Why only one pseudo force for a revolving frame? Pg 348, Kleppner and Kolenkow.
Earth in a free fall towards sun is shown.
Authors write

The earth accelerates toward the sun at rate $\mathrm{A}=\mathrm{G}_{0}$
...
The apparent force to an earthbound observer is
$\mathbf{F}^{\prime}=\mathbf{F}-m \mathbf{A}=m\left[\mathbf{G}(\mathbf{r})-\mathbf{G}_{0}\right]$


What is the proof that the pseudo force required in this case is and only is $-mA$?
That would have been true had the accelerating frame been moving in a straight line ,but here we've a revolving coordinate system that maybe needs Coriolis as well.
Can anyone please help me understand this?
 A: I can think of two reasons to ignore the Coriolis force but include the centrifugal force, as has been done here for our "earthbound observer".
First, we could be considering an observer who orbits the Sun along with the Earth, but does not rotate in space at all: she always faces the same star, with the Sun on her left half the year and on her right the other half. The acceleration, $\mathbf{A}$, of this frame is constant everywhere in space and rotates with time.  The pseudo-force that arises in this frame (an Euler force) is $-m\mathbf{A}$, and there is no Coriolis force. The magnitude of $\mathbf{A}$ is $\Omega^2 r_s$, where $\Omega$ is $2 \pi$ per year.
Second, we could be considering a frame that rotates around the Sun, with angular speed $\Omega$. So, the observer is not rotating daily with the Earth, but instead keeps the Sun always on her left, say, as if the Earth's rotation was locked to its orbit about the Sun the way the Moon's is about the Earth.  In this case, you're right, $-m\mathbf{A}$ is the centrifugal force, and the Coriolis force is present but being neglected.  But neglecting it might be pretty fair: the magnitude of the centrifugal acceleration, $\Omega^2 r_s$, is $6 \text{ mm/s}^2$. The Coriolis acceleration depends on the speed and direction of the object, but for something moving at $v=15 \text{ m/s}$, the maximum magnitude, $2\Omega v$, is 1000 times smaller ($0.006 \text{ mm/s}^2$).
As a side note, the pseudo-force that arises in the frame of something in free fall can usually be ignored: it cancels out with the gravity.  You can leave out the gravity that's causing the falling motion and also the pseudo-force, and everything is simpler.  The exception is if you're dealing with the tidal force experienced by an extended body.
A: The word apparent may be misleading and give the impression that fictitious forces are involved. If $\bf G(r)$ is the acceleration due to the Sun and Earth's combined gravitational fields on a freely falling body at a displacement $\bf r$ (from the centre of the Earth), and $\bf G_0$ is the acceleration due to the Sun's gravity field on a freely falling body at the centre of the Earth, then when a freely falling body is at $\bf r$ its acceleration relative to the Earth will be $\bf G(r)-G_0$. So the force on it measured by a (non-rotating) observer on Earth will be $m[\bf G(r)-G_0\bf]$. Hence the equation.
This is without considering any effects of rotating frames of reference. Any such effects would need to be allowed for separately.
