i tried this problem... but im not getting the answer.. can u please solve it and explain the answer
A space traveller takes off from earth and moves at speed 0.99c [c is sped of light] towards the star Vega,which is 26light years distant.How much older will Earth Observers calculate thetaveler to be(meaured from her frme) when she reaches Vega than she was when she started the trip.
Ans. is not 52 yrs nor 26 yrs
2006-08-10 06:38:24 · 8 answers · asked by Prakash 4 in Science & Mathematics ➔ Astronomy & Space
the answers is 3.7yrs and i need the working and this is not my home work
question is based on relativity of time
2006-08-10 06:47:38 · update #1
T(stationary) = 1/SQRT(1-v^2/c^2) * T(moving)
Plug in 26 years for T(stationary), 1 for c, 0.99 for v and solve for T(moving)
26 = 1/SQRT(1-0.99^2/1^2) * T
T = 26 * SQRT(1-.99^2) = 3.7 years.
2006-08-10 07:29:17 · answer #1 · answered by campbelp2002 7 · 0⤊ 0⤋
The time dilation equation is
t = t(0) / sqrt (1 - v^2/c^2)
t(0) is 26 years, v=0.99c and c is a constant
So 26 years/ sqrt(1-(0.99c)^2/c^2) = 26 years / sqrt (1-0.9801)
= 26 years/0.1411 = 184.31 years
Subtract the 26 subjective years your traveller experiences, and she will be 158.31 years older than her observer on Earth.
2006-08-10 07:06:02 · answer #2 · answered by theyuks 4 · 0⤊ 0⤋
If the Earth Observer calculates assuming time dilation. Then
(1-.99)^2 * 26 = .0026 years
2006-08-10 06:46:17 · answer #3 · answered by rscanner 6 · 0⤊ 0⤋
v= 0.99c
distance x= 26light years
1 light year=9.461×1015 m
therefore 26 light years = 26 x 9.461×1015 m
time taken by the astraunaut in frame of reference of earths observer is x/v= 84065.92* 10 to power (-8) sec older.
2006-08-10 14:57:17 · answer #4 · answered by Mysterious 3 · 0⤊ 0⤋
When you travel at 0c it means that your time is at par with everybody else(even if you're moving in a bus or something that wud only be maybe 0.000000001c !)
If you travel at 1c(the speed of light), it means that your time will not move ahead at all, (with reference to a static observer's point of view).
Thus, when you travel at 0.99c, it means that for every 0.01secs of your time, others will experience 1 second of normal time.
So, she will reach the star in
26*0.99 = 25.74 normal years.
Thus, to Earthlings, she will have aged 25.74years.
But now remember that for every 0.01secs of her time, others will experience 1 second of normal time.Thus, we have:
If 1= 25.74
0.01= ?
Solving, we get 0.2574years i.e. she has aged wrt her time only .2574 years!
Thus the difference in age, as perceived by the timeframes
=Age in Earth timeframe -Age in her timeframe = 25.74 - 0.2574
= 25.4826 years!
2006-08-10 07:05:50 · answer #5 · answered by DichloroDiphenyl 5 · 0⤊ 0⤋
the observers will be dead at that time
2006-08-10 06:44:35 · answer #6 · answered by Anonymous · 0⤊ 0⤋
I think rscanner is right...
2006-08-10 07:08:31 · answer #7 · answered by Anonymous · 0⤊ 0⤋
There are a few grammatical mistakes that is making me confuse. Please rectify them.
2006-08-10 06:58:54 · answer #8 · answered by tuhinrao 3 · 0⤊ 0⤋