A long fall

By conservation of energy, when the Earth is stopped, $\frac{m}{2} \left( \frac{dr}{dt} \right)^2 - \frac{GMm}{r} = \frac{GMm}{R},$ where symbols have standard meaning. For the Earth moving towards the Sun, $\frac{dr}{dt} = - \sqrt{\frac{2GM}{R}} \sqrt{\frac{R}{r}-1}$ Then the time of fall is $T = \int_0^T dt = \int_R^0 \frac{dr}{dr/dt} = \sqrt{\frac{R}{2GM}} \int_0^R \sqrt{\frac{R}{R-r}} dr = \frac{\pi}{2} R \sqrt{\frac{R}{2GM}}.$

Because the mass of the Sun is very big, in compare with the mass of the Earth, we can assume that only the Earth moves and the Sun stays.

If the Earth falls straight towards the Sun, we can pretend that the orbit of the Earth is an ellipse with minor axis by 0 and major axis 2a=r, so $a=\frac{r}{2}$ .

According to Kepler's third law, we have: $\frac{T^2}{a^3}=\frac{4 \pi^2}{GM}$ .

Therefore, time for the Earth to reach the Sun is: $t=\frac{T}{2}=\pi \sqrt{\frac{r^3}{8GM}}=5.56.10^6(s)$ .

Consider the fall to the sun as a degenerate elliptical orbit. Its semimajor axis is 0.5 AU. From Kepler's third law, we immediately have $(2t)^2 = (0.5)^3$ , where t is measured in years (and is multiplied by 2 because t is only half of the entire period of the orbit). Therefore, $t=0.177$ years, or $5.56*10^6$ seconds.

Given is a bunch of data about the masses of the sun and earth, the radius of the earth's orbit, and Newton's universal constant of gravitation. But, you do not need any of that stuff—all you need to know is Kepler's first and third laws and the fact that the period of earth's (approximately circular) orbit is one year.

Kepler's first law states that the orbit of a planet is an ellipse with a semi major axis a and with the sun at one focus of the ellipse. A circular orbit has a semi major axis equal to the radius of the circle, so $a_1$ = $RE$ where RE is the radius of the earth's orbit; the eccentricity of a circle is $0$ . The other extreme is an ellipse with eccentricity $1$ which is a straight line from the sun to the earth and so the semi major axis for a "dropped earth" is $a_2$ = $RE/2$ . Of course, this is not an orbit you will actually see in nature because there is no such thing as a point mass and the speed of the earth when the two point masses coincide would be infinite, but if we can cleverly deduce the period of this orbit, one-half that period will be the answer to this question.

Kepler's third law states that the square of the period T of an orbit is proportional to the cube of its semi major axis, so

$T_1^{2}$ $/$ $T_2^{2}$ = $a_1^{3}$ $/$ $a_2^{3}$ = $RE^{3}$ $/$ $(RE/2)^{3}$ = $8$ .

So, $T_2$ = $T_1$ $/$ $\surd8$ = $0.354$ years. So, the time to go half a period is $0.177$ years= $64.6$ days, which when converted to seconds give $5.56 \times 10^{6}$

If the distance between Sun (mass $M$ ) and Earth (mass $m$ ) is $r$ , then the acceleration of the Sun is $\frac{-mG}{r^2}$ , and the Earth's is $\frac{-MG}{r^2}$ , so the total acceleration is (\frac{-(m+M)G}{r^2}).

$\displaystyle \ddot{r}=\frac{-(m+M)G}{r^2}=\frac{- \gamma}{r^2}$

$\displaystyle \int_{0}^{t} \dot{r} \ddot{r} dt= \int_{0}^{t} \frac{- \gamma}{r^2} \dot{r} dt$

$\displaystyle \int_{0}^{t} \dot{r} d \dot{r}= \int_{0}^{t} \frac{- \gamma}{r^2} dr$

$\displaystyle \frac{\dot{r}^2}{2}-\frac{\dot{r}(0)^2}{2}=\frac{\dot{r}^2}{2}= \frac{ \gamma (r_0-r)}{r_0r}$ Taking the negative square root as the velocity is negative: $\displaystyle \dot{r}= -\sqrt{\frac{ 2 \gamma }{r_0}} \sqrt{\frac{ r_0-r}{r}}$

