Effect of Coriolis Force on Bodies falling Vertically Downward on Earth

Let a body of mass \(m\) is falling downward from a height \(h\) due to gravity. After \(t\) seconds its velocity will be

\[ \boxed{ \vec{v}' = -\hat{k}'gt } \qquad ...(1) \]

where \(\hat{k}'\) is the unit vector vertically upward.

If angle of latitude at this place is \(\lambda\) and the angular velocity of rotational reference frame is \(\vec{\omega}\), then

\[ \boxed{ \vec{\omega} = \hat{j}'\omega\cos\lambda + \hat{k}'\omega\sin\lambda } \qquad ...(2) \]

Thus, Coriolis force acting on the body is

\[ \vec{F}_c = -2m \left( \vec{\omega}\times\vec{v}' \right) \]
\[ = -2m \begin{vmatrix} \hat{i}' & \hat{j}' & \hat{k}'\\ 0 & \omega\cos\lambda & \omega\sin\lambda\\ 0 & 0 & -gt \end{vmatrix} \]
\[ \boxed{ \vec{F}_c = \hat{i}'\,2mgt\omega\cos\lambda } \qquad ...(3) \]

Since the direction of \(\hat{i}'\) is towards east, therefore the body falling downward in the northern hemisphere moves toward east.

Thus from Newton second law the equation of motion of a body towards east will be—

\[ m\frac{d^2x'}{dt^2} = 2mgt\omega\cos\lambda \] \[ \text{or}\qquad \frac{d^2x'}{dt^2} = 2gt\omega\cos\lambda \]

Integrating with respect to time \(t\), we have

\[ \frac{dx'}{dt} = (g\omega\cos\lambda)t^2+C \]

where \(C\) is an integral constant.

Since \(\frac{dx'}{dt}=0\) at \(t=0\), therefore \(C=0\).

\[ \therefore\quad \frac{dx'}{dt} = (g\omega\cos\lambda)t^2 \]

Again integrating,

\[ x' = \frac{1}{3} (g\omega\cos\lambda)t^3+C' \]

where \(C'\) is another integral constant.

Let the coordinates of point P at \(t=0\) is \((0,0,h)\).

Hence \(x'=0\) at \(t=0\). ∴ \(C'=0\).

\[ \boxed{ x' = \frac{1}{3} (g\omega\cos\lambda)t^3 } \qquad ...(4) \]

If the body is falling from a height \(h\), then time taken by the body in falling is—

\[ \boxed{ t= \left( \frac{2h}{g} \right)^{1/2} } \qquad \left[ \text{By }h=(0)t+\frac{1}{2}gt^2 \right] \qquad ...(5) \]

Substituting equation (5) in equation (4), we have

\[ x' = \frac{1}{3} (g\omega\cos\lambda) \left( \frac{2h}{g} \right)^{3/2} \]
\[ \boxed{ x' = \frac{2}{3} (h\omega\cos\lambda) \sqrt{\frac{2h}{g}} } \qquad ...(6) \]