Effect of Coriolis Force on Bodies Thrown Vertically Upward from Earth

If a body of mass \(m\) is thrown vertically upward with a velocity \(u\) from a certain place of the earth.

After \(t\) seconds the velocity of a body will be—

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

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

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

Coriolis force acting on the body is

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

Thus the displacement of the body due to Coriolis force acting toward west in the northern hemisphere is toward west.

From Newton second law,

\[ m\frac{d^2x'}{dt^2} = -2m(u-gt)\omega\cos\lambda \]
\[ \text{or}\qquad \frac{d^2x'}{dt^2} = -2 \left( \frac{u}{g}-t \right) g\omega\cos\lambda \qquad ...(4) \]

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

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

where \( C \) is an integral constant.

Since

\(\dfrac{\mathrm{d}x'}{\mathrm{d}t} = 0\)

at \( t = 0 \)
therefore \( C = 0 \).

\[\therefore \quad \dfrac{\mathrm{d}x'}{\mathrm{d}t} = -2(g\omega\cos\lambda)\left(\dfrac{ut}{g} - \dfrac{t^{2}}{2}\right)\]

Again integrating,

\( x' = -(g\omega\cos\lambda)\left(\dfrac{ut^{2}}{g} - \dfrac{t^{3}}{3}\right) + C' \)

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

\[\because \quad x' = 0 \quad \text{at} \quad t = 0\]

\[\therefore \quad C' = 0\]

The displacement towards west

\( x' = -(g\omega\cos\lambda)\left(\dfrac{ut^{2}}{g} - \dfrac{t^{3}}{3}\right) \) ......(5)

If a body reaches the ground in \( t \) second, then from equation

\( s = ut - \dfrac{1}{2}gt^{2} \)

\( 0 = ut - \dfrac{1}{2}gt^{2} \)

or

\( t = \dfrac{2u}{g} \)

Substituting the value of \( t \) in equation (5),

\( x' = -\dfrac{4}{3}(g\omega\cos\lambda)\left(\dfrac{u}{g}\right)^{3} \) ......(6)

If a body reaches a height \( h \), then from equation

\( v^{2} = u^{2} + 2as \)

\( 0 = u^{2} - 2gh \)

or

\( u = \sqrt{2gh} \)

Substituting this equation in equation (6), we get the equation for the displacement of the body on the ground toward west as

\( x' = -\dfrac{4}{3}(g\omega\cos\lambda)\left(\dfrac{2h}{g}\right)^{3/2} \) ......(7)