Two Interesting Questions related to Free Fall

This post involves two questions related to free falling. The first is about the terminal velocity of human and a plane under turbulent flow, and the second is about what happens if a human drops to the center of the earth. The results are highly fasicinating!

Free Fall of objects

Inspired by a conversation in Shenzhen, I am trying to compare the terminal velocity of a human and a plane under turbulent flow:

Consider \(a=\dot{v(t)}\), we have \(F=m\dot{v}\). Now consider downwards as positive. Assume air resistance is \(kv^2\), where \(k \in \mathbb{R}\). Then we have:

\[\begin{aligned} m\dot{v}&=mg-kv^2 \\ m \frac{dv}{dt}&=mg-kv^2 \\ mdv&=(mg-kv^2)dt \\ \frac{mdv}{mg-kv^2}&=dt \\ \int_{0}^{v} \frac{mdv}{mg-kv^2} &= \int_{0}^{t}dt \\ t&=\int_{0}^{v} \frac{mdv}{mg-kv^2} \end{aligned}\]

Where we assumed \(v(0)=0\)

\[\begin{equation} \int \frac{d v}{m g-k v^2} =\frac{1}{m g} \int\frac{d v}{1-\frac{k}{m g} v^2} \\ \end{equation}\]

Evaluate the indefinite integral first, if we assume \(u=\sqrt\frac{k}{mg} v\), hence \(\sqrt\frac{mg}{k} du=dv\), we have:

\[\begin{aligned} &\frac{1}{\sqrt{m g k}} \int \frac{d u}{1-u^2} \\ & =\frac{1}{\sqrt{m g k}} \tanh ^{-1}(u) \\ &=\frac{1}{\sqrt{m g k}} \tanh ^{-1}\left(v \sqrt{\frac{k}{m g}}\right) \end{aligned}\]

The reader can easily check that the definite integral is the same, i.e.

\[t=\frac{1}{\sqrt{m g k}} \tanh ^{-1}\left(v \sqrt{\frac{k}{m g}}\right)\]

After this we can do some algebraic manipulations to get : \begin{equation} v=\sqrt {\frac{mg}{k}}\tanh[(\sqrt{mgk})(t)] \end{equation}

As \(\tanh(x)\) approaches \(1\), \(v_\text{Terminal}=\sqrt{\frac{mg}{k}}\)

Next, we have to find the value of \(k\): Using drag equation, assuming turbulent flow:

\(F_{\mathrm{d}}=\frac{1}{2} \rho c_{\mathrm{d}} A v^2\) Hence:

\[\begin{aligned} v_\text{Terminal}&=\sqrt{\frac{mg}{k}} \\ &=\sqrt{\frac{2mg}{\rho c_d A}} \end{aligned}\]

As \(m_\text{human}\)«<\(m_\text{plane}\), the difference in \(c_d\) and \(A\) can’t compensate, and hence terminal velocity of planes are much higher during free fall.

Exercises for the reader

• Why is the ODE separable (University Year 1)
• Develop the theory if the air resistance is linearly proportional to velocity. (Difficulty: University Year 1-2)
• What are the terminal velocies for the plane and the human if there is no air resistance? Why? (Difficulty: HKDSE F4)

Dropping to center of earth

How long does it take to drop to the center of earth and what will happen after that?

Force acting on the ball:

\[F=G\frac{M'm}{r^2}\]

where:

\[M'=M\frac{r^3}{R^3}\] \[\begin{aligned} m\ddot{r}&=G\frac{Mm}{R^3}r \\ \ddot{r}&=G\frac{M}{R^3}r \end{aligned}\]

Any physics student should recognise this as a simple harmonic motion with frequency:

\[\omega=\sqrt{\frac{GM}{R^3}}\]

This corresponds to a period of:

\[T=2 \pi \sqrt{\frac{R^3}{GM}}\]

Dubious Assumption?

The astute reader will notice that this assumes the gravitational force is only contributed from the enclosed mass. Why is this true?

Gauss’ Law for Gravitation

Gauss’ Law for electromagnetism (Maxwell’s First Law) is very well known, but his law for gravitation is equally important.

Gauss’ Law for Gravitation states that:

\[\nabla \cdot \boldsymbol{g} = -4\pi G \rho\]

This can be shown using the steps below. The details are left as an exercise for the readers (see below)

\[\begin{aligned} \nabla \cdot\mathbf{g}(\mathbf{r})&=-G \int \rho(\mathbf{s}) \left [\nabla \cdot\frac{(\mathbf{r}-\mathbf{s})}{|\mathbf{r}-\mathbf{s}|^3} \right] d^3 \mathbf{s} \\ &=-4 \pi G \int \rho(\mathbf{s}) \delta(\mathbf{r}-\mathbf{s}) d^3 \mathbf{s} \end{aligned}\]

As gravitational force on a particle is proportional to the gravitational field, and gravitaitonal field is only proportional to the mass enclosed, we can concude that gravitational force is only contributed from the enclosed mass.

Exercises for the reader

\(\nabla \cdot \mathbf{A}=\frac{1}{r^2} \frac{\partial\left(r^2 A_r\right)}{\partial r}+\frac{1}{r \sin \theta} \frac{\partial}{\partial \theta}\left(A_\theta \sin \theta\right)+\frac{1}{r \sin \theta} \frac{\partial A_{\varphi}}{\partial \varphi}\) can be assumed while attempting the exercises.

• Show that \(\nabla \cdot (\frac{\mathbf{r}}{\lVert\mathbf{r}\rVert})^3=\left(\frac{1}{r^2}\right)(0)=0\) for \(r \neq 0\). (Difficulty: University Year 1-2)
• Use divergence theorem to show what happens at \(r=0\). (Difficulty: University Year 1-2)
• Show: \(\oiint_{\partial V} \mathbf{g} \cdot d \mathbf{A}=-4 \pi G M\) (Difficulty: University Year 1-2)