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.
Best Answer
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