Difficult falling body problem?

Question

how long , T(tot), does a mass m with initial velocity =0, take to fall to the surface of a planet (of mass M and radius r) from a height H above the surface , in a non-constant field given by GMm/(r+s)^2?

(where s is the height above the surface at any time t)

this is a lot more difficult than it looks!

full points for an elegant solution.

(full working details if you can thanks)

in terms of H,r,m and M of course.

KEITH P
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d = 1/2 a t^2 is only true for the first infinitesimal d element where the initial velocuty is 0. after that the equation is
d=U(t)*t +(1/2) at^2 where the function
U(t) = the integral from t=0 to t of the velocity, which integral must then be included in the calculation of each infinitesimal d element

Answer 1

Starting with a simplified case of dt/dr, we make r (and therefore acceleration a) invariant over a small distance and apply the classic equation:
d = 1/2 a t^2
2d = a t^2
2d / a = t^2
t = √[ 2d/a ]

Since
f = ma, then
a = f/m
Substituting for a gives us
t = √[ 2dm/f ]
Further, f (force, in this case gravitational) can be computed according to classical Newtonian mechanics as given:
f = GMm/(r+s)^2
... and substituting for f gives us

t = √ [ {2dm (r+s)^2} / GMm ]

We can extract the square/square root portion and combine terms:

t = (r+s) √ [ 2d / GM ]
Note that m (mass of the falling body) factors out here (as it should!) And remembering that d, distance, is the same as the infinitesimal radius vector dr in this case, we have
t = (r+s) √[ 2dr / GM ]

Now we just integrate t over the interval r=H to r=s

T(tot) = (r=H)∫(r=s) {(r+s) √ [ 2dr / GM ] } