# Question c668f

Apr 24, 2017

Final calculations are left for the reader.

#### Explanation:

In the instant case I could not find applicability of [Kepler's In the instant case I could not find applicability of Kepler's Laws of motion. As such expressions applicable to satellites have been used and Law of Conservation of energy is used.

Let a body of mass $m$ with initial velocity $= 0$ situated at a distance R=1.496×10^11 m, from the sun. It falls towards the sun, under the influence of gravitational attraction of the sun. Effect of gravitational attraction due to all planets is ignored.

Gravitational PE of this body$= - G \frac{M m}{R}$
where M=1.989 × 10^30 kg is the mass of the sun and ${R}_{s} = 6.957 \times {10}^{8} m$ is radius of sun.
Total initial energy $= K E + P E = - G \frac{M m}{R}$ ......(1)

Let the body be at a distance $x$ from the sun's center and have a velocity $v$
Total energy $= \frac{1}{2} m {v}^{2} + \left(- G \frac{M m}{x}\right)$ .....(2)
Using law of conservation of energy we get
$\frac{1}{2} m {v}^{2} + \left(- G \frac{M m}{x}\right) = - G \frac{M m}{R}$
Rearranging we get
$\frac{1}{2} m {v}^{2} = \left(G \frac{M m}{x}\right) - G \frac{M m}{R}$
$\implies v = \pm \sqrt{2 G \frac{M}{R} \left[\frac{R - x}{x}\right]}$
$\implies v = \pm \frac{1}{k} {\left[\frac{R - x}{x}\right]}^{\frac{1}{2}}$ .....(3)
where 1/k=sqrt(2GM/(R)

If the body takes time $\mathrm{dt}$ to cover an infinitesimal distance $\mathrm{dx}$ we have
$\mathrm{dt} = \frac{\mathrm{dx}}{v}$
Using (3) we get
$\mathrm{dt} = \pm k {\left[\frac{x}{R - x}\right]}^{\frac{1}{2}} \mathrm{dx}$
Time taken to reach the sun's surface is time integral of LHS from $t = 0$ to $t = t$
Which is ${\int}_{0}^{t} \mathrm{dt} = t$
and total distance traveled by the body is distance integral of RHS from $x = R$ to $x = {R}_{s}$
we have
$t = \pm {\int}_{R}^{{R}_{s}} k {\left[\frac{x}{R - x}\right]}^{\frac{1}{2}} \mathrm{dx}$
Using online integral calculator we get

t=+-k[Rtan^-1(sqrt(x/(R−x)))+sqrt(x/(R−x))(x−R)]_R^(R_s)
=>t=+-k[Rtan^-1(sqrt(R_s/(R−R_s)))+sqrt(R_s/(R−R_s))(R_s−R)-(Rtan^-1(sqrt(R/(R−R)))+sqrt(R/(R−R))(R−R))]
=>t=+-k[Rtan^-1sqrt(R_s/(R−R_s))+sqrt(R_s(R−R_s))-Rtan^-1oo]
=>t=+-k[Rtan^-1sqrt(R_s/(R−R_s))+sqrt(R_s(R−R_s))-Rpi/2]
-.-.-.-.-.-.-.-.-.-.-.

Choose the appropriate root as time can't be negative.
Insert value of $k$ in the final expression to obtain value of $t$

Velocity as the body reaches surface of the sun, i.e., at $x = {R}_{s}$
$= \sqrt{2 G \frac{M}{R} \left[\frac{R - x}{x}\right]}$
$\implies \sqrt{2 G \frac{M}{R} \left[\frac{R - {R}_{s}}{{R}_{s}}\right]}$
Inserting values we get

=>sqrt(2(6.67408 × 10^-11)(1.989 × 10^30)/(1.496×10^11)[((1.496×10^11)-(6.957xx10^8))/(6.957xx10^8)])#
$\implies \sqrt{\left(1.775 \times {10}^{8}\right) \times \left(214\right)}$
$\implies 1.949 \times {10}^{5} m {s}^{-} 1$

Even though we know that velocity increases steeply, lets find rough average velocity$= \frac{\text{initial velocity"+"final velocity}}{2} = \frac{0 + 1.949 \times {10}^{5}}{2} = 97450 {m}^{-} 1$
Hence time taken $t = \frac{R - {R}_{s}}{97450} \approx 17.7$ earth days.
Actual time would be much less.

Most of the bodies would have evaporated much earlier due to temperature at sun's surface estimated as 5,778 K.
Compare this temperature with highest bp - Tungsten 5,555 °C
and bp - Uranium 4,131 °C