Chapter 1, Problem 34P

Two powerless rockets are on a collision course. The rockets are moving with speeds of 0.800c and 0.600c and are initially 2.52 × 10¹² m apart as measured by Liz, an Earth observer, as shown in Figure P1.34. Both rockets are 50.0 m in length as measured by Liz. (a) What are their respective proper lengths? (b) What is the length of each rocket as measured by an observer in the other rocket? (c) According to Liz, how long before the rockets collide? (d) According to rocket 1, how long before they collide? (e) According to rocket 2, how long before they collide? (f) If both rocket crews are capable of total evacuation within 90 min (their own time), will there be any casualties?

Figure P1.34

To determine

The proper length of the two rockets.

Answer to Problem 34P

The proper length of rocket 1 is $83.5\text{ m}$ and the proper length of rocket 2 is $62.5\text{ m}$.

Explanation of Solution

Write the formula for length contraction

    $L=L_p\sqrt{1-{\left(\frac{v}{c}\right)}^2}$        (I)

Here, $L$ is the distance observed by Liz on Earth, $L_p$ is the proper distance, $v$ is the velocity of the rocket and $c$ is the speed of light.

Rearrange equation (I) and solve for $L_p$

$L_p=\frac{L}{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}$        (II)

Substitute $0.8c$ for $v_1$, $50\text{ m}$ for $L_1$ in equation (II) and solve for $L_{p1}$

$\begin{array}{c}{L}_{p1}=\frac{L_1}{\sqrt{1-{\left(\frac{v_1}{c}\right)}^2}}\\ =\frac{50\text{ m}}{\sqrt{1-{\left(\frac{0.8c}{c}\right)}^2}}\\ =\frac{50\text{ m}}{\sqrt{1-{\left(0.8\right)}^2}}\\ =83.5\text{ m}\end{array}$

Thus, the proper length of rocket 1 is $83.5\text{ m}$.

Substitute $-0.6c$ for $v_2$, $50\text{ m}$ for $L_2$ in equation (II) and solve for $L_{p2}$

$\begin{array}{c}{L}_{p2}=\frac{L_2}{\sqrt{1-{\left(\frac{v_2}{c}\right)}^2}}\\ =\frac{50\text{ m}}{\sqrt{1-{\left(\frac{-0.6c}{c}\right)}^2}}\\ =\frac{50\text{ m}}{\sqrt{1-{\left(0.6\right)}^2}}\\ =62.5\text{ m}\end{array}$

Thus, the proper length of rocket 2 is $62.5\text{ m}$.

Conclusion:

Thus, the proper length of rocket 1 is $83.5\text{ m}$ and the proper length of rocket 2 is $62.5\text{ m}$.

To determine

The length of each rocket as measured by an observer in the other rocket.

Answer to Problem 34P

The contracted length of rocket 2 measured by an observer in rocket 1 is $27.1\text{ m}$ and the contracted length of rocket 1 measured by an observer in rocket 2 is $20.3\text{ m}$.

Explanation of Solution

Write the expression of Lorentz velocity transformation

    $u=\frac{v+u^{\prime }}{1+v{u}^{\prime }/c^2}$        (III)

Here, $u$ is velocity of rocket with respect to the other rocket, $v$ is the velocity of rocket 1 relative to Earth and $u\text{'}$ is the velocity of rocket 2.

Substitute $-0.8c$ for $v$ and $0.6c$ for $u\text{'}$ in equation (III)

$\begin{array}{c}{u}_{12}=\frac{\left(-0.8c\right)+\left(-0.6c\right)}{1+\left(-0.8c\right)\left(-0.6c\right)/c^2}\\ =\frac{-1.4}{1.48}\\ =-0.946c\end{array}$

Thus, the velocity of rocket 2 with respect to rocket 1 is $-0.946c$. And similarly, the velocity of rocket 1 with respect to rocket 2 is $0.946c$.

Substitute $83.5\text{ m}$ for $L_{p1}$ and $0.946c$ for $v_1$ in equation (I) and solve for $L_1^{\text{'}}$

$\begin{array}{c}{L}_1^{\text{'}}=L_{p1}\sqrt{1-{\left(\frac{v_1}{c}\right)}^2}\\ =\left(83.5\text{ m}\right)\sqrt{1-{\left(\frac{0.946c}{c}\right)}^2}\\ =\left(83.5\text{ m}\right)\sqrt{1-{\left(0.946\right)}^2}\\ =27.1\text{ m}\end{array}$

The contracted length of rocket 2 measured by an observer in rocket 1 is $27.1\text{ m}$.

Substitute $62.5\text{ m}$ for $L_{p2}$ and $-0.946c$ for $v_2$ in equation (I) and solve for $L_2^{\text{'}}$

$\begin{array}{c}{L}_2^{\text{'}}=L_{p2}\sqrt{1-{\left(\frac{v_2}{c}\right)}^2}\\ =\left(62.5\text{ m}\right)\sqrt{1-{\left(\frac{-0.946c}{c}\right)}^2}\\ =\left(62.5\text{ m}\right)\sqrt{1-{\left(0.946\right)}^2}\\ =20.3\text{ m}\end{array}$

The contracted length of rocket 1 measured by an observer in rocket 2 is $20.3\text{ m}$.

