Question: The force of gravitational attraction on a mass mat distance r fromthe center of the earth( r > radius of the earth) is mgR² / r². Then the differential equation of motion of a mass m projected radially outward from the surface of the earth, with initial velocity v₀ , is$\mathit{m}{\mathit{d}}^{\mathbf{2}}\mathit{r}\mathbf{/}\mathit{d}{\mathit{t}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathit{m}\mathit{g}{\mathit{R}}^{\mathbf{2}}\mathbf{/}{\mathit{r}}^{\mathbf{2}}\mathbf{.}$

Use method(c)above to find v as a function of r if v = v₀ initially (that is, when r = R ). Find the maximum value of r for a given v₀, that is, the value of r when v = 0. Find the escape velocity, that is, the smallest value of v₀ for which r can tend to infinity.

The force of gravitational attraction on a mass m at distance rfrom the center of the earth ( r > radius R of the earth) is mg R²/ r² . Then the differential equation of motion of a mass m projected radially outward from the surface of the earth, with initial velocity v₀, is$m{d}^{2}r/d{t}^2=-mg{R}^2/r^2.$

Definition of escape velocity:
A moving body (such as a rocket) must have a minimum velocity in order to leave the gravitational field of a celestial body (such as the earth) and go outward into space.

The given equation is in the form of $m{d}^{2}r/d{t}^2=-mg{R}^2/r^2$and after dividing it by m the equation is in the form of r " +f(r) = 0 with denoting differentiation w.r.t to time t .To see how this can be cast into the more convenient shape, multiply it by r$r^{\text{'}}{r}^{\text{'}\text{'}}+r^{\text{'}}f\left(r\right)=0$.
Now, multiply the equation by and exploit the table integral $\int xdx=x^2+$const. $r^{\text{'}}d{r}^{\text{'}}+f\left(r\right)dr=0\Rightarrow \frac{1}{2}{r}^{\text{'}2}+\int f\left(r\right)dr=\text{const}$
To the given differential equation, it is immediately evident that it is already in aforementioned form with$f\left(r\right)=\frac{g{R}^2}{r^2}$ .
$$\begin{gathered} \int f\left(r\right)dr=\int \frac{g{R}^2}{r^2}dr \\ =-\frac{g{R}^2}{r^2}+\text{const} \end{gathered}$$
Upon using the table integral$\int \frac{dx}{x^2}=-\frac{1}{x}+$ const again. And after denoting the integration constant -A as for convenience. Insert this result back into the equation
$\frac{1}{2}{r}^{\text{'}2}+\frac{g{R}^2}{r}=A$
Now it is possible to obtain the integration constant, since would be given the initial conditions in the form of . Remembering that since the definition of radial velocity is time derivative of radial coordinate i.e. r = R:
$$\begin{gathered} \frac{1}{2}{r}^{\text{'}2}(r=R)-\frac{g{R}^2}{R}=A \\ \frac{1}{2}{v}_0^2-gR=A \\ A=\frac{1}{2}{v}_0^2-gR \end{gathered}$$

Set r (r_max ) = 0 to obtain the maximum radius of given trajectory. Exploiting this consideration, upon inserting $\mathrm{r}\left({\mathrm{r}}_{\max }\right)=0$into equation (2) and recalling the expression for A:
$$\begin{gathered} -\frac{g{R}^2}{r_{\max }}=A \\ =\frac{1}{2}{v}_0^2-gR \\ \Rightarrow r_{\max }=\frac{g{R}^2}{gr-\frac{1}{2}{v}_0^2} \end{gathered}$$
After dividing the denominator and numerator with gR, finally obtain:
$r_{\max }=\frac{R}{1-\frac{v^{2}0}{2gR}}$
It is manifestly nonnegative; its minimum value is zero.
$$\begin{gathered} =\frac{1}{2}{v_0}^2\min -gR \\ {v}_{0\min }=\sqrt{2gR} \end{gathered}$$
Thus, the solution of the maximum and minimum velocity$r_{\max }=\frac{R}{1-\frac{v^{2}0}{2gR}}$and$v_{0\min }=\sqrt{2gR}.$