Chapter 7.3, Problem 54E

i.

To determine

To Prove: The initial velocity, $v_0=\sqrt{\frac{2gRh}{R+h}}$ .

Explanation of Solution

Given:

  $m\frac{dv}{dt}=-\frac{mg{R}^2}{{(x+R)}^2}$ , Where $x$ is the objects distance above the surface, $R$ is earth’s radius, $m$ mass and $g$ is acceleration due to gravity.

Calculation:

Let $h$ be maximum height reached by the object and $v_0$ be the initial velocity.

Since velocity,

  $v=\frac{dx}{dt}$ .

By chain rule we can write,

  $mv\frac{dv}{dx}=m\frac{dx}{dt}\frac{dv}{dx}=m\frac{dv}{dt}$ .

Therefore from the given information,

  $mv\frac{dv}{dx}=-\frac{mg{R}^2}{{(x+R)}^2}$

  $vdv=-\frac{g{R}^2}{{(x+R)}^2}dx$

Integrating both sides,

  ${\displaystyle {\int }_{v_0}^{0}vdv}=-{\displaystyle \underset{0}{\overset{h}{\int }}\frac{g{R}^2}{{(x+R)}^2}dx}$

  ${\left[\frac{v^2}{2}\right]}_{v_0}^0=-{\left[\frac{g{R}^2}{{(x+R)}^2}\right]}_0^h$

  $-\frac{v_0^2}{2}=\frac{g{R}^2}{h+R}-\frac{g{R}^2}{R}$

  $-\frac{v_0^2}{2}=-\frac{g{R}^{2}h}{(h+R)R}$

  $v_0^2=\frac{2gRh}{h+R}$

  $v_0=\sqrt{\frac{2gRh}{h+R}}$

i.e, the initial velocity is,

  $v_0=\sqrt{\frac{2gRh}{h+R}}$

ii.

To determine

The escape velocity $v_e.$

Answer to Problem 54E

The escape velocity is, $v_e=\sqrt{2gR}.$

Explanation of Solution

Given:

  $v_e={\lim }_{h\to \infty }{v}_0$

Calculation:

The initial velocity is given by,

  $v_0=\sqrt{\frac{2gRh}{R+h}}$

As $h\to \infty$ , we can write,

  $v_e={\lim }_{h\to \infty }{v}_0={\lim }_{h\to \infty }\sqrt{\frac{2gRh}{R+h}}$

Using the L-hospital rule for limit, Taking derivative with respect to $h$ on both the numerator and denominator,

  $v_e={\lim }_{h\to \infty }\sqrt{\frac{2gR}{1}}$

Therefore, the escape velocity is,

  $v_e=\sqrt{2gR}.$

iii.

To determine

The escape velocity $v_e.$

Answer to Problem 54E

  $v_e=47.5 mi/\sec$.

  ${\text{v}}_e\text{=250800} \text{ft/sec}$

Explanation of Solution

Given:

  $R=3960 mi$ and $g=32 ft/s^2$ .

Calculation:

The escape velocity is, $v_e=\sqrt{2gR}.$

From the given information,

  $g=32 ft/s^2=0.006 mi/s^2$ .

Therefore,

  $\begin{array}{l}{v}_e=\sqrt{2gR}=\sqrt{2\times 0.006\times 3960}\\ v_e=47.5 mi/\sec \end{array}$

In foot,

  ${\text{v}}_e\text{=250800} \text{ft/sec}$ .