Falling body displacement¶
Suppose a reference frame \(S'\) is fixed to a moving object \(A\) (e.g., Earth) and some body \(B\) moving freely (i.e. the sum of external non-gravitational forces acting on it is zero). In the case of an inertial frame of reference, the displacement of body \(B\) from the starting position would follow the usual rule \(\vec s = {\vec v}_0 t + \frac{1}{2} {\vec g} t^2\). But in the case of non-inertial frames, we have to take the Coriolis and the centrifugal force into account as well, which results into the series shown below.
Notes:
Note that the series is truncated at the fifth term. More terms can be obtained by plugging the result into the equation of motion \(\vec a = \vec g + \left[ \vec v, \vec \omega \right]\) and integrating it over time.
Conditions:
The sum \(\vec F\) of all other, non-gravitational forces acting on body \(B\) is \(0\).
\(\vec g\) is independent of coordinates.
- Symbol:
s- Latex:
\({\vec s}\)
- Dimension:
length
- Symbol:
t- Latex:
\(t\)
- Dimension:
time
- Symbol:
v_0- Latex:
\({\vec v}_{0}\)
- Dimension:
velocity
- angular_velocity¶
Pseudovector of the angular velocity of body \(B\). See
angular_speed.
- Symbol:
w- Latex:
\({\vec \omega}\)
- Dimension:
angle/time
- acceleration_due_to_gravity¶
Vector of the acceleration due to gravity of body \(B\).
- Symbol:
g- Latex:
\({\vec g}\)
- Dimension:
acceleration
- law¶
s = v_0 * t + t^2 * (g / 2 + cross(v_0, w)) + t^3 / 3 * (cross(g, w) + 2 * cross(cross(v_0, w), w)) + t^4 / 6 * cross(cross(g, w), w)- Latex:
- \[{\vec s} = {\vec v}_{0} t + t^{2} \left(\frac{{\vec g}}{2} + \left[ {\vec v}_{0}, {\vec \omega} \right]\right) + \frac{t^{3}}{3} \left(\left[ {\vec g}, {\vec \omega} \right] + 2 \left[ \left[ {\vec v}_{0}, {\vec \omega} \right], {\vec \omega} \right]\right) + \frac{t^{4}}{6} \left[ \left[ {\vec g}, {\vec \omega} \right], {\vec \omega} \right]\]