What happens to the velocity of a photonic rocket as it approaches $c$?

Question

Imagine you have a photonic rocket (powered by something like antimatter). If it accelerates at a constant $9.8m/s^2$, it'll reach a velocity of $c$, the speed of light, in around 350 days. What happens when to the velocity when it's approaching at at this speed? If it's throwing out photons at say, $0.5c$ do they convey the same thrust at $0c$, less, or more?

Answer 1

I we ignore practicalities (eg fuel running out, collisions with dust particles etc), to the crew of the rocket, their ship appears to accelerate at the same rate for ever. In the original rest frame of the ship, the acceleration seems to approach zero as the ship's speed approaches c. Photons are ejected from the rocket at the same speed, c, at all times.

Answer 2

No, it will not reach $c$. According to special relativity acceleration is dependent on current speed :

$$ a = a_0 \sqrt{1- \frac {v^2}{c^2}} \tag 1,$$

where $a_0$ is initial acceleration when object has only rest mass $m_0$, i.e. when object starts accelerating from rest.

Limit of (1) expression shows that :

$$ \lim_{v \to c} ~a_0 \sqrt{1- \frac {v^2}{c^2}} = 0, \tag 2$$

so when object speed approaches light speed $c$,- acceleration on object approaches $0$, in affect canceling any acceleration and hence,- stopping body from reaching $c$.

In addition, your calculations what time period is required to reach approximate speed $c$ is also wrong. Time required to accelerate from $0$ to $v$ is :

$$ \Delta t = \frac {v}{a_0 \sqrt{1- \frac {v^2}{c^2}}} \tag 3$$

Let's say we want to reach $v=0.99c$, substituting this final speed into (3) expression and your initial acceleration $9.8 m/s^2$, gives that we will reach this speed only after $\approx 6.8~\textbf{years}$. Not even close to 350 days. Btw, as (3) formula shows, one needs infinite amount of time to reach exact speed $c$.