Question

At what speed, in m/s, would a moving clock lose 4.5 ns in 1.0 day according to experimenters on the ground

Answer 1

Answer:

the speed at which a moving clock lose 4,5ns in 1.0 day is 98.58 m/s

Explanation:

Let's first find the relation for the time dilation by using the formula:

$\Delta t' = \Delta t \gamma$

Making $\gamma$ the subject of the formula, we have:

$\gamma= \dfrac{\Delta t'}{\Delta t}$

Number of day(s) = 1

1 day in seconds = 24 × 60 × 60

= 84600 seconds

$\gamma= \dfrac{86400 \ s}{86400 \ s - 4.5 \ ns}$

$\sqrt{1- \dfrac{v^2}{c^2} }= \dfrac{86400 \ s - 4.5 \ ns}{86400 \ s}$

$\sqrt{1- \dfrac{v^2}{c^2} }= 1 - 5.4 \times 10^{-14}$

From the above equation, if we apply binomial expansion to it, we have the following,

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

⇒$1 -\dfrac{1}{2}(\dfrac{v^2}{c^2})$

So;

$1 -\dfrac{1}{2}(\dfrac{v^2}{c^2}) = 1 - 5.4 \times 10^{-14}$

$\dfrac{1}{2}(\dfrac{v^2}{c^2}) = 5.4 \times 10^{-14}$

$\dfrac{v^2}{c^2} = 2 \times 5.4 \times 10^{-14}$

$\dfrac{v^2}{c^2} = 1.08 \times 10^{-13}$

$v^2 = 1.08 \times 10^{-13} \times c^2$

where ;

$c = 3 \times 10^8 \ m/s$

$v^2 = 1.08 \times 10^{-13} \times (3 \times 10^8 \ m/s)^2$

$v = \sqrt{ 1.08 \times 10^{-13} \times (3 \times 10^8 \ m/s)^2}$

$v = \sqrt{ 1.08 \times 10^{-13} } \times (3 \times 10^8 \ m/s)$

$v =3.286 \times 10^{-7} \times (3 \times 10^8 \ m/s)$

$\mathbf{v =98.58 \ m/s}$

Therefore, the speed at which a moving clock lose 4,5ns in 1.0 day is 98.58 m/s