Question

As a spaceship flies past with speed v, you observe that 1 hr elapses on the ship's clock in the same time that 7 years elapses on Earth. How fast is the ship traveling, relative to the Earth?]

Answer 1

According to special relativity, if an observer at rest measures a time interval Δt, then a clock traveling at speed v woud measure a time Δt' given by:

`\Delta t=\frac{\Delta t^(\prime)}{\sqrt{1-(v^2)/(c^2)}}`

Notice that the denominator of the expression is always lower or equal to 1, which means that Δt (the time measured by a clock at rest) is greater than Δt'. This can be interpreted as if the moving clock was ticking slower than the clock at rest.

Isolate v from the expression:

`\begin{gathered} \Rightarrow\sqrt{1-(v^2)/(c^2)}=(\Delta t^(\prime))/(\Delta t) \\ \\ \Rightarrow1-(v^2)/(c^2)=\left((\Delta t^(\prime))/(\Delta t)\right)^2 \\ \\ \Rightarrow(v^2)/(c^2)=1-\left((\Delta t^(\prime))/(\Delta t)\right)^2 \\ \\ \Rightarrow(v)/(c)=\sqrt{1-\left((\Delta t^(\prime))/(\Delta t)\right)^2} \\ \\ \Rightarrow v=c\sqrt{1-\left((\Delta t^(\prime))/(\Delta t)\right)^2} \end{gathered}`

Since 7 years elapses on Earth while only 1 hour elapses on the ship, then, replace Δt'=1h and Δt=7y, as well as c=3*10^8m/s:

`\begin{gathered} v=(3*10^8(m)/(s))*\sqrt{1-\left((1h)/(7y*365.25(d)/(y)*24(h)/(d))\right)^2} \\ \\ =(3*10^8(m)/(s))*\sqrt{1-\left((1h)/(61,362h)\right)^2} \end{gathered}`

Since the quotient (1h/61,362h)^2 is much smaller than 1, we can approximate the value of the square root as follows:

`√(1-x)\approx1-(1)/(2)x`

Then:

`\sqrt{1-\left((1)/(61,362)\right)^2}\approx1-(1)/(2)\left((1)/(61,362)\right)^2=0.999999999867`

Then, the speed of the spaceship is:

`v=(3*10^8)(0.999999999867)=2.9999999996*10^8(m)/(s)`

Therefore, the speed of the spaceship is 2.9999999996*10^8 meters per second, which is equivalent to 0.99999999987 times the speed of light.