From a standstill. you begin jogging at$5m/s$ directly toward the galaxy Centaurus A. which is on the horizon $2\times {10}^{23}m$away.

(a) There is a clock in Centaurus A. According to you. How Will readings on this clock differ before and after you begin jogging? (Remember: You change
frames.)

(b) The planet Neptune is between Earth and Centaurus $4.5\times {10}^{9}m$is from Earth- How much readings a clock there differ?

(c)What would the time differences if had instead begun jogging in the opposite direction?

(d) What these results tell you about the observations of a traveling twin who accelerates toward his Earth-bound twin? How would these observation depend on the distance the twins?

The velocity of the frame,$v=5m/s$

The position of the event in frame,$x=2\times {10}^{23}\text{m}$

The expression to convert the time from the frame to frame is given as follows.

$t\text{'}={\gamma }_v\left(t-\frac{v}{c^2}x\right)$ ...(i)

Here, t is the time in frame S, t is the time in frame S,${\gamma }_v$ is the Lorentz factor for the velocity, C is the speed of the light.

Let assume you and Centaurus were in the same frame S , therefore the clock of Earth and Centaurus reads the same. When you move into the frame S, your clock reads 0 as you start.

Calculate the time of the clock in Centaury,

Substitute 0 for t into equation (i).

$\begin{array}{rcl}0& =& {\gamma }_v\left(t-\frac{v}{c^2}x\right)\\ 0& =& t-\frac{v}{c^2}x\\ t& =& \frac{v}{c^2}x\\ & & \end{array}$ ...(ii)

Substitute $5m/s$for V, $3\times {10}^{8}m/s$for C and $2\times {10}^{23}\text{m}$for X into equation (ii)

$\begin{array}{rcl}t& =& \frac{5m/s}{{\left(3\times {10}^{8}m/s\right)}^2}2\times {10}^{23}\text{m}\\ t& =& \frac{10\times {10}^{23}}{9\times {10}^{16}}sec\\ t& =& 1.11\times {10}^7\text{sec}\end{array}$

Convert the time from seconds to days,

$$\begin{gathered} t=1.11\times {10}^7\times \frac{1}{3600\times 24} \\ t=128.47\text{days} \end{gathered}$$

Hence the time of the clock in the Centaury is $128.47$days.

The position of the event in frame $x=4.5\times {10}^9\text{m}$,

Calculate the time on Neptune.

Substitute $5m/s$for v ,$3\times {10}^{8}m/s$for C and $4.5\times {10}^9\text{m}$for x into equation (ii)

$$\begin{gathered} t=\frac{5m/s}{{\left(3\times {10}^{8}m/s\right)}^2}4.5\times {10}^9\text{m} \\ t=\frac{22.5\times {10}^9}{9\times {10}^{16}}\text{sec} \\ t=2.5\times {10}^{-7}\text{sec} \\ t=250\text{ns} \end{gathered}$$

Hence the time on clock of Neptune is $250\text{ns}$.

If had began instead jogging in opposite direction then all Centaury and Neptune would be in behind you and everything will be remaining the same accept the distances. Now use the negative distance. Therefore, clock will jump by the same amount but in the opposite direction.

From the above discussion, if the twin is accelerated toward you, the twin will see the time on your clock jump ahead, and it is depending upon the distance between the twin and you.