Armageddon. Earth revolves around the sun in an approximately circular orbit with radius r = a, completing a revolution in the time $\mathrm{T}=2\mathrm{\pi }{\left(\frac{{\mathrm{a}}^3}{\mathrm{GM}}\right)}^{\frac{1}{2}}$, which is one Earth year; here M is the mass of the sun and G is the universal gravitational constant. The gravitational force of the sun on Earth is given by $\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$, where m is the mass of Earth. Therefore, if Earth “stood still,” losing its orbital velocity, it would fall on a straight line into the sun in accordance with Newton’s second law: $\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{r}}{{\mathrm{dt}}^2}=-\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$.

If this calamity occurred, what fraction of the normal year T would it take for Earth to splash into the sun (i.e., achieve r = 0)? [Hint: Use the energy integral lemma and the initial conditions$\mathrm{r}\left(0\right)=\mathrm{a},\mathrm{r}\text{'}\left(0\right)=0$ .]

The Energy Integral Lemma:

Let y(t) be a solution to the differential equation $\mathrm{y}=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y’ or the independent variable t. Let F(y) be an indefinite integral of$\mathrm{f}\left(\mathrm{y}\right)$ , that is,$\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$ . Then the quantity $\mathrm{E}\left(\mathrm{t}\right):=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$is constant; i.e.$\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$

Given that,Newton’s second law of equation is $\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{r}}{{\mathrm{dt}}^2}=-\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$ … (1)

Then, $\mathrm{r}=-\frac{\mathrm{GM}}{{\mathrm{r}}^2}$. So, $\mathrm{f}\left(\mathrm{r}\right)=-\frac{\mathrm{GM}}{{\mathrm{r}}^2}$

Now find the F(r).

$$\begin{gathered} \mathrm{f}\left(\mathrm{r}\right)=\frac{\mathrm{d}}{\mathrm{dr}}\mathrm{F}\left(\mathrm{r}\right)-\frac{\mathrm{GM}}{{\mathrm{r}}^2} \\ =\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{\mathrm{GM}}{\mathrm{r}}\right) \\ \mathrm{F}\left(\mathrm{r}\right)=\frac{\mathrm{GM}}{\mathrm{r}} \end{gathered}$$

Then, find the quantity.

$$\begin{gathered} \mathrm{E}=\frac{1}{2}{\left(\mathrm{r}\text{'}\left(\mathrm{t}\right)\right)}^2-\mathrm{F}\left(\mathrm{r}\left(\mathrm{t}\right)\right) \\ =\frac{1}{2}{\left(\mathrm{r}\text{'}\left(\mathrm{t}\right)\right)}^2-\frac{\mathrm{GM}}{\mathrm{r}} \end{gathered}$$

$\mathrm{E}=\frac{1}{2}{\left(\mathrm{r}\text{'}\left(\mathrm{t}\right)\right)}^2-\frac{\mathrm{GM}}{\mathrm{r}}$.......(2)

Given that initial conditions are $\mathrm{r}\left(0\right)=\mathrm{a},\mathrm{r}\text{'}\left(0\right)=0$

Then,

$$\begin{gathered} \mathrm{E}=\frac{1}{2}{\left(\mathrm{r}\text{'}\left(0\right)\right)}^2-\frac{\mathrm{GM}}{\mathrm{r}\left(0\right)} \\ =-\frac{\mathrm{GM}}{\mathrm{a}} \end{gathered}$$

Now, substitute the value of E in equation (2).

$$\begin{gathered} \mathrm{E}=\frac{1}{2}{\left(\mathrm{r}\text{'}\left(\mathrm{t}\right)\right)}^2-\frac{\mathrm{GM}}{\mathrm{r}} \\ -\frac{\mathrm{GM}}{\mathrm{a}}=\frac{1}{2}{\left(\mathrm{r}\text{'}\left(\mathrm{t}\right)\right)}^2-\frac{\mathrm{GM}}{\mathrm{r}} \\ \frac{1}{2}{\left(\mathrm{r}\text{'}\left(\mathrm{t}\right)\right)}^2=\frac{\mathrm{GM}}{\mathrm{r}}-\frac{\mathrm{GM}}{\mathrm{a}} \\ =\mathrm{GM}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{a}}\right) \\ {\left(\mathrm{r}\text{'}\left(\mathrm{t}\right)\right)}^2=2\mathrm{GM}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{a}}\right) \\ \mathrm{r}\text{'}\left(\mathrm{t}\right)=\pm \sqrt{2\mathrm{GM}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{a}}\right)} \end{gathered}$$

Since the Earth is losing its orbital velocity. That is, r is decreasing function. So,

$\mathrm{r}\text{'}\left(\mathrm{t}\right)=-\sqrt{2\mathrm{GM}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{a}}\right)}$

Then, use the above condition to find the value of t.

$$\begin{gathered} \frac{\mathrm{dr}}{\mathrm{dt}}=-\sqrt{2\mathrm{GM}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{a}}\right)} \\ -\mathrm{dt}=\frac{\mathrm{dr}}{\sqrt{2\mathrm{GM}\left(\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{a}}\right)}} \\ =\frac{1}{\sqrt{2\mathrm{GM}}}\times \frac{\mathrm{dr}}{\sqrt{\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{a}}}} \\ \sqrt{2\mathrm{GM}}\mathrm{t}=-\mathrm{a}\sqrt{\frac{1}{\mathrm{a}}-{\mathrm{a}}^{-\mathrm{r}}}-{\mathrm{a}}^{\frac{3}{2}}\arctan \left(\sqrt{\mathrm{a}}\sqrt{\frac{1}{\mathrm{r}}-\frac{1}{\mathrm{a}}}\right) \end{gathered}$$

Put r = 0,

$$\begin{gathered} \sqrt{2\mathrm{GM}}\mathrm{t}=\frac{{\mathrm{a}}^{\frac{3}{2}}\mathrm{\pi }}{2} \\ \mathrm{t}=\frac{{\mathrm{a}}^{\frac{3}{2}}\mathrm{\pi }}{2\sqrt{2\mathrm{GM}}} \end{gathered}$$

Given that,

$\mathrm{T}=2\mathrm{\pi }{\left(\frac{{\mathrm{a}}^3}{\mathrm{GM}}\right)}^{\frac{1}{2}}$

Now find the ratio.

$$\begin{gathered} \frac{\mathrm{t}}{\mathrm{T}}=\frac{\frac{{\mathrm{a}}^{\frac{3}{2}}\mathrm{\pi }}{2\sqrt{2\mathrm{GM}}}}{2\mathrm{\pi }{\left(\frac{{\mathrm{a}}^3}{\mathrm{GM}}\right)}^{\frac{1}{2}}} \\ =\frac{1}{4\sqrt{2}} \\ \mathrm{t}=\frac{\mathrm{T}}{4\sqrt{2}} \end{gathered}$$

Hence proved.