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Non-constant gravity

Gravitational acceleration depends on the distance from the centre of the source of gravity, so there is some variation in gravitational acceleration between the base and summit of hills and mountains although it is minor. In this section, the equation of motion will be derived while accounting for this variability.

Formulating the Lagrangian

The kinetic energy we use when studying non-constant gravity is the same as before. Consequently, Equation (??) is also true for this Lagrangian.

Potential energy

If we cannot assume constant gravity, then: $\begin{aligned} V &=\dfrac{GMm}{(r_{\mathrm{base}} + h(s))} - \dfrac{GMm}{r_{\mathrm{summit}}}. \end{aligned}$ Factorizing: $\begin{aligned} V &= GMm\left((r_{\mathrm{base}} + h(s))^{-1} - r_{\mathrm{summit}}^{-1}\right). \end{aligned}$ Where $G$ is the gravitational constant and $M$ is the mass of the Earth.

Substituting Equation (??) into Equation (2) yields:

Final Lagrangian

The Lagrangian of our system is therefore:

Non-conservative forces

If gravity cannot be assumed to be constant then:

Substituting Equation (2) into Equation (??) yields: $\begin{aligned} F_r &= \dfrac{GMmC_{rr}\cos{\theta(s)}}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} \end{aligned}$

Computing Equation (??)

The conservative force term of the Euler-Lagrange equation is therefore: $\begin{aligned} \dfrac{\partial L}{\partial s} &= \dfrac{GMm\sin{\theta(s)}}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2}. \end{aligned}$ Hence the functional derivative of $S$ with respect to $s$ is therefore:

The time derivative of generalized momentum is the same as above. As for $Q_s$ it is: $\begin{aligned} Q_s &= -\dfrac{GMmC_{rr}\cos{\theta(s)}}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} - \dfrac{\rho C_D A \dot{s}^2}{2}. \end{aligned}$ Substituting Equations (4) and (5) into Equation (??) yields: Rearranging and factorizing: Dividing by $m$: $\begin{aligned} \ddot{s} &= \dfrac{GM}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} \left[\sin{\theta(s)}-C_{rr}\cos{\theta(s)}\right] - \dfrac{\rho C_D A\dot{s}^2}{2m}. \end{aligned}$

Simplification to first order

$\ddot{s} = \dot{s} \dfrac{d\dot{s}}{ds}$ hence Equation (6) becomes:

Letting $u=\dot{s}^2$, then $\dfrac{du}{ds} = 2\dot{s}\dfrac{d\dot{s}}{ds}$ and so: Multiplying both sides by 2 yields: Adding the $u$ term to the left-hand side: $\begin{aligned} \dfrac{du}{ds} + \dfrac{\rho C_D A}{m}u &= \dfrac{2GM}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2} \left[\sin{\theta(s)}-C_{rr}\cos{\theta(s)}\right]. \end{aligned}$

Solution

We first seek an integration factor, $I(s)$, so that we can re-write Equation (7) as:

Hence $I'(s)$ must equal $I(s)\dfrac{\rho C_D A}{m}$, which therefore implies that We can replace the left-hand side with $\dfrac{d}{ds}\left[I(s)u(s)\right]$ hence: Therefore Equation (7) becomes: $\begin{aligned} \dfrac{d}{ds}\left[e^{\frac{\rho C_D As}{m}}u(s)\right] &= 2GMe^{\frac{\rho C_D As}{m}}\dfrac{\sin{\theta(s)}-C_{rr}\cos{\theta(s)}}{\left(r_{\mathrm{summit}} - \int_0^s \sin{\theta(\sigma)}d\sigma\right)^2}. \end{aligned}$ Integrating both sides of Equation (8) from 0 to $s$ with respect to $s$ (calling it a new dummy variable $\sigma$; $\sigma$ in our denominator will be renamed $\xi$) yields: $\begin{aligned} e^{\frac{\rho C_D As}{m}}u(s)-e^0 u(0) &= 2GM\int_0^s e^{\frac{\rho C_D A\sigma}{m}}\dfrac{\sin{\theta(\sigma)}-C_{rr}\cos{\theta(\sigma)}}{\left(r_{\mathrm{summit}} - \int_0^{\sigma} \sin{\theta(\xi)}d\xi\right)^2} d\sigma. \end{aligned}$ The trike's initial velocity is assumed to be zero, hence $u(0)=0$: Multiplying by $e^{-\frac{\rho C_D As}{m}}$ yields: Square rooting $u$ to obtain velocity yields: $\begin{aligned} v(s) &= \sqrt{2GMe^{-\frac{\rho C_D As}{m}}\int_0^s e^{\frac{\rho C_D A\sigma}{m}}\dfrac{\sin{\theta(\sigma)}-C_{rr} \cos{\theta(\sigma)}}{\left(r_{\mathrm{summit}}-\int_0^\sigma \sin{\theta(\xi)} d\xi\right)^2} d \sigma}. \end{aligned}$ Below are the velocities achieved in the above model for various $C_D A$ and $C_{rr}$ values when descending Castle Hill. The simulation did not end at 0 metres above sea level; it ended at about 36 metres above sea level, which is approximately at the high end of Gregory Street.