Answer by JD for RG, index: 1612280709:

@ Charles

 1. The distance from the center $r(t) = R+h(t)$ at instant $t$ is governed by the following differential equation of order 2:

$ (1)  r''(t)= - \frac{ g_E \cdot R^2 }{ r(t)^2 },\; \mbox{ within the interval }\; r>0,$

 where g_E =0.00981\, km/s^2 \right]$, and the initial conditions are

$(2)   r(0) = R + h(0), \quad r'(0) = 0, \quad \mbox{ where } h(0) = h = 20184, \; R = 6378. $

The size and the mass of BB does not matter in the assumed scale.

2. The general solution of equation (1) is obtainable by the following trick: Multiply both sides by $2 \cdot r'(t)$; then by the chain rule we get, that the left hand side, which is equal to $[(r'(t))^2]'$, has to equal the derivative $ \left[ \frac{2\cdot g_E \cdot R^2 }{ r(t) }\right]'$.

3. Since the two derivatives are equal, by the theorems from calculus, the differentiated functions have to differ by a constant; call it $A$. Thus:

$ (r'(t))^2 = \frac{2 \cdot g_E \cdot R^2 }{ r(t) } +A $

4. By the second initial condition of  (2) one gets $ A = - \frac{2 \cdot g_E \cdot R^2 }{ r(0) } $, which implies

$ (r'(t))^2 = \frac{2 \cdot g_E \cdot R^2 }{ r(t) } - \frac{2 \cdot g_E \cdot R^2 }{ r(0) }$

5. The next step is to integrate the differential equation:

$(3) r'(t) =\pm \sqrt{ \frac{2 \cdot g_E \cdot R^2 }{ r(t) } - \frac{2 \cdot g_E \cdot R^2 }{ r(0) } },  $

valid only for $ r(t) \le r(0)$

Possibly, this would be what you are asking for:)

Good luck. Give response if there are errors in my outline, please.

PS: Afterwards, you can consult your results with

https://en.wikipedia.org/wiki/Equations_for_a_falling_body

@ Joachim

Thanks for your reply, though it's a notch above my pay grade. I thought the math would be less esoteric. I'll ask a colleague to guide me through your response. Again, thanks for your efforts.

Answer by JD for RG, index: 1612281301:

@ Charles

OK, a direct answer to your questions are as follows, where 

\sqrt{ s }  = "the_square_root_of_expression_s"

r(0) = R + h

g_E = acceleration at the surface of the Earth

4. The omitted steps should lead you to the following formula for the  instant t_E when the falling ball reaches the Earth, i.e. then r(t_E) = R:

(4)    t_E =[  \sqrt{R/ r(0)} * \sqrt{ 1 - R / r(0) } + arccos ( \sqrt{ R/ r(0)} ) ] / B.

where   B= .\sqrt{  2 * g_E * R / r(0)2 }

5.  . . .  and from equation (3) one can get a formula for the hitting speed v_E of the falling down BB from the hight $h$ above the Earth surface, observed at the surface of the Earth, i.e. at instant  t_E:

(5)                     v_E =  \sqrt{ 2 * g_E * h * R / R+h  }.

Obviously, since this material is prepared  as an advice only, it might contain mistakes. Thus it is a good idea to consult with other people, as well as the  page of WIKI indicated in  my first answer (there are given the formulas without derivation, and in different notation).

Hopefully this will work. By the way, the average speed equals  h/ t_E, doesn't it?

@ Joachim

Joachim, I've attempted to work this out using your formula. Does my result appear correct?

v_E = impact Velocity

Using you formula: v_E = sqrt ( 2 * g_E * R * ( R/R+h))

We have:

v_E = sqrt (2 * 9.8 * 6500000 * (6500000 / 6500000 + 26700000))

v_E = sqrt (127400000 * (0.19578))

v_E = sqrt (24942372)

v_E = 4994.27382 m/s

v_E = 4.99427382 km/s

I think it's ok. Joachim

@ Charles

RED ALERT: I have not noticed that you have misused   R  instead of  h in one place.

I need time for more exact answer :(    J.D.

Yes, my error, you are correct...

I've corrected the formula and recalculated. The formula you presented was...

v_E = sqrt ( 2 * g_E * h * ( R/R+h))

v_E = sqrt (2 * 9.8 * 20184000 * (6500000 / 6500000 + 20184000))

v_E = sqrt (395606400 * (0.24344))

v_E = sqrt (9696422)

v_E = 9813.5835 m/s

v_E = 9.814 km/s

@ Charles

OK, but now the approximation should be put in a  desired order: either  all quantities are exact, or all are rounded. More precisely:

In  your question the quantities are qual  to 20,184 km  and  6,378 km, and now the first number is unchanged and the secon rounded to  6,500, which causes unnececessary doubts: why the first is more exact? is the result then reliable? especially - in view of the 4 digits of the final number. Simple argument allow to guess, that the final result should be then rounded to at most 3 digits.

Regards

PS.  The time elapsed till the impact given at point 4. of my answer (answer  no. 3 of this post) has an error: 

instead of   B=\sqrt{ 2 * g_E * R / r(0)2 } 

it should be  B=\sqrt{ 2 * g_E * R^2 / r(0)^3 } 

Indeed, you need a comparison to the WIKI page referred to in my first answer, please.

Joachim

@ Joachim

Good point on precision and thanks for your correction and your help.

Thanks for thanks! 

Joachim

Gravity g varies as 1/r2

so starting gravity is g1 = g2 ( r2 / r1)2   =  9.81 m/sec2 *( 6378 km / 26,562 km )2

or 0.566 m/sec2

dv/dt = g = g1 (r1/r)2

vertical velocity

v =- dr / dt

v1 = zero

v dv = - g1 (r1 /r )2 dr

v22 = 2 g1r1 ( r1 / r2 - 1)  = 2 * 0.566 * 26,562* 1000 m/km *( 26,562/ 6378 - 1)

or 9751 meters per second

It differs from CW only in the precision of distances used in Excel Spread Sheet. Also G and M are not needed since they are implied by the other data.

This method is taught in high school physics and ignores a lot of non idealities.

I like Professor Decker,s  way of deriving the ODE  for  v  as a function of  the path   r (hiding the other fashion of a function of  time  t ).  Let me  add, that fortunalety this leads immediately to the energy conservation law.  Consequently one can get the impact speed formula for arbitrary initial speed. The problems then still is not beyond the high school math, unless the starting speed is greater or equal the first cosmic one (appropriate to the initial height). . . 

. . .  OR the starting direction should ensure return to the earth - and this is already over elementary one dimensional calculus.

Professor Decker, thanks for pointing at this nice and simple way of  solving the problem of this question.

Best regards, Joachim Domsta

Jerry, appreciate you sharing this methodology.

I considered not this problem, but a related for aquaplane. I did not check the codes or regulations yet, but if large accelerations, a weighting of such may also give action of impact.