Prove $v = \sqrt{\frac{2gRh}{R + h}}$ from Given (below)

Question

Given:

$$F = \frac{mgR^2}{(x + R)^2}$$

$m = \text{mass}$

$g = \text{Acceleration due to gravity}$

$x = x(t)$ is the object's distance above the surface at time $t$.

I believe this is the Universal Law of Gravitation (correct me if I am wrong)

Also by Newton's Second Law, $F = ma = m\left(\dfrac{dv}{dt}\right).$

The question reads:

Suppose a rocket is first vertically upward with an initial velocity $v$. Let $h$ be the maximum height above the surface reached by the object. Show that:

$$v = \sqrt{\frac{2gRh}{R + h}}.$$

At the bottom of the problem it says: Hint: by the chain rule $$m\frac{dv}{dt} = mv\frac{dv}{dx}$$

If it isn't too much trouble, how did the textbook writers get this result using chain rule?

Thank you for your time.

Answer 1

Since $$v=\frac{dx}{dt}$$ we have $$\frac{dv}{dt}=\frac{dx}{dt}\frac{dv}{dx}$$

Answer 2

Since $v=dx/dt$, by the chain rule, we get the following separable differential equation $$mv\frac{dv}{dx}=m\frac{dv}{dx}\frac{dx}{dt}=m\frac{dv}{dt}=ma = F = -\frac{mgR^2}{(x + R)^2}$$ (note that minus sign, the gravitation force is attractive!). Then by integrating we get $$\int_{v_0}^0v{dv}=-\int_0^{h}\frac{gR^2}{(x + R)^2}{dx}$$ that is $$\left[\frac{v^2}{2}\right]_{v_0}^0 =\left[\frac{gR^2}{x + R}\right]_0^{h}\implies -\frac{v_0^2}{2}=\frac{gR^2}{(h + R)}-\frac{gR^2}{R}\implies v_0 = \sqrt{\frac{2gRh}{R + h}}$$ where $v_0$ is the initial velocity and $h$ the maximum height above the surface reached by the object (at that point the velocity of the object is zero).

P.S. Note that for $h\to +\infty$ the above result gives the formula of the escape velocity, i.e. $v_e=\sqrt{2gR}$.