Chapter 26, Problem 56AP

An alien spaceship traveling 0.600c toward Earth launches a landing craft with an advance guard of purchasing agents. The lander travels in the same direction with a velocity 0.800c relative to the spaceship. As observed on Earth, the spaceship is 0.200 light-years from Earth when the lander is launched. (a) With what velocity is the lander observed to be approaching by observers on Earth? (b) What is the distance to Earth at the time of landcr launch, as observed by the aliens on the mother ship? (c) How long does it take the lander to reach Earth as observed by the aliens on the mother ship? (d) If the lander has a mass of 4.00 × 10⁵ kg, what is its kinetic energy as observed in Earth’s reference frame?

To determine

The speed of lander approaching observed from Earth.

Answer to Problem 56AP

The speed of lander approaching observed from Earth is $0.946c$.

Explanation of Solution

Given Info:

The velocity of the ship relative to Earth is $0.600c$.

The velocity of the lander relative to the ship is $0.800c$.

Formula to calculate the velocity of the lander relative to Earth is,

$v_{LE}=\frac{v_{LS}+v_{SE}}{1+\frac{v_{LS}{v}_{SE}}{c^2}}$

• $v_{LE}$ is the velocity of the lander relative to Earth
• $v_{LS}$ is the lander relative to ship
• $v_{SE}$ is the ship relative to Earth
• $c$ is the speed of light

Substitute $0.600c$ for $v_{SE}$ and $0.800c$ for $v_{LS}$ to find $v_{LE}$.

$\begin{array}{c}{v}_{LE}=\frac{\left(0.800c\right)+\left(0.600c\right)}{1+\frac{\left(0.800c\right)\left(0.600c\right)}{c^2}}\\ =\frac{1.40c}{1.48}\\ =0.946c\end{array}$

Thus, the speed of lander approaching observed from Earth is $0.946c$.

Conclusion:

The speed of lander approaching observed from Earth is $0.946c$.

To determine

The distance to Earth when the lander launched as measured by aliens.

Answer to Problem 56AP

The distance to Earth when the lander launched as measured by aliens is $0.160\text{ly}$.

Explanation of Solution

Given Info:

The proper length between the ship and Earth is $0.200\text{ly}$.

Formula to calculate the contracted distance measured by observers at rest relative to the spaceship is,

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

• $L$ is the contracted distance measured by observers at rest relative to the spaceship
• $L_p$ is the proper length between the ship and Earth

Substitute $0.200\text{ly}$ for $L_p$ and $0.600c$ for $v_{SE}$ to find $L$.

$\begin{array}{c}L=\left(0.200\text{ly}\right)\sqrt{1-{\left(\frac{0.600c}{c}\right)}^2}\\ =0.160\text{ly}\end{array}$

Thus, The distance to Earth when the lander launched as measured by aliens is $0.160\text{ly}$.

Conclusion:

The distance to Earth when the lander launched as measured by aliens is $0.160\text{ly}$.

To determine

The time taken by the lander to reach Earth as observed by aliens.

Answer to Problem 56AP

The time taken by the lander to reach Earth as observed by aliens is $0.114\text{yr}$.

Explanation of Solution

Formula to calculate time taken by the lander to reach Earth as observed by aliens is,

$t=\frac{L}{v_{LS}+v_{SE}}$

• $t$ is the time taken by the lander to reach Earth as observed by aliens

Substitute $0.800c$ for $v_{LS}$, $0.600c$ for $v_{SE}$ and $0.160\text{ly}$ for $L$ to find $t$.

$\begin{array}{c}t=\frac{0.160\text{ly}}{\left(0.800c\right)+\left(0.600c\right) }\\ =\frac{\left(0.160\text{ly}\right)\left(\frac{1c\text{yr}}{\text{ly}}\right)}{1.40c}\\ =0.114\text{yr}\end{array}$

Thus, the time taken by the lander to reach Earth as observed by aliens is $0.114\text{yr}$.

Conclusion:

The time taken by the lander to reach Earth as observed by aliens is $0.114\text{yr}$.

To determine

The kinetic energy of lander as observed in Earth’s reference frame.

Answer to Problem 56AP

The kinetic energy of lander as observed in Earth’s reference frame is $7.51\times {10}^{22}\text{J}$.

Explanation of Solution

Given Info:

The mass of lander is $4.00\times {10}^5\text{kg}$.

The speed of light is $3.00\times {10}^8\text{m/s}$.

Formula to calculate The kinetic energy of lander as observed in Earth’s reference frame is,

$K.E=\left(\frac{1}{\sqrt{1-{\left(\frac{v_{LE}}{c}\right)}^2}}-1\right)m{c}^2$

• $m$ is the mass
• $K.E$ is kinetic energy

Substitute $4.00\times {10}^5\text{kg}$ for $m$, $3.00\times {10}^8\text{m/s}$ for $c$ and $0.946c$ for $v_{LE}$ to find $K.E$.

$\begin{array}{c}K.E=\left(\frac{1}{\sqrt{1-{\left(\frac{0.946c}{c}\right)}^2}}-1\right)\left(4.00\times {10}^5\text{kg}\right){\left(3.00\times {10}^8\text{m/s}\right)}^2\\ =7.51\times {10}^{22}\text{J}\end{array}$

Thus, kinetic energy of lander as observed in Earth’s reference frame is $7.51\times {10}^{22}\text{J}$.

Conclusion:

Kinetic energy of lander as observed in Earth’s reference frame is $7.51\times {10}^{22}\text{J}$