# Orbit change after a astronaut jumps toward to earth from spaceship [closed]

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, $$E_{\text{jump}}=E_0+\frac{m_{as}\dot{r}^2}{2}$$

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?

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

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• 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. – user191954 Aug 18 '18 at 16:58
• 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. – David Hammen Aug 18 '18 at 17:10
• @david-hammen Thanks for mentioned that, I have corrected it. – rig Aug 18 '18 at 18:46
• @chair Regarding to the off-topic here, I'm sorry about that, the mods may like to close this thread if necessary. – rig Aug 18 '18 at 18:46