We continue on with the same situation as before. After a time T, the position of the spaceship can be written as \(r=\frac{1}{2}gT^2\)
This means that the time elepsed when the spaceship has reached a distance \(r\) from event E can be written as:
\(r(t) = \frac{1}{2}gt^2 \longleftrightarrow t = \sqrt{\frac{2r}{g}}\)
We insert this into the expression for the relation between a time interval in the spaceship frame and a time interval in the planet frame, \(\Delta T' = \sqrt{1-g^2t^2}\Delta T\), and get:
\(\Delta T' = \sqrt{1-2gr}\Delta T\)
where r is still the position after a time T.
How do we relate this to acceleration and gravity? Well, we know that the gravitational acceleration is given by \(g=\frac{GM}{r^2}\). Let's see what happens if we insert this for g:
\(\Delta T' = \sqrt{1-2\frac{GM}{r^2}r}\Delta T= \sqrt{1-\frac{2GM}{r}}\Delta T\)
Seemingly, you cannot judge whether you are in an accelerated frame or in a gravitational field! We see that the form of the expressions for time dilation in an accelerated frame, or in a gravitational field take the same form. We would actually have gotten the same results if the astronaut had been in a gravitational field for some time rather than being accelerated. The formula shows us that time will run slower for the astronaut being either accelerated or in a gravitational field compared to observers in the frame not being accelerated/in a gravitational field.
This formula will show up again, so keep it in mind!