I want to check if I am reasoning a deduction correctly and am not missing some elementary part of the math.

I have a rocket that I launch at $10,000\; m/s$. Earth's escape velocity is $11,200\; m/s$, so this rocket isn't going to escape, it's coming back or staying in orbit at the very least. Let's say my rocket's mass is 200 kg. I want to see how far away it will get.

By the way, I am assuming that my rocket accelerated effectively instantaneously; maybe I have a space drive or something that pulls an insane amount of $g$s so I reach that $10,000 \;m/s$ within a few seconds relative to my orbital period (this is to simplify my model here).

So, I look for where my gravitational PE equals my $KE = \frac{1}{2}mv^2$ so $KE = \frac{1}{2}(200kg)(10^4 m/s)^2 = 10^{10}J$. My potential is $$U = G\frac{m_1m_2}{r} = (6.674 \times 10^{-11}) \frac{(5.972 \times 10^{24}kg)(200kg)}{(r)},$$ so that setting them equal implies $$10^{10}J = \frac{7.97143 \times 10^{16}}{r} \rightarrow r=7.9714 \times 10^6m.$$

That's as far as my rocket will get. By that time it's velocity has dropped to zero as well, so that gets me the maximum distance from Earth. Earth (or it's center anyway) will be one of the foci of the orbit. And the maximum orbital velocity is $10,000 \;m/s$.

Originally I thought that this gets me the semi-major axis plus the distance from the center to the focus of the ellipse.

I was thinking of the Vis-viva equation and using that, but then I realized that by the time the rocket gets way out there it's velocity relative to the Earth is zero, so I would get $$0 = GM(\frac{2}{r}-\frac{1}{a}) \rightarrow 0 = \frac{2GM}{r}-\frac{GM}{a},$$ $$\frac{GM}{a}=\frac{2GM}{r},$$ $$r=2a.$$ But this seemed wrong since the distance $2a$ would be the distance between the edges of the ellipse, rather than a distance to the focus.

Then I thought 'well, wait a minute, a rocket launching straight up isn't launching straight up, really, it's got a velocity vector tangent to the Earth's rotation, and that's $460\; m/s$'.

So if I plug that into Vis-viva I get $\frac{460^2}{GM} =\frac{2}{r}-\frac{1}{a}$, which gets me an $a = 3.99 \times 10^6 m$ or thereabouts. Now I can use $r_{max}=a(1+e)$ to get a value for eccentricity, as I know what $r_{max}$ is. and I can plug that back in and use $r_{min}=a(1-e)$ to get the perigee distance. And Kepler's Third Law (using the Earth's mass as a proportionality constant, since the mass of the rocket is so tiny) will get me the period.

I am curious if I made any grievous errors here. My sense is that when I initially didn't count the Earth's rotation I ended up with basically 1D motion and the rocket just falls straight back down, which is why I got the result I did. But I also feel that I may have over-complicated things a bit.

And if I approached this the right way, do let me know that too.


1 Answer 1


The major problem I'm seeing is that you are assuming that the gravitational potential energy for the rocket when launching from the surface of the Earth is 0 when performing your energy calculation, to find the maximum distance it achieves. This is not true, you'll need to put the gravitational potential at launch on the left side of your second equation.

A second point is that, if your spacecraft is not going straight up relative to the center of the Earth, the velocity at apogee will not be zero.

How I'd approach this problem (assuming three significant figures for all quantities involved). I'm assuming you're launching straight up from approximately the equator, with a surface- relative velocity of $10\,000 \mathrm{m/s}$ on an airless Earth, using the numbers you've provided:

  • Surface Tangential Velocity: $v_h=480 \mathrm{ m/s}$
  • Surface-Relative Launch Velocity: $v_v=10\,000 \mathrm{ m/s}$
  • Orbital Velocity at the surface of Earth: $v = \sqrt{v_h^2 + v_v^2} = 1.00\times10^4\mathrm{m/s}$
  • Radius of the Earth: $ r=6.37 \times 10^6 \mathrm{m}$
  • Gravitational Parameter of Earth: $\mu= 3.99\times10^{14}\mathrm{m^3/s^2}$

The shape of any Keplerian orbit around a body is characterized by two quantities that remain constant for an object any point on the object. Orbital Energy, and Orbital Angular momentum. Given that the proposed mass of your rocket is negligible compared to the mass of the planet, its contribution to the gravitational parameter is negligible, and we can divide out the mass in the calculations for orbital energy and orbital angular momentum.

