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:

$\Delta t=\frac{\Delta t^{\prime}}{\sqrt{1-\frac{v^2}{c^2}}}$

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:

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

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:

$\begin{gathered} v=(3\times10^8\frac{m}{s})\times\sqrt{1-\left(\frac{1h}{7y\times365.25\frac{d}{y}\times24\frac{h}{d}}\right)^2} \\ \\ =(3\times10^8\frac{m}{s})\times\sqrt{1-\left(\frac{1h}{61,362h}\right)^2} \end{gathered}$

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

$\sqrt{1-x}\approx1-\frac{1}{2}x$

Then:

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

Then, the speed of the spaceship is:

$v=(3\times10^8)(0.999999999867)=2.9999999996\times10^8\frac{m}{s}$

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.