An alien spaceship traveling at 0.620 c toward the Earth launches a landing craf

Question

An alien spaceship traveling at 0.620c 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 on the Earth, the spaceship is 0.250 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?
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

show work please

Explanation / Answer

(a) What speed do the Earth-based observers measure for the approaching landing craft? c
u' = (u - v)/(1 - uv/c2)
0.780 = (u - 0.620)/(1 - u*0.620)
u = 0.9436
==> u = 0.9436 c

(b) What is the distance to the Earth at the moment of the landing craft's launch as measured by the aliens? ly
gamma = 1/(1-(v/c)^2)^0.5 = 1/(1-(0.620)^2)^0.5 = 1.2745
L = Lp/gamma = 0.250/1.2745 = 0.19615 = 0.196 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
t = 0.19615/0.620 = 0.3163 yr

(d*)Since we know the distance from Earth to the spaceship as seen from Earth's reference frame (0.250 ly) and we know the velocity of the landing craft as observed by Earth, all we need to do solve for time in the t=d/v equation.

Thus, we have

t= d/v = 0.250cy/ 0.9436 c = 0.2649y

(e*)Here we will treat Earth as the stationary inertial reference frame. We note that the velocity (as observed from Earth) of the landing craft is already known to be 0.9436c. We begin by finding the distance to Earth as measured by aliens on the landing craft (note: we are no longer considering the mothership's distance to Earth). To find this we begin by finding ? for the two inertial reference frames (Earth and landing craft):

y= 1/ sqrt (1-( ?^2)) = 3.02

here I used ?=0.9436

Earth's ovserver measures the distance do be .250ly, but since the landing craft is moving, the crew aboard sees a contraction in distance. We now use the length contraction equation (note: this does not always work for distances -- refer to the Twin Paradox):

L= Lp / y = 0.250cy / 3.02 = 0.0827cy

this is the distance to Earth as observed by the landing craft

Now it is obvious that we use t=d/v again, therefor we have:

t=d/v

= 0.0827cy / 0.9436 c = 0.0877y

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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