# Calculate escape velocity without using energy? [closed]

Is it practically doable to calculate the escape velocity on the earth not by using the principle of energy conservation, but by calculating the velocity of the particle as a function of time using Newton's equation: $$\ddot x_t=-\frac {Gm_e} {x_t^2}$$

Where $$x_t$$ is the distance from the centre of the earth.

Is it doable to solve this equation for $$x_t$$ and $$\dot x_t$$ and then find $$x_0$$ by assuming that $$\lim\limits_{t\to \infty}v_t=0$$?

## closed as off-topic by Aaron Stevens, Jon Custer, ZeroTheHero, Gert, glSAug 20 at 19:46

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• There is an error in the equation. – my2cts Aug 17 at 11:30
• It would probably be easier to just find the critical point at which the solutions are no longer periodic. – Aaron Stevens Aug 17 at 11:41

It is possible, and is easier than you might think with the right trickery.

I'm going to choose my unit system so that $$G m_e = 1$$ to make the algebra easier, if you want to reproduce the standard formula you'll have to follow the equations with the constants in.

To start with, we're going to define a velocity to turn this from a 1D second order equation into a 2D first order equation:

$$v = \dot x$$.

We then have

$$\dot x = v$$, $$\dot v = - \frac{1}{x^2}$$.

We can now think of this as a problem in phase space, where the solutions we're interested in will start at high $$v$$ and low $$x$$ and move towards the $$v$$ axis. The crucial behaviour is that some of these will then cross the axis at return to $$x=0$$, while others will asymptote to infinite $$x$$ and constant $$v$$.

To find out where the divide is, let's look at $$\frac{dx}{dv}$$ along the solutions:

$$\frac{dx}{dv} = \frac{dx}{dt} \frac{dt}{dv} = \frac{\dot x}{\dot v} = -v x^2$$.

This equation is solvable, and little bit of algebra gives us

$$x = \frac{2}{v^2 + c}$$,

where $$c$$ is a constant of integration given by $$c = \frac{2}{x_0} - v_0^2$$, with $$x_0$$ and $$v_0$$ the starting values.

We can now see that for $$c \leq 0$$, the solutions will tend to $$x = \infty$$ and $$v^2 = - c$$, while for $$c > 0$$, the solutions will cross the $$v$$ axis then decay to $$x=0$$. So, our limiting criterion for escape velocity is given by

$$0 \geq c = \frac{2}{x_0} - v_0^2$$,

$$v_0^2 \geq \frac{2}{x_0}$$,

$$v_0 \geq \sqrt{\frac{2}{x_0}}$$,

which is the result we wanted. I hope that's helpful, and that you'll be able to do that with constants included if you need to.

Sort of. The standard method to solve that equation (any equation that has $$\ddot x = f(x),$$ really) is to multiply both sides by $$\dot x_t$$ so that you can integrate both sides separately... But then the right hand side is $$\frac12 \dot x^2 - \frac12 x_0^2$$ and you are looking at an energy equation again! Similarly James Cochran in his answer comes out to $$x=2/(v^2+c)$$ which is a thinly veiled energy balance as well; he does not try to recover the full trajectory and then inspect it.

Aaron Stevens gives a more interesting suggestion in comments which might better serve your needs. I don’t see a great way to do a Fourier analysis on $$\ddot x = -GM/x^2$$ but we can certainly use established methods to get the actual trajectory from conservation of angular momentum, and then classify it.

# How to derive trajectories for central forces

So you are going to have a 2D trajectory, often described in polar coordinates $$r,\theta$$ and we have $$x=r\cos\theta,y=r\sin\theta$$ and as a result the acceleration takes the form $$\vec a = \begin{bmatrix}\ddot x\\\ddot y\end{bmatrix}= \begin{bmatrix}\ddot r \cos\theta - 2\dot r\dot\theta \sin\theta - r \dot \theta^2\cos\theta \\ \ddot r \sin\theta + 2\dot r \dot \theta\cos\theta - r \dot\theta^2\sin\theta \end{bmatrix}$$ which clearly shows a radial component $$\vec a\cdot\hat r=\ddot r-r\dot\theta^2 = -\frac{GM}{r^2}$$and thus using angular momentum $$mr^2\dot\theta=L=m\ell$$ we can rewrite this as $$\ddot r-\frac{\ell^2}{r^3} = -\frac{GM}{r^2}.\tag{A}$$ We can then derive a time-independent trajectory from a force law; let $$r(t)=s(\theta(t))$$ and let primes be partial derivatives with respect to $$\theta$$ , then the full time derivative will look like$$\dot r=s'(\theta(t))~\dot\theta(t)=\frac{s'(\theta(t))~\ell} {[s(\theta(t))]^2},$$ but the last expression can be written in some sense as $$-\ell~(1/s)',$$ suggesting a change of intermediate functions to $$k(\theta)=1/s(\theta)$$ when you have conservation of angular momentum.

