}{K}_{1i}=0\\ K_{2i}=0\\ U_i=0\end{array}$        (V)

Use equation (V) in (IV) to solve for the conservation of energy.

    $0=K_{1f}+K_{2f}+U_f$        (VI)

Write the expression for the conservation of momentum.

    ${\left(P_1+P_2\right)}_i={\left(P_1+P_2\right)}_f$        (VII)

Here, $P_{1i}$ is the initial momentum of planet $1$, $P_{2i}$ is the initial momentum of planet $2$, $P_{1f}$ is the final momentum of planet $1$, $P_{2f}$ is the final momentum of planet $2$.

Use equation (V) in (VII) to solve for the conservation momentum.

    $0={\left(P_1+P_2\right)}_f$        (VIII)

Write the expression for the kinetic energy.

    $K=\frac{1}{2}m{v}^2$        (IX)

Here, $m$ is the mass, $v$ is the velocity.

Write the expression for the momentum.

    $P=mv$        (X)

Write the expression for the $U_f$.

    $U_f=-\frac{G{m}_1{m}_2}{d}$        (XI)

Here, $G$ is the gravitational constant, $d$ is the separation between the planets, $m_1$ is the mass of planet $1$, $m_2$ is the mass of planet $2$.

Use equation (IX) and (XI) in (VI) and it becomes,

    $\begin{array}{c}\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2-\frac{G{m}_1{m}_2}{d}=0\\ \frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=\frac{G{m}_1{m}_2}{d}\\ m_1{v}_1^2+m_2{v}_2^2=\frac{2G{m}_1{m}_2}{d}\end{array}$        (XII)

Use equation (X) in (VIII) to solve for $v_1$.

    $\begin{array}{c}0=m_1{v}_1-m_2{v}_2\\ m_1{v}_1=m_2{v}_2\\ v_1=\left(\frac{m_2}{m_1}\right)v_2\end{array}$        (XIII)

Use equation (XIII) in (XII) to solve for $v_2$.

    $\begin{array}{l}{m}_2\left(\frac{m_1}{m_2}{v}_1^2+v_2^2\right)=\frac{2G{m}_1{m}_2}{d}\\ \frac{m_1}{m_2}{v}_1^2+v_2^2=\frac{2G{m}_1{m}_2}{d}\end{array}$        (XIV)

Substitute the value of $v_1$ with equation (XIII),

    $\begin{array}{c}\frac{m_1}{m_2}{\left(\frac{m_2}{m_1}\left(v_2\right)\right)}^2+v_2^2=\frac{2G{m}_1}{d}\\ \frac{m_2}{m_1}{v}_2^2+v_2^2=\frac{2G{m}_1}{d}\\ v_2^2\left(\frac{m_2+m_1}{m_1}\right)=\frac{2G{m}_1}{d}\end{array}$        (XV)

Use equation (XV) to solve for $v_2$.

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

Use equation (XVI) in (XIII) to solve for $v_1$.

    $\begin{array}{c}{v}_1=\frac{m_2}{m_1}\left(m_1\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}\right)\\ =m_2\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}\end{array}$        (XVII)

Write the expression for the relative velocity.

    $v_r=v_1-\left(v_2\right)$        (XVIII)

Here, $v_r$ is the relative velocity of the two planets.

Use equation (XVI) and (XVII) in (XVIII) to solve for $v_r$.

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

Conclusion:

Therefore, the expression for the speed of planet $1$ is $\underset{\_}{m_2\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}}$, planet $2$ is $\underset{\_}{m_1\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}}$ and for their relative speed is $\underset{\_}{\sqrt{\frac{2G\left(m_1+m_2\right)}{d}}}$.

(b)

To determine

The kinetic energy of the each planet just before the collision.

Answer to Problem 57P

The kinetic energy of the planet $1$ is $\underset{\_}{1.07\times {10}^{32}\text{J}}$ and the planet $2$ is $\underset{\_}{2.67\times {10}^{31}\text{J}}$ just before the collision.

Explanation of Solution

Write the expression for the $d$.

    $d=r_1+r_2$        (XX)

Here, $r_1$ is the radius of planet $1$, $r_2$ is the radius of planet $2$.

Use equation (XVI) in (IX) to solve for $K_1$.

    $K_1=\frac{1}{2}{m}_1{m}_2\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}$        (XXI)

Here, $K_1$ is the kinetic energy of planet $1$ just before the collision.

Use equation (XVII) in (IX) to solve for $K_2$.

    $K_2=\frac{1}{2}{m}_2{m}_1\sqrt{\frac{2G}{d\left(m_1+m_2\right)}}$        (XXII)

Here, $K_2$ is the kinetic energy of planet $2$ just before the collision.

Conclusion:

Substitute $3.00\times {10}^6\text{m}$ for $r_1$, $5.00\times {10}^6\text{m}$ for $r_2$ in equation (XX) to find $d$.

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

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $2.00\times {10}^{24}\text{kg}$ for $m_1$, $8.00\times {10}^{24}\text{kg}$ for $m_2$, $8.00\times {10}^6\text{m}$ for $d$ in equation (XXI) to find $K_1$.

    $\begin{array}{c}{K}_1=\frac{1}{2}\left(2.00\times {10}^{24}\text{kg}\right)\left(8.00\times {10}^{24}\text{kg}\right)\sqrt{\frac{2\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)}{8.00\times {10}^6\text{m}\left(2.00\times {10}^{24}\text{kg}+8.00\times {10}^{24}\text{kg}\right)}}\\ =\frac{1}{2}\left(2.00\times {10}^{24}\text{kg}\right)\left(1.03\times {10}^4\text{m}/\text{s}\right)\\ =1.07\times {10}^{32}\text{J}\end{array}$

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $2.00\times {10}^{24}\text{kg}$ for $m_1$, $8.00\times {10}^{24}\text{kg}$ for $m_2$, $8.00\times {10}^6\text{m}$ for $d$ in equation (XXII) to find $K_2$.

    $\begin{array}{c}{K}_2=\frac{1}{2}\left(8.00\times {10}^{24}\text{kg}\right)\left(2.00\times {10}^{24}\text{kg}\right)\sqrt{\frac{2\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)}{8.00\times {10}^6\text{m}\left(2.00\times {10}^{24}\text{kg}+8.00\times {10}^{24}\text{kg}\right)}}\\ =\frac{1}{2}\left(2.00\times {10}^{24}\text{kg}\right)\left(2.58\times {10}^3\text{m}/\text{s}\right)\\ =2.67\times {10}^{31}\text{J}\end{array}$

Therefore, the kinetic energy of the planet $1$ is $\underset{\_}{1.07\times {10}^{32}\text{J}}$ and the planet $2$ is $\underset{\_}{2.67\times {10}^{31}\text{J}}$ just before the collision.