An alien spaceship traveling at 0.560 c toward the Earth launches a landing craf

Question

An alien spaceship traveling at 0.560c toward the Earth launches a landing craft. The landing craft travels in the same direction with a speed of 0.780c relative to the mother ship (as measured by aliens on the mother ship). As measured on the Earth, the spaceship is 0.290 ly from the Earth when the landing craft is launched.

(a) What speed do the Earth-based observers measure for the approaching landing craft?
c

(b) What is the distance to the Earth at the moment of the landing craft's launch as measured by the aliens (mother ship)?
ly

(c) What travel time is required for the landing craft to reach the Earth as measured by the aliens on the mother ship?
yr

(d*) What travel time is required for the landing craft to reach the Earth as measured by Earth-based observers?
yr

(e*) What travel time is required for the landing craft to reach the Earth as measured by those on the landing craft?
yr

Explanation / Answer

Ans:-

For relativity problems, the first step is to set up coordinate systems and clearly define all the velocities and the reference frames the velocity is relative to.   

A] In this part, set the earth being the rest frame (O frame), the mother ship as O' frame. Then v=0.560c and the landing craft relative to the mother ship is u'=0.780c. To find the landing craft's velocity relative to the earth, we can use the Lorentz transformation on velocity:

                                u = (u’+ v)/(1+vu’/c^2)

                                   = 0.780c + 0.560c/(1+0.560c*0.780c/c^2)

                                u =0.9326c

(b) From the mother ship's point of view, the earth is moving toward's mother ship at v=-0.560c, therefore, we have

x=x/

Here x is the length measured in the earth frame.

Gamma for 0.560c is 1/sqrt (1-(0.560)^2)= 1.207

x’ = 0.290/1.207 = 0.240ly

(c) From the mother ship, the distance between the earth and the landing craft is x, the landing craft is moving in +x direction with u'=0.780c relative to the mother ship and the earth is moving towards the -x direction with u'e=-0.560c.

The time for the landing craft to meet the earth calculated by the observer on the mother ship would be

t=x/uue

= 0.240/0.780-0.560 =1.091yr

(d) From the earth based observer, the landing craft is traveling at a velocity u found in part (a). The distance between the landing craft and the earth is given as 0.28 ly when the landing craft is released. Therefore, the time takes the landing craft to meet the earth calculated in earth frame is

t=x/u

= 0.290/0.9326

= 0.311yr

(e) For those on the landing craft, the earth's velocity is v=u. Here u is the velocity found in part a. Now we need to find the distance between the earth and the landing craft as observed by the landing craft.

We can use this velocity to calculate for the observer on the landing craft to observe earth events. Then we can calculate the distance with x=x/. Then the time observed on the landing craft is

t=x/v

x’’ = 0.240ly/1.207 =0.1988ly

t= 0.1988ly/ 0.560c = 0.355yr

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.