From there you can simplify our $$1/r^2$$ force equation (A) above by computing the second full-time derivative analogously, $$\ddot r=-\ell ~k''(\theta(t))~\dot\theta(t)=-k^2~\ell^2~k''$$ And then (A) becomes the time-independent $$k^2 ~k''+ k^3 = k^2/\lambda$$ for $$\lambda=\ell^2/(GM).$$ Dividing by $$k^2$$ gives a harmonic oscillator equation for the $$1/r^2$$ force law, $$k'' + k = 1/\lambda,$$solved by (WLOG choosing our $$+x$$-axis as the closest approach) an expression $$k = \lambda^{-1} + \epsilon \lambda^{-1}\cos\theta,$$so that we have $$r(\theta) = \frac{\lambda}{1 + \epsilon \cos\theta},$$proving as Newton did long ago that Kepler’s orbits of “ellipses with the sun at one focus” are precisely the orbits belonging to this $$1/r^2$$ force law.

# The critical trajectory and escape velocity

So if we were at a point of closest approach we would hold $$r_0 = \lambda/(1 + \epsilon)$$ constant while varying the velocity and hence both $$\ell$$ and $$\epsilon$$. Eventually we would have $$\epsilon = 1$$ and the trajectory would escape to 1/0 = ∞ at θ = 180°. If you go to higher $$\epsilon$$ then you get there at a larger range of angles, and some interesting stuff happens in those angles (you get a negative radius describing the hyperbola that is out past your current trajectory; hyperbolas are disconnected around a symmetry axis).

For the threshold case of $$\epsilon = 1$$ we therefore have the equations $$r_0 = \lambda/2$$ and $$r = 2 r_0 / (1 + \cos\theta) = 2 r_0 / (1 + x/r)$$ Solving for $$r(x)$$ and then using $$r^2 = x^2 + y^2$$ we can derive an equation $$x = r_0 - y^2/(4 r_0)$$ and the conic section in question is a parabola.

But of course there is the point you are waiting for, $$\lambda = 2 r_0 = \frac{\ell^2}{GM} = \frac{r_0^2 v_0^2}{GM},$$ and hence $$v_0 = \sqrt{2GM/r_0}.$$ Of course this is using the fact that $$\ell = (\vec r\times\vec v)\cdot\hat z = r~v~\sin\phi$$ where $$\phi$$ is the angle between $$\vec r$$ and $$\vec v$$; in this case that angle is $$90^\circ$$.

# Extending to other angles

We can then maybe extend that argument to be more general, and to not just consider the special case where ϕ = 90°, though “energy considerations” say that that’s good enough. So if we are at some position $$\theta$$ along the curve then we have $$r=\lambda/(1 + \cos\theta)$$ at that position, but the above does not give $$|\vec v|$$ but merely $$v_\theta$$ and we need to figure out this angle $$\phi$$. For this it might be best to write $$1+\cos\theta$$ as $$2\cos^2(\theta/2)$$ maybe. Then a little change $$\delta\theta$$ induces a change in $$r$$, $$\delta r = \frac{\lambda}{2}\frac{\sin(\theta/2)}{\cos^3(\theta/2)} \delta\theta= r~\delta\theta~\tan(\theta/2),$$ so that we should have that $$\sin\phi = \cos(\theta/2).$$

Once that is determined the argument is the same: if you used a lower velocity than the parabolic orbit which happens to go through that point $$(r, \theta)$$ with that specific tangent vector, you would have a lower eccentricity and your orbit would fall back to being elliptical, so the parabolic velocity is the escape velocity; but the parabolic orbit is given by $$r = \frac{(r~v~\sin\phi)^2}{2GM\cos^2(\theta/2)}$$ and once we substitute in $$\sin \phi$$ our $$\theta$$-term drops out and we are left with $$v = \sqrt{\frac{2GM}{r}}.$$