Question

3. A spacecraft is moving relative to the Earth and an observer on Earth finds that between exactly 1pm-2pm according to her clock, 3601s elapse on the spacecraft's clock. What is the spacecraft's speed relative to Earth?

Answer 1

Spacecraft's speed relative to Earth is 0.14c .

Step-by-step explanation:

Let v be the speed of the spacecraft with respect to Earth's frame. According to special theory of relativity, there is time dilation i.e. given by the relation :

t = t₀γ

Here t is time measured in moving frame, t₀ is time measured in rest frame and γ is constant.

We know that γ =
`\frac{1}{\sqrt{1-\ (v^(2) )/(c^(2) ) } }`

Here c is the speed of light.

So, t =
`\frac{t_(0) }{\sqrt{1-\ (v^(2) )/(c^(2) ) } }` .......(1)

According to the problem, the time measure in Earth's frame is :

t₀ = 1 hr = 60 min =60 x 60 s = 3600 s

The time measured in the space craft frame is :

t = 3601 s

Substitute t and t₀ in equation (1) :

3601 =
`\frac{3600}{\sqrt{1-\ (v^(2) )/(c^(2) ) } }`

`{\sqrt{1-\ (v^(2) )/(c^(2) ) } } = (3600)/(3601)`

`1 - (v^(2) )/(c^(2) ) = 0.99^(2)`

`(v^(2) )/(c^(2) ) = 1 - 0.98`

v = 0.14 c