Conclusion:

Thus, the contracted length of rocket 2 measured by an observer in rocket 1 is $27.1\text{ m}$ and the contracted length of rocket 1 measured by an observer in rocket 2 is $20.3\text{ m}$.

To determine

The time takes for the rockets to collide according to Liz.

Answer to Problem 34P

The time takes for the rockets to collide according to Liz is $1.67\text{ hours}$.

Explanation of Solution

The velocity of the rocket 1 is $0.8c$ and the velocity of rocket 2 is $0.6c$. Thus, the distance between the rocket is decreasing at a rate of $1.4c$.

Write the formula to find the time with respect to distance

    $t=\frac{d}{v}$        (IV)

Here, $t$ is the time, $d$ is the distance and $v$ is the total velocity.

Substitute $2.52\times {10}^{12}\text{ m}$ for $d$ and $1.4c$ for $v$ in the above equation to find the value of $t$

$\begin{array}{c}t=\frac{2.52\times {10}^{12}\text{ m}}{\text{1}\text{.4}\left(\text{3}\times {\text{10}}^{\text{8}}\text{m}/\text{s}\right)}\\ =6000\text{ s}\\ =1.67\text{ hours}\end{array}$

Conclusion:

Thus, the time takes for the rockets to collide according to Liz is $1.67\text{ hours}$.

To determine

The time takes for the rockets to collide according to rocket 1.

Answer to Problem 34P

The time takes for the rockets to collide according to rocket 1 is $1.47\text{ hours}$.

Explanation of Solution

Substitute $2.52\times {10}^{12}\text{ m}$ for $L_p$ and $0.8c$ for $v$ in equation (I) and solve for $L_1$

$\begin{array}{c}{L}_1=L_p\sqrt{1-{\left(\frac{v}{c}\right)}^2}\\ =\left(2.52\times {10}^{12}\text{ m}\right)\sqrt{1-{\left(\frac{0.8c}{c}\right)}^2}\\ =\left(2.52\times {10}^{12}\text{ m}\right)\sqrt{1-{\left(0.8\right)}^2}\\ =1.51\times {10}^{12}\text{ m}\end{array}$

Substitute $0.946c$ for $u_{12}$ and $1.51\times {10}^{12}\text{ m}$ for $L_1$ in equation (IV) to find the value for $t_1$

$\begin{array}{c}{t}_1=\frac{L_1}{u_{12}}\\ =\frac{1.51\times {10}^{12}\text{ m}}{\text{0}\text{.946}\left(\text{3}\times {\text{10}}^{\text{8}}\text{m}/\text{s}\right)}\\ =5321\text{ s}\\ =1.47\text{ hours}\end{array}$

Conclusion:

Thus, the time takes for the rockets to collide according to rocket 1 is $1.47\text{ hours}$.

To determine

The time takes for the rockets to collide according to rocket 2.

Answer to Problem 34P

The time takes for the rockets to collide according to rocket 2 is $1.97\text{ hours}$.

Explanation of Solution

Substitute $2.52\times {10}^{12}\text{ m}$ for $L_p$ and $0.6c$ for $v$ in equation (I) and solve for $L_2$

$\begin{array}{c}{L}_2=L_p\sqrt{1-{\left(\frac{v}{c}\right)}^2}\\ =\left(2.52\times {10}^{12}\text{ m}\right)\sqrt{1-{\left(\frac{0.6c}{c}\right)}^2}\\ =\left(2.52\times {10}^{12}\text{ m}\right)\sqrt{1-{\left(0.6\right)}^2}\\ =2.02\times {10}^{12}\text{ m}\end{array}$

Substitute $0.946c$ for $u_{12}$ and $2.02\times {10}^{12}\text{ m}$ for $L_2$ in equation (IV) to find the value for $t_2$

$\begin{array}{c}{t}_2=\frac{L_2}{u_{12}}\\ =\frac{2.02\times {10}^{12}\text{ m}}{\text{0}\text{.946}\left(\text{3}\times {\text{10}}^{\text{8}}\text{m}/\text{s}\right)}\\ =7104\text{ s}\\ =1.97\text{ hours}\end{array}$

Conclusion:

Thus, the time takes for the rockets to collide according to rocket 2 is $1.97\text{ hours}$.

To determine

Whether there will be any casualties in the rockets during the collision.

Answer to Problem 34P

There will be no casualties in rocket 2 during collision while there will be some casualties left in rocket 1 during collision.

Explanation of Solution

Both the rocket crews are capable of total evacuation with $90\text{ min}=1.5\text{ hours}$.

From part (d), the time takes for the rockets to collide according to rocket 1 is $1.47\text{ hours}$.

From part (e), the time takes for the rockets to collide according to rocket 2 is $1.97\text{ hours}$.

Conclusion:

Therefore, there will be no casualties in rocket 2 during collision while there will be some casualties left in rocket 1 during collision.