Question

 Earth rotates around the Sun with time period = T

Due to some cosmic event the earth's rotation and revolution,Sun pulls Earth toward its center 
i)Find the velocity v when it's distance is x from sun's  center 
ii)Find the Time taken to reach the velocity v and The time taken to reach the sun's center.

Answer 1

As the Earth is revolving around the Sun under the influence of a centripetal acceleration "a", with an angular velocity "ω" at a time period T.  Let the radius of Earth's orbit be R, Ms be the mass of Sun and the mass of Earth be Me. Let G be the universal Gravitational constant.Centripetal force is supplied by the gravitational attraction force. Hence,

$F_g=M_ea =\frac{M_e v^2}{R} = M_e R \omega^2 = \frac{G M_s M_e}{R^2}\\ \\\omega^2 =\frac{G M_s}{R^3},\ \ \ \ =>\ \ \ (\frac{2 \pi }{T})^2 = \frac{G M_s}{R^3}\\ \\T^2 = \frac{4 \pi^2 R^3}{ G M_s},\ \ \ \ T=\frac{2\pi}{\sqrt{GM_s}} R^{\frac{3}{2}},\ \ \ \ \ GM_s=\frac{4\pi^2R^3}{T^2}$

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Let us assume that the gravitational force of attraction on the Earth pulls it towards the Sun and the Earth starts from an initial velocity 0 at t=0.  Let us assume that Earth starts from rest. (no rotational or other kinetic energy).

Force on the Earth at a distance R from the Sun = Fg = G Ms Me / R²
Work done by F for moving Earth to a distance x from R = Wg =

$= \int\limits^x_R {F\ .} \, ds = \int\limits^x_R {\frac{GM_sM_e}{s^2}} \, ds\\ \\=GM_sM_e \int\limits^x_R {\frac{1}{s^2}} \, ds= -GM_sM_e [\frac{1}{s}]_R^x\\\\=-GM_sM_e[\frac{1}{x}-\frac{1}{R}]$

Kinetic energy gained by Earth = Change in gravitational PE = Wg

$=\frac{1}{2} M_ev^2=GM_sM_e [ \frac{1}{x} - \frac{1}{R} ]\\ \\v^2=2GM_s [ \frac{1}{x} - \frac{1}{R} ]=2\frac{4\pi^2R^3}{T^2}[ \frac{1}{x} - \frac{1}{R} ]\\ \\.\ \ \ \ \ \ v=\frac{2\sqrt2 \pi R^{\frac{3}{2}}}{T}\sqrt {[ \frac{1}{x} - \frac{1}{R} ]}.\ \ \ \ v=\frac{2\sqrt2 \pi x_0^{\frac{3}{2}}}{T}\sqrt {[ \frac{1}{x} - \frac{1}{x_0} ]}$
 
       Where x₀ = R = distance of Earth from Sun at t = 0.

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To derive expression for t in terms of x and v.

$Let\ x=x_0\ Cos^2 \theta =\frac{x_0}{2}(1+Cos\ 2\theta)\\dx=-\frac{x_0}{2}2Sin\ 2\theta\ d\theta=-x_0Sin\ 2\theta\ d\theta\\v=K\sqrt{\frac{x_0}{x}-1}=K\sqrt{sec^2\theta-1}=K\ tan\theta,\ \ \ where\ K=\frac{2\sqrt2\pi x_0}{T}$

$\frac{dx}{dt}= -x_0Sin2\theta\ \frac{d\theta}{dt} = -v = -K tan\theta\\\\\frac{K}{2x_0}dt=Cos^2\theta\ d\theta=\frac{1}{2}(1+Cos2\theta) d\theta,\ \ \ \ \ \theta=0\ when\ t=0.\\ \\\frac{K}{x_0} \int\limits^t_{0} {} \, dt = \int\limits^\theta_0 {(1+Cos2\theta)} \, d\theta\\ \\\frac{K}{x_0}t=\theta+\frac{1}{2}Sin2\theta\\ \\t=\frac{x_0}{K}\ [\ tan^{-1} \sqrt{\frac{x_0}{x}-1}+\sqrt{\frac{x}{x_0}-\frac{x^2}{x_0^2}}\ \ ]\\ \\When\ x=0,\ \ t=\frac{x_0}{K}[\frac{\pi}{2}+0]=\frac{T}{4\sqrt2}$

So time to reach the Sun is T/4√2
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Time duration for Earth to reach velocity v.

$\sqrt{\frac{x_0}{x}-1}=\frac{v}{K}=\frac{vT}{2\sqrt2\pi x_0}\\ \\t=\frac{T}{2\sqrt2\pi}[ Tan^{-1}\frac{Tv}{2\sqrt2\pi x_0} + \frac{Kv}{K^2+v^2}]\\ \\t=\frac{T}{2\sqrt2\pi}[ Tan^{-1}\frac{Tv}{2\sqrt2\pi x_0} + \frac{2\sqrt2 \pi x_0 Tv}{8\pi^2x_0^2+T^2v^2}]$