Gravitational Force and initial velocity

[FLms]

Homework Statement

What is the motion of a body thrown upwards from the Earth's surface, with escape velocity as it's initial velocity. Disregard the air resistance.

Homework Equations

$$v_e = \sqrt{\frac{2 G M}{x}}$$
$$F_g = \frac{G M m}{x^2}$$

The Attempt at a Solution

I though this was a simple problem, by just applying Newton's 2nd Law of Motion.

$$m \frac{dv}{dt} = \frac{-G M m}{x^2}$$

However, as the force F depends on the position, x(t) can be determined by solving the integral:

$$\int_{x_0}^{x} \frac{dx}{\pm \sqrt{(E + \frac{G M m}{x})}} = \sqrt{\frac{2}{m}} t$$

I'm really lost here. How do I solve this?
And, by the way, shouldn't the Energy (E), in this case, be zero?

Any help appreciated.


PS: The correct answer is:

$$x (t) = (x_0^\frac{3}{2} + \frac{3}{2} \sqrt{2 G M} t)^\frac{2}{3}$$

 

[FLms]

That integral came up wrong.
$x_0$ and $x$ should be the limits of integration.

 

[olivermsun]

You seem to be most of the way there, except that I don't understand exactly how you set up your integral.

One way I can think of to do it is, starting from your Newton's 2nd law:
$$m \frac{dv}{dt} = \frac{-G M m}{x^2}$$

Then using the fact dv/dt = dv/dx dx/dt = v dv/dx, which you probably also did, separate variables and integrate both sides:
$$\int v dv = \int -\frac{GM}{x^2} dx$$

to get
$$\frac{1}{2} v^2 = \frac{GM}{x}$$
This expression just expresses conservation of energy (-d(KE) = d(PE)) and satisfies the "escape velocity" condition of KE + PE = 0 at x0 (with no extra constant of integration).

Isolating and substituting v = dx/dt:
$$v = \sqrt{2GM} x^{-1/2} = dx/dt$$

Now separate variables and integrate again, and determine your integration constant at x0.

 

[FLms]

I can't believe I missed a simple separated variables equation.
Thanks for your help.

You seem to be most of the way there, except that I don't understand exactly how you set up your integral.

The integral comes from the conservation of energy principle.

$$T + V(x) = E$$
$$\frac{1}{2}m v^2 + V(x) = E$$
$$v = \frac{dx}{dt} = \sqrt{\frac{2}{m}} [E - V(x)]^\frac{1}{2}$$

Then, you can find $x(t)$ solving:

$$\sqrt{\frac{m}{2}} \int_{x_0}^{x} \frac{dx}{\sqrt{E - V(x)}} = t - t_0$$

In this case, the potential energy $V(x) = \frac{G M m}{x}$

 

[olivermsun]

Ah, right, it's the same integral, but you have chosen to retain little m and moved everything to the dx side. Cool. :)