Question

An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same direction with a speed of 0.800 c relative to the mother ship. As measured on the Earth, the spaceship is 0.200 ly from the Earth when the landing craft is launched.(d) If the landing craft has a mass of 4.00 × 10⁵ kg , what is its kinetic energy as measured in the Earth reference frame?

Answer 1

The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

What is its kinetic energy as measured in the Earth reference frame?

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

                  $V_x'=0.8c\\V=0.6c\\m=4*10^5kg$

• Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

                  $KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

• So, to $V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016$find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
• We have an expression from Lorents transformation for relativistic law of addition of velocities as,

                      $V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V$

• Substituting values, we get,

          $V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c$

• Thus, the KE will be,

              $KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J$

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

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