Sirius is about 9.0 light-years from Earth.
To reach the star by spaceship in 10 years (ship time), how fast must you travel?
How long would the trip take according to an Earth-based observer?
How far is the trip according to you?
To calculate the spaceship's required speed, we need to convert the given distance of 9.0 light-years into a distance traveled within 10 years according to the spaceship's onboard clock.
Since the spaceship is moving near the speed of light, we will use the time dilation formula from special relativity:
t(ship) = t(earth) / γ
Where:
t(ship) is the time experienced on the spaceship
t(earth) is the time experienced on Earth
γ (gamma) is the Lorentz factor, defined as 1 / sqrt(1 - (v^2 / c^2)), where v is the velocity of the spaceship, and c is the speed of light.
Let's calculate the velocity (v) needed for the spaceship to travel 9.0 light-years in 10 years (according to the spaceship's clock):
Distance = Velocity × Time
9.0 light-years = Velocity × 10 years
To convert years into seconds, we'll multiply both sides by 365 days/year × 24 hours/day × 60 minutes/hour × 60 seconds/minute:
9.0 light-years = Velocity × 10 years × (365 days/year × 24 hours/day × 60 minutes/hour × 60 seconds/minute)
Now we can plug in the numbers:
9.0 light-years = Velocity × (10 × 365 × 24 × 60 × 60) seconds
Rearranging the equation to solve for Velocity:
Velocity = 9.0 light-years / (10 × 365 × 24 × 60 × 60) seconds
Calculating this, we find:
Velocity ≈ 0.0009476 light-years/second
To find the time it would take according to an Earth-based observer, we divide the distance by the spaceship's velocity:
Time = Distance / Velocity = 9.0 light-years / 0.0009476 light-years/second
Calculating this, we find:
Time ≈ 9476 years
According to an Earth-based observer, it would take approximately 9476 years to complete the trip.
Lastly, to determine the distance as measured by someone on the spaceship, we use the time dilation formula mentioned earlier:
Distance(ship) = Velocity × t(ship)
Plugging in the values:
Distance(ship) = 0.0009476 light-years/second × 10 years
Calculating this, we find:
Distance(ship) ≈ 0.009476 light-years
According to someone on the spaceship, the trip would be approximately 0.009476 light-years.
Your question spurred me to look up the topic, being totally unfamiliar with it at this point.
I found it stated that Einstein’s special theory of relativity states that the time scale of an object moving at high speed is shortened in accordance with
……………t/t(o) = 1/sqrt[1 – (v^2/c^2)] where
t = the travel time to Sirius in seconds as observed by a person on Earth
t(o) = the travel time to Sirius as observed on a clock on the spaceship (10 years)
c = the speed of light = 186,300 m/s
v = the velocity of the spaceship as viewed by an observer on Earth
v(o) = the velocity of the spaceship as viewed by an observer on the spaceship
Let d = the distance from Earth to Sirius in miles = 9 light years
Given the information you offered, you should be able to derive the answers you seek.
Note that many scientists regard the contraction of time as a real phenomenon while others maintain that it is merely apparent.
A google search will undoubtedly provide you with much more information.