Question

Question: Falling Body In Problem 23,suppose r=R+s, where s is the distance from the surface of the Earth to the falling body. What does the differential equation obtained in Problem 23 become when is very small in comparison to R ? [Hint: Think binomial series for $\mathbf{(}\mathbf{R}\mathbf{+}\mathbf{s}{\mathbf{)}}^{\mathbf{-}\mathbf{2}}\mathbf{=}{\mathbf{R}}^{\mathbf{-}\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{s}\mathbf{/}\mathbf{R}{\mathbf{)}}^{\mathbf{-}\mathbf{2}}$].

Short answer

The differential equation becomes when s is very small is $\frac{d^{2}s}{d{t}^2}+g=0$.

Step-by-step solution

Define a derivative of the function.

The derivative of a function of a real variable in mathematics describes the sensitivity of the function value (output value) to changes in its argument (input value).

Calculus uses derivatives as a fundamental tool. When a derivative of a single-variable function exists at a given input value, it is the slope of the tangent line to the function's graph at that point.

Determine the differential equation.

Let the differential equation that was obtained be, $\frac{d^{2}r}{d{t}^2}+\frac{g{R}^2}{r^2}=0$

For $r=R+s$, then $\frac{d^{2}r}{d{t}^2}=\frac{d^{2}s}{d{t}^2}$.

As is a constant, and ${\mathit{r}}^{\mathbf{-}\mathbf{2}}\mathbf{=}\mathbf{(}\mathbf{R}\mathbf{+}\mathbf{s}{\mathbf{)}}^{\mathbf{-}\mathbf{2}}\mathbf{=}{\mathit{R}}^{\mathbf{-}\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{s}\mathbf{/}\mathbf{R}{\mathbf{)}}^{\mathbf{-}\mathbf{2}}$, then the DE becomes,$\frac{d^{2}s}{d{t}^2}+\frac{g}{{(1+s/R)}^2}=0$ .
If $s<<Rs/R\approx 0$, then the DE becomes,$\frac{d^{2}s}{d{t}^2}+g=0$ .