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

Answer:

Spacecraft's speed relative to Earth is 0.14c .

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

Here c is the speed of light.

So, t = $\frac{t_{0} }{\sqrt{1-\ \frac{v^{2} }{c^{2} } } }$        .......(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 = $\frac{3600}{\sqrt{1-\ \frac{v^{2} }{c^{2} } } }$

${\sqrt{1-\ \frac{v^{2} }{c^{2} } } } = \frac{3600}{3601}$

$1 - \frac{v^{2} }{c^{2} } = 0.99^{2}$

$\frac{v^{2} }{c^{2} } = 1 - 0.98$

v = 0.14 c