I am a high school student in AP Physics C. We are currently on our gravitation unit, and one of my homework questions goes like this:

Show that the escape speed, $v_e$, from a planet is related to the speed of a circular orbit just about the surface of the planet, $v_c$, according to the following law: $v_e = \sqrt 2 v_c$.

I know that when moving from an orbital state to an unbounded state, no external work should be done on the object-planet system, so I should be able to use conservation of energy:

$$\begin{align} E_1 &= E_2 \\ K_c + U_{Gc} &= K_e \end{align}$$

I decided to let the radius of the bounded orbit equal $r_c$, the mass of the object equal $m$, and the mass of the planet equal $M$.

$$\frac{1}{2}mv_c^2 - \frac{GmM}{r_c} = \frac{1}{2}mv_e^2$$

As expected, the mass $m$ of the object itself is irrelevant:

$$\frac{1}{2}v_c^2 - \frac{GM}{r_c} = \frac{1}{2}v_e^2$$

This is where I immediately get stuck, so to get $M$ and $r_c$ out of the equation, I add in a second equation that interprets the force of gravity while the object is in a bound circular orbit as a centripetal force:

$$\begin{align} F_c = ma_c &= F_G \\ m\frac{v_c^2}{r_c} &= \frac{GmM}{r_c^2} \\ v_c^2 &= \frac{GM}{r_c} \end{align}$$

How wonderful! I think. I should be able to substitute this back in and solve for $v_e = \sqrt 2 v_c$. . . .

$$\begin{align} \frac{1}{2}v_c^2 - v_c^2 &= \frac{1}{2}v_e^2 \\ -\frac{1}{2}v_c^2 &= \frac{1}{2}v_e^2 \\ -v_c^2 &= v_e^2 \\ \sqrt{-v_c^2} &= v_e \\ \sqrt{-1}v_c &= v_e \end{align}$$

Well, isn't that lovely? The classic non-real answer. My first suspicion is that I made an arithmetical error, but I can't find one. Now I'm thinking that my most probable error would involve the signs of the forces since I'm not using unit vectors to keep track of directions. (My formula chart reads “$a_c = v^2/r = \omega^2r$” and “$\left\vert \vec{F}_G \right\vert = Gm_1m_2/r^2$.”)

Does anyone have any ideas of what error(s) I made or what step(s) I failed to produce? If you could, please provide a few lines of equation showing your thought process and the substitutions you make.

  • 2
    $\begingroup$ For the record: people asking us not to flag their question as off topic is usually a big red flag that the question is, in fact, off topic. Fortunately, this seems to be one of the exceptions. (But it's something to keep in mind for the future that asking that your question not be put on hold is more harm than help.) $\endgroup$
    – David Z
    Jan 23, 2017 at 22:43
  • $\begingroup$ I don't really get what you are doing in the first equations. The escape velocity is the velocity you need to throw the particle (from the surface) such it reaches infinity with zero velocity. The energy conservation in this case reads $mv_e^2/2-GmM/r_c^2=0$. $\endgroup$
    – Diracology
    Jan 23, 2017 at 22:57

2 Answers 2


The velocity to escape can be found by the equation $E= K+ U = 0$ so $K=-U$ at the end you obtain this $$ v_e =\sqrt{\frac{2GM}{R}} $$ To find orbital velocity you use this $$\frac{GmM}{R}=\frac{mv^2}{R}$$ so $$v_c=\sqrt{\frac{GM}{R}}$$ so it's easy to notice the relation between the two

  • $\begingroup$ I've been trying this for three days and your explanation made it work! $\endgroup$ Jan 24, 2017 at 21:27

There are two issues in your calculation. First the energy state for the escaping case. You are using the kinetic energy at the beginning of the escape path and the potential energy at the final state. You need both to be in the final state, And in the final state, the object will have 0 velocity. So there is only potential energy. The other issue is the sign of the potential energy. The potential energy grows as the distance grows. So at the earth's surface there is zero potential energy and as the object raises, that potential energy increases. These 2 changes will give you the correct result.

  • $\begingroup$ Okay, thank you so much! I will try that out right now and see if I can work the problem out. I appreciate the help. $\endgroup$ Jan 23, 2017 at 23:00
  • $\begingroup$ I trued, but I an still having some difficulty. Here's what I have: $\endgroup$ Jan 23, 2017 at 23:44
  • $\begingroup$ $$F_G=F_c$$$$\frac{GMm}{R^2}=\frac{mv_c^2}{R}$$$$\frac{GM}{R}=v_c^2$$ Where $M$ is the mass of the planet and $R$ is the radius of the planet $\endgroup$ Jan 23, 2017 at 23:46
  • $\begingroup$ $$E_{surface}=E_{unbound}$$$$K+U_G=0$$$$\frac{1}{2}mv_c^2+\frac{1}{2}mv_e^2-\frac{GMm}{R}=0$$$$\frac{1}{2}\left( v_c^2+v_e^2\right)=v_c^2$$$$v_c^2+v_e^2=2v_c^2$$$$v_e^2=v_c^2$$ $\endgroup$ Jan 23, 2017 at 23:49
  • $\begingroup$ Any ideas of where to go from here? $\endgroup$ Jan 23, 2017 at 23:50

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