I would like to ask a stupid question here.
If a body 'b' moving downward with a velocity $v$ in a rotating frame of reference with angular velocity $\omega$, and $\omega$ and $v$ not being parallel and anti parallel. We all know that the body 'b' experiences a Coriolis force.
If it gets deflected in its trajectory then according to newtons third law shouldn't it push some body with equal force?
what if the frame of reference was moon where there is no atmosphere? will the body be deflected?
Nothing feels a reaction force to the Coriolis force because Newton's laws only apply in an inertial frame. If we imagine a frame with some sort of wild, random acceleration, then in that frame at a given moment it would look like, in a room with completely stationary tables and chairs and things, everything was simultaneously accelerating the same direction, and there would be huge "fictitious forces" needed to explain this. There are no third-law pairs going on, though. The fictitious forces were necessary to make $F=ma$ work out in this accelerating frame, but by introducing them we sacrificed the third law (at least in regards to those kinematic forces).
Imagine dropping a ball off a tower. It gets deflected east.
If we analyze this scenario from a frame rotating along with the Earth, there is no conflict with Newton's third law because it does not hold for the Coriolis force or other kinematic forces.
If we analyze the scenario from an inertial frame in which the Earth is rotating, the trajectory simply is not deflected. The ball lands east of where it was dropped because its velocity in a tangential direction, which comes from the rotation of the Earth, stayed the same, but as it fell its angular velocity $v/r$ increased because $r$ decreased. That led it to "deflect east" when it was really moving in a perfectly ordinary parabola.
I am not sure (!), but I think in the next way.
First, consider a bead sliding along a rod without a friction in a inertial frame $Oxyz$. The rod rotates around axis $Oz$ with given function for angular velocity: $$\omega = \omega(t)$$ Let write Newton's equation of motion for the bead in a frame $Ox'y'z'$ which rotates with the rod ($Ox'$ is parallel to the rod, $z=z'$), using formulas for inertial forces: $$m\textbf{a}'=\boldsymbol N+m[\textbf{r}',\frac{d\boldsymbol{\omega}}{dt}]+2m[\boldsymbol{v}',\boldsymbol{\omega}]+m[\boldsymbol{\omega},[\boldsymbol{r}',\boldsymbol{\omega}]]=\boldsymbol N+\boldsymbol F_{Euler}+\boldsymbol F_{Coriolis}+\boldsymbol F_{centrifugal}$$ $$\boldsymbol{v}'=\dot{\boldsymbol{r}}',\boldsymbol a'=\ddot{\boldsymbol{r}}',\boldsymbol{r}'=(x',y',z')$$
A force $\boldsymbol N$ is a total (unknown) force of reaction from the rod to the bead. Without friction the force $\boldsymbol N$ is orthogonal to $\boldsymbol r'$. The bead is sliding along the rod so: $$\boldsymbol r'=(x',0,0)$$ $$\boldsymbol N=(0,N_{y'},0)$$
Let $\omega(t)=const$, then $\boldsymbol F_{Euler}=0$. After all ($\boldsymbol F_{Coriolis}$ is also orthogonal to $\boldsymbol r'$) Newton's equation of motion gives a system of equations:
$$\begin{cases}\ddot x'=m\omega^2x'\\0=N_{y'}-2m\dot x'\omega\end{cases}$$ So when $\omega=const$ the reaction force $\boldsymbol N$ is just opposite to Coriolis force (from $Ox'y'z'$) and $\boldsymbol N= \boldsymbol N(v_r(t),\omega)$ in both $Oxyz$ and $Ox'y'z'$ frames, where $v_r(t)$ is a radial velocity of the bead.
Secondary, consider a planet which rotates around a star by circular orbit with a constant angular velocity in inertial frame $Oxyz$. A using of the rotated frame $Ox'y'z'$ gives us null values for Euler and Coriolis forces. In this case only centrifugal is here. And centrifugal force just is opposite to Newton Gravity force. $$\boldsymbol r'(t)=(x'(t),0,0)=(R,0,0)=const$$ $$m\ddot x'=-GmM/{x'}^2+F^{centrifugal}_{x'}=0$$ where $R$ is the orbit radius. Thus $$-GmM/{x'}^2+m{\omega}^2x'=0$$
In the case of an elliptic orbit centrifugal force isn't equal to Gravity force and the radius $R=R(t)=x'(t)$ is changed. The both Euler and Coriolis forces are not null, still orthogonal to radius-vector $$\boldsymbol r'(t)=(x'(t),0,0)=(R(t),0,0)$$ $$\boldsymbol F^{Euler}=(0,F^{Euler}_{y'},0)$$ $$\boldsymbol F^{Coriolis}=(0,F^{Coriolis}_{y'},0)$$ and their sum must be zero $$F^{Euler}_{y'}+F^{Coriolis}_{y'}=0$$ Probably the last formula gives an equation for radial velocity, tangential velocity and distance for the planet to the star $$-mx'\dot\omega+2m\dot x'\omega=0$$ $$v_r=\dot x',v_\tau=\omega x',\dot\omega=\frac{d}{dt}(v_\tau/x')$$
