3
$\begingroup$

I confused with a question from the past-paper of Lagrangian and Hamiltonian mechanics. A Lagrangian (plane polar coordinate) for the spaceship (mass is $m$) under influence of central force directed towards the centre of Earth is $$L=\frac{1}{2} m \left( \dot{r}^2 +r^2 \dot{\phi}^2\right)+\frac{k}{r}$$

Where the $k$ is a constant of the central field, $l$ is the magnitude of angular momentum which is a conserved quantity, and the total energy of this system is given by:

$$ m r^2 \dot\phi =l \quad \quad E=\frac{m\dot{r}^2}{2}+\frac{l^2}{2mr^2}-\frac{k}{r}$$

I could find the effective potential $$V^{\text{eff}}(r)=\frac{l^2}{2mr^2}-\frac{k}{r} $$ The radius $R$ and period $T$ when spaceship moves on a circular orbit with a constant magnitude of angular momentum $l$, cloud be written in terms of $l$, $k$ and $m$ :

$$ R=\frac{l^2}{mk} \quad \quad T=\frac{2 \pi l^3}{ mk^2} $$

When the spaceship moves on a circular orbit with radius $R = 7000$ km and velocity $V = 2\pi R/T = 8 $km/s. The astronaut on the spaceship jumps directly towards the Earth with velocity $v = 8 $m/s. I'm asked to calculate the minimum distance of the astronaut from the centre of the Earth, with a hint that $v \ll V $ so I could expand the effective potential near its minimum.

My basic idea is below:

Before the astronaut jumps, the energy of the spaceship $E_0$ is equal to the effective potential $V^{\text{eff}}(r_0)$: $$E_0=V^{\text{eff}}(r_0)=\frac{l^2}{2mr_0^2}-\frac{k}{r_0}$$

After the astronaut jumps, the energy of is $$E_{\text{jump}}= \frac{m_{as}\dot{r}^2}{2}+ \frac{l^2}{2m_{as}r^2}-\frac{k}{r}$$

Where $m_{as}$ is the mass of astronaut. Because $v \ll V$, the radius change is small, therefore the potential change is small, as, $$E_{\text{jump}}=E_0+\frac{m_{as}\dot{r}^2}{2}$$

Effective potential

As the curve of effective potential shown above, I would like to use a approximation of $E_{\text{jump}}$, $$E_{\text{jump}}=E_0 + \frac12 \frac{\partial^2 V^{\text{eff}}(r_0)}{\partial r^2}(r-R)^2$$

Therefore, $$E_{\text{jump}}- E_0= \frac{m_{as}\dot{r}^2}{2} =\frac12 \frac{\partial^2 V^{\text{eff}}(r_0)}{\partial r^2}(r-R)^2 $$

But I found problem from here, $m_{as}$ is not given, therefore I'm not able to find the numerical solution from the above equation. Is there anything wrong with my idea?

$\endgroup$

closed as off-topic by Jon Custer, Kyle Kanos, Emilio Pisanty, AccidentalFourierTransform, ZeroTheHero Aug 24 '18 at 4:31

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Jon Custer, Kyle Kanos, Emilio Pisanty, AccidentalFourierTransform, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Hi, welcome to Physics SE! Unfortunately, check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions. Can you try making a question about some concepts that you'd need to solve this problem? Also mention any thoughts you currently have about the solution, and your assessment of their usefulness. $\endgroup$ – user191954 Aug 18 '18 at 16:58
  • $\begingroup$ Regarding I'm not sure whether this is correct. It's not. Look at the units. Your $E= \cdots-\frac k r$ means that your $k$ has units of mass * length^3 / time^2. Your angular momentum $l$ has units of mass * length^2 / time, so your $\frac l{mk}$ has units of time/(mass*length). These are not the units of velocity, so it's not correct. $\endgroup$ – David Hammen Aug 18 '18 at 17:10
  • $\begingroup$ @david-hammen Thanks for mentioned that, I have corrected it. $\endgroup$ – rig Aug 18 '18 at 18:46
  • $\begingroup$ @chair Regarding to the off-topic here, I'm sorry about that, the mods may like to close this thread if necessary. $\endgroup$ – rig Aug 18 '18 at 18:46