Specific orbital energy($\epsilon$) is the sum of the kinetic energy and the gravitational potential energy of the orbiting body, divided by the mass of the orbiting body. $$\begin{aligned}\epsilon &=\epsilon _{k}+\epsilon _{p} \\&={\frac {v^{2}}{2}}-{\frac {\mu }{r}} \\&={\frac {(1.00\times10^4\mathrm{m/s})^{2}}{2}}-{\frac {3.97\times10^{14}\mathrm{m^3/s^2} }{6.37 \times 10^6 \mathrm{m}}} \\&=-1.25\times10^7 \mathrm{J/kg} \end{aligned}$$

The semi-major axis of the orbit ($a$) can be calculated from specific orbital energy. $$\begin{aligned} a&=-\frac{\mu}{2\epsilon} \\&=-\frac{3.99\times10^{14}\mathrm{m^3/s^2}}{2(-1.25\times10^7 \mathrm{J/kg})} \\&=1.59\times10^7\mathrm{m} \end{aligned}$$

Specific angular momentum (${\vec {h}}$) is a vectory quantity representing the per-mass-unit angular momentum of the orbiting body. It's the cross product of the radial distance vector and the velocity vector: $${\vec {h}}={\vec {r}}\times {\vec {v}}$$

But what we want here is the magnitude of the specific angular momentum, $h$ $${\displaystyle h=\left\|{\vec {h}}\right\|} =rv\sin\theta$$

Where $\theta$ is the angle between the radial distance vector and the velocity vector. You can also think of it as the product of two values: First, the radial distance and second, the component of the velocity at right angles to the direction from the center of the body. We already have the second value as a given in this problem: It's the tangential velocity, $v_h$

$$h=rv \sin\theta = r v_h =(6.37 \times 10^6 \mathrm{m})(480\mathrm{m/s}) \\=3.06 \times 10^9\mathrm{m^2/s}$$

And with those two, we can find the orbital eccentricity, $e$:

$$\begin{aligned} e &={\sqrt {1+{\frac {2\epsilon h^{2}}{\mu ^{2}}}}} \\&={\sqrt {1+{\frac {2(-1.23\times10^7 \mathrm{J/kg}) (3.06 \times 10^{9}\mathrm{m^2/s})^{2}} {(3.99\times10^{14}\mathrm{m^3/s^2}) ^{2}}}}} \\&=0.999 \end{aligned}$$

And from there, we can find the apogee distance of the rocket, $Q$ $$\begin{aligned} Q &= a(1+e) \\&= (1.59\times10^7\mathrm{m})(1+0.999) \\&= 3.19\times10^7 \mathrm{m} \end{aligned}$$

Or about $31\,900\,\mathrm{km}$ from the center of the Earth, and about $25\,500\,\mathrm{km}$ from the surface.

For completeness, we can use the vis-viva equation to calculate what the velocity would be at apogee ($v_Q$): $$\begin{aligned} v_Q&=\sqrt{\mu\left(\frac{2}{Q}-\frac{1}{a}\right)} \\&=\sqrt{3.99\times10^{14}\mathrm{m^3/s^2}\left(\frac{2}{3.19\times10^7 \mathrm{m}}-\frac{1}{1.59\times10^7\mathrm{m}}\right)} \\&=96.0 \mathrm{m/s} \end{aligned}$$

  • $\begingroup$ Thanks very much; I was thinking of working out an example for high school students, and we were talking about how to get some maximal values and working out what the limits are. And if I understand you correctly, the way I could have used the KE part of this was not to set it equal to gravitational potential at whatever distance, but equal to the difference so really I should have had $G\frac{m_1m_2}{r_{Earth}} - G\frac{m_1m_2}{r_{max}}$ -- is that correct? $\endgroup$
    – Jesse
    Apr 17, 2021 at 19:47
  • $\begingroup$ @Jesse If the rocket is going directly up and down relative to Earth's center, yes. Otherwise you'd have $\frac{m_2v^2}{2}-\frac{Gm_1m_2}{r_{Earth}}=\frac{m_2(v_Q)^2}{2}-\frac{Gm_1m_2}{r_{max}}$ and have both the apogee velocity $v_Q$ and the apoapsis distance $r_{max}$ as unknowns in the energy balance equation. $\endgroup$
    – notovny
    Apr 17, 2021 at 20:52
  • $\begingroup$ thanks again, BTW I saw the "homework like question" flag and wanted to say this isn't a homework problem -- it's a problem I would be assigning to students, if anything. But I didn't know how to make that clear; the underlying concept I was trying to get at is inherent in the problem. $\endgroup$
    – Jesse
    Apr 19, 2021 at 23:59

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