Question

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×107 light-years from Earth. If the lifetime of a human is taken to be 70.0 years, a spaceship would need to achieve some minimum speed min to deliver a living human being to this galaxy. How close to the speed of light would this minimum speed be? Express your answer as the difference between min and the speed of light .

Answer 1

`\Delta V=3.43*10^(18)(km)/(year)`

Step-by-step explanation:

A light-year is the distance that the light travels in vacuum in one year. If we take the light velocity as 299792458 m/s, we would be able to know how many km a light-year is:

The light displacement is taken as a velocity constant displacement, where the distance 'x', time 't' and velocity 'v' are related as:

`x=v*t`

We have the velocity (299792458 m/s) and the time (1 year), so we can calculate the distance, converting first the velocity to the proper units:

`299792458(m)/(s) *(1km)/(1000m) *(3600s)/(1h) *(24h)/(1day) *(365days)/(1year)=9.45425496*10^(12) (km)/(year)`

The conversion was made by multiplying by conversion factors, from meter to km, and from seconds, to hours, then to days, and then to years. This data is just the velocity of light in km/year. This means that a light-year (the distance that the velocity travels in one year) is
`9.45425496*10^(12)km`.

Andromeda Galaxy is
`2.54*10^7` light years; it is the distance that the light would travel in
`2.54*10^7` years, so that distance is:

`x=v*t`

`x=2.54*10^7-years*(9.45425496*10^(12))/(year)=2.40*10^(20)km`

Now, as we want to deliver a living human to this galaxy, we would need to cover that distance (
`2.40*10^(20)km`) in 70.0 years. In order to find the velocity, consider again:

`x=v*t\\v=(x)/(t)`

We have the distance and the time, so it is possible to calculate the velocity required:

`v=(2.40^(20)km)/(70.0years)=3.43*10^(18)(km)/(year)`

Then, let's see how much is the difference between this velocity and the speed of light:

`\Delta V=v_(min)-v_(light)\\\Delta V=3.43*10^(18)(km)/(year) -9.45*10^(12)(km)/(year) \\\Delta V=3.43*10^(18)(km)/(year)`

As you see, the required velocity is as big as the light velocity is negligible compared to that.