Question

1.By Newton’s universal law of gravitation the free-fall acceleration a of a body, such as the satellite shown in Figure 1.3.20, falling a great distance to the surface is not the constant g. Rather, the acceleration a is inversely proportional to the square of the distance from the center of the Earth, $\mathit{a}\mathbf{=}\mathit{k}\mathbf{/}{\mathit{r}}^{\mathbf{2}}$, where k is the constant of proportionality. Use the fact that at the surface of the Earth $\mathbf{r}=\mathbf{R}$and $\mathbf{a}=\mathbf{g}$to determine k. If the positive direction is upward, use Newton’s second law and his universal law of gravitation to find a differential equation forthe distance r.

Short answer

The DE for the distance r to the center of the Earth$\frac{d^{2}r}{d{t}^2}+\frac{g{R}^2}{r^2}=0$

Step-by-step solution

Definition of Newton’s law of cooling with mathematical statement.

Newton’s law of cooling/warming translates into the mathematical statement

$\frac{dT}{dt}\propto T-T_m\text{or}\frac{dT}{dt}=k\left(T-T_m\right)$

where k is a constant of proportionality. In either case, cooling or warming, if$T_m$is a

constant, it stands to reason that $k<0$.

Compute Acceleration.

Acceleration of a freely falling body is directly proportional to the distance of the body from the center of the Earth,

$a=\frac{k}{r^2}$

where a is acceleration, k is the constant of proportionality and r is distance of the body from the center of the earth.

Then, from the question

$r=R\text{and}a=g$

which implies that

role="math" $g=\frac{k}{R^2}$

Using the Newton's Second Law, we get:

$\begin{array}{rcl}F& =& \frac{d}{dt}(m·v)\\ & =& m·\frac{dv}{dt}+v·\frac{dm}{dt}\\ & =& m·\frac{dv}{dt}\\ \frac{dm}{dt}& =& 0\end{array}$

Compute the Differential equation.

$F=m·a\Rightarrow a=\frac{dv}{dt}\Rightarrow a=\frac{d^{2}r}{d{t}^2}$

Taking upward direction as positive,

$\begin{array}{rcl}-a& =& \frac{d^{2}r}{d{t}^2}\\ -\frac{k}{r^2}& =& \frac{d^{2}r}{d{t}^2}\\ -\frac{g{R}^2}{r^2}& =& \frac{d^{2}r}{d{t}^2}\\ -g& =& \frac{k}{R^2}\to k=-g{R}^2\\ \frac{d^{2}r}{d{t}^2}+\frac{g{R}^2}{r^2}& =& 0\end{array}$

Which is the DE for the distance r to the center of the Earth.