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## Main Question or Discussion Point

Heey everyone,

I was wondering how long it takes for a small mass to reach the earth from infinity . The trajectory is completely radial, so it's a relatively simple equation.

$$r''(t ) =\frac{ - GM} {r(t)^2}. $$

But what if we multiplied both sides by ##2r'(t) ##. Then it follows,

$$2r'(t) r''(t ) =\frac{ - 2GM} {r(t)^2}r'(t).$$ $$ \implies \frac{d} {dt} (r'(t) ^2) =\frac{ - 2GM} {r(t)^2}r'(t).$$

Integrate with respect to t from t=t to t=##\infty##. and $$r'(\infty) =0 $$ and ##r(\infty) =\infty##.

$$ r'(\infty) ^2 - r'(t) ^2 =\frac{2GM} {r(\infty)} -\frac{2GM} {r(t)} $$, leaving with $$r(t)r'(t) ^2 =2 GM$$

Which has general solution:

$$r(t) =(3/2) ^{(2/3)} (C+\sqrt{2GM}t)^{(2/3)} $$.

Letting ##r(0)=R## we can determine C, that is ##C=2/3R^{(3/2)}##.

So our trajectory from earth to infinity is given by:

$$r(t) =(3/2)^{(2/3)} [(2/3)R^{(3/2)}+\sqrt{2GM}t] ^{(2 /3)}$$

Differentiating gives:

$$r'(t) =(3/2)^{(2/3)}\frac{2 \sqrt{2GM}}{3(\sqrt{2GM}t+2R^{(3/2) } /3)^{(1/3)}} $$ $$ r'(0)= \sqrt{\frac{2GM}{R}}$$. Which validates the escape velocity.

I approached the problem by shooting the mass from the earth to infinity and now I have come to the conclusion that it takes infinite amount of time.. Oops.

I suppose dropping it from any given height will give:

$$r'(t) ^2 =\frac{2GM}{r(t)}-\frac{2GM}{b} $$ where b is the height we are dropping it from.

Then $$b\pi/2 -\sqrt{2GM/b}t=b\tan^{-1}(\sqrt{\frac{r(t)}{b-r(t)}})-\sqrt{r(t)(b-r(t))}$$.

Then for ##t=t_{final} ## we must have ##r(t_{final}) =R##. Then the time can be determined how long the fall would take from b to R.

Thanks

I was wondering how long it takes for a small mass to reach the earth from infinity . The trajectory is completely radial, so it's a relatively simple equation.

$$r''(t ) =\frac{ - GM} {r(t)^2}. $$

But what if we multiplied both sides by ##2r'(t) ##. Then it follows,

$$2r'(t) r''(t ) =\frac{ - 2GM} {r(t)^2}r'(t).$$ $$ \implies \frac{d} {dt} (r'(t) ^2) =\frac{ - 2GM} {r(t)^2}r'(t).$$

Integrate with respect to t from t=t to t=##\infty##. and $$r'(\infty) =0 $$ and ##r(\infty) =\infty##.

$$ r'(\infty) ^2 - r'(t) ^2 =\frac{2GM} {r(\infty)} -\frac{2GM} {r(t)} $$, leaving with $$r(t)r'(t) ^2 =2 GM$$

Which has general solution:

$$r(t) =(3/2) ^{(2/3)} (C+\sqrt{2GM}t)^{(2/3)} $$.

Letting ##r(0)=R## we can determine C, that is ##C=2/3R^{(3/2)}##.

So our trajectory from earth to infinity is given by:

$$r(t) =(3/2)^{(2/3)} [(2/3)R^{(3/2)}+\sqrt{2GM}t] ^{(2 /3)}$$

Differentiating gives:

$$r'(t) =(3/2)^{(2/3)}\frac{2 \sqrt{2GM}}{3(\sqrt{2GM}t+2R^{(3/2) } /3)^{(1/3)}} $$ $$ r'(0)= \sqrt{\frac{2GM}{R}}$$. Which validates the escape velocity.

I approached the problem by shooting the mass from the earth to infinity and now I have come to the conclusion that it takes infinite amount of time.. Oops.

I suppose dropping it from any given height will give:

$$r'(t) ^2 =\frac{2GM}{r(t)}-\frac{2GM}{b} $$ where b is the height we are dropping it from.

Then $$b\pi/2 -\sqrt{2GM/b}t=b\tan^{-1}(\sqrt{\frac{r(t)}{b-r(t)}})-\sqrt{r(t)(b-r(t))}$$.

Then for ##t=t_{final} ## we must have ##r(t_{final}) =R##. Then the time can be determined how long the fall would take from b to R.

Thanks

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