A very fast spaceship with a speed of 0.80c travels to the nearby star Omega Cen

Question

A very fast spaceship with a speed of 0.80c travels to the nearby star Omega Centauri, which is 12 light years away from Earth. If the pilot turned 21 years old on the day she left Earth, find the following:
1) How much Earth time will have elapsed by the time she arrives Omega Centauri?
2) What is the biological age of the pilot when she arrives at Omega Centauri? A very fast spaceship with a speed of 0.80c travels to the nearby star Omega Centauri, which is 12 light years away from Earth. If the pilot turned 21 years old on the day she left Earth, find the following:
1) How much Earth time will have elapsed by the time she arrives Omega Centauri?
2) What is the biological age of the pilot when she arrives at Omega Centauri?
1) How much Earth time will have elapsed by the time she arrives Omega Centauri?
2) What is the biological age of the pilot when she arrives at Omega Centauri?

Explanation / Answer

Solution: Speed of the spaceship v = 0.80c; where c = 3*108 m/s = speed of light

Distance of the Omega Centauri from the Earth d = 12 light years = 12 ly.

Number of seconds in one year is given by,

t = [(365days/yr)*(24hr/day)*(60min/hr)*(60sec/min)]

t = 3.1536*107 sec

1 ly = The distance covered when travelling at the speed c in one year .

1 ly = c*t

1 ly = (3*108 m/s)*( 3.1536*107 s)

1 ly = 9.4608*1015 m

The pilot turns 21 years on the day she lives the Earth.

Part (a) From the frame of reference based on Earth, speed of the space ship is v = 0.80c,

Distance to Omega Centauri d = 12 ly

tEarth = d/v

tEarth = 12 ly / (0.8c)

tEarth = 12*(9.4608*1015 m) / (0.8*3*108 m/s)

tEarth = 4.7304*108 sec

tEarth = (4.7304*108 sec)*(1min/60sec)*(1hr/60min)*(1day/24hr)*(1yr/365day)

tEarth = 15 yr

Thus 15 years of Earth time will have elapsed by the time she arrives Omega Centauri.

Hence answer is 15 years.

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Part (b) For the people on the Earth, the age of the pilot when she arrives will be (21yr + 15 yr) = 36 yr.

For people on Earth, time runs slower in the spaceship as it is moving with speed v = 0.8c. The time dilation is given by,

to = tEarth/

Where to = proper time with respect to pilot. (time measured by pilot in the moving ship)

= Lorentz factor.

Lorentz factor is given by,

= 1 / (1 – v2/c2)

= 1 / (1 – (0.8c)2/c2)

= 1 / (1 – 0.82)

= 1.66666666…..

= 1.6667

tEarth = Time interval between spaceship leaving the Earth and arriving at the Omega Centauri for people on the Earth.

tEarth = 15 years

Thus, proper time measured by the pilot t is given by,

t = tEarth /

t = (15 yr)/1.6667

t = 9 years.

It means that for 15 years of Earth time elapsed, only 9 years elapses in the moving spaceship.

It also means that the biological age of the pilot increases by only 9 years as she too is in the moving ship (moving frame of reference)

Thus 9 years of the biological age of the pilot when she arrives at the Omega Centauri.

Hence answer is 9 years.

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