$\displaystyle -\sqrt{\frac{ 2 \gamma }{r_0}} \int_{0}^{t} dt= -\sqrt{\frac{ 2 \gamma }{r_0}} \Delta t= \int_{0}^{t} \sqrt{\frac{r}{r_0-r}} dr$

Let $r=r_0 \cos(u)$ :

$\displaystyle -\sqrt{\frac{ 2 \gamma }{r_0}} \Delta t= \int_{0}^{t} \frac{\cos(u)}{\sin(u)} -2r_0\cos(u)\sin(u)du$

$\displaystyle \sqrt{\frac{ 2 \gamma }{r_0^3}} \Delta t= 2 \int_{0}^{t} \cos^2(u)du$

$\displaystyle \sqrt{\frac{ 2 \gamma }{r_0^3}} \Delta t= \int_{0}^{t} 1+\cos(2u)du$

$\displaystyle \sqrt{\frac{ 2 \gamma }{r_0^3}} \Delta t= [u+\frac{1}{2} \sin(2u) ]_{0}^{t}$

$\displaystyle \sqrt{\frac{ 2 \gamma }{r_0^3}} \Delta t= u(t)-u(0)+[\sin(u)\cos(u) ]_{0}^{t}$

Noting that $\sin(u)=\sqrt{\frac{r_0-r}{r}}$ ,

$\displaystyle \sqrt{\frac{ 2 \gamma }{r_0^3}} \Delta t= \arccos(\sqrt{\frac{r}{r_0}})-\arccos(1)+[\frac{\sqrt{r(r_0-r)}}{r_0} ]_{0}^{t}$

$\displaystyle \Delta t= \sqrt{\frac{ 2 r_0^3 }{(m+M)G}} [\arccos(\sqrt{\frac{r}{r_0}})+[\frac{\sqrt{r(r_0-r)}}{r_0} ]$

Letting $r=0$ ,

$\displaystyle \Delta t= \sqrt{\frac{ 2 r_0^3 }{(m+M)G}} \frac{\pi}{2}$

By using Kepler's Third Law, we may obtain equation as below:

$\frac{T^2}{a^3} = \frac{4π^2}{GM}$

If the Earth's orbit is an ellipse, hence the time required is half a period then the length of its major axis is $a=\frac{R}{2}$

$t = \frac{T}{2}$

$t = \frac{π}{2} \sqrt{\frac{R^3}{2GM}}$

$t = 5.56 \times 10^6$

We can assume the path of Earth falling to the Sun equivalent to an ellipse with semiminor axis nearly zero and semimajor axis nearly: a= \frac{R}{2} where R is the distance from Sun to Earth. Use the third formular of Doppler: \frac{a^3}{T^2} = \frac{GM}{a\pi^2} where M is mass of the Sun Then we can calculate that T= 5.56E+6 seconds

None of the solutions above have taken into account the Sun's kinetic energy. Let's try doing so. If $E$ is the total mechanic energy, then

$E = - \frac{GmM}{d} = \frac{mv^2}{2} + \frac{MV^2}{2} - \frac{GmM}{r},$

where $v$ and $V$ are the Earth's and the Sun's velocities when they are separated by a distance $r$ and $d$ is their distance at time $t = 0$ . Momentum must be constant therefore we may write

$v^2 = \frac{2GM}{1 + \frac{m}{M}} \left( \frac{1}{r} - \frac{1}{d} \right).$

But $v \neq r'(t)$ since $v$ is Earth's velocity and $r'$ would be the RELATIVE velocity between the Earth and the Sun. Knowing that it is easy to find the right relation:

$\frac{\text{d}r}{\text{d}t} = \sqrt{ 2GM \left( 1 + \frac{m}{M} \right) \left( \frac{1}{r} - \frac{1}{d} \right) }.$

Under integration from $r = 0$ to $r = d$ and finding $t$ :

$t = \frac{ \pi d^{3/2} }{2 \sqrt{2GM \left( 1 + \frac{m}{M} \right)}}.$

sir i have used calculatin and newtons constant formula by doing some maths in mass of sun and earth and using clue and reading the related books and some help of my maths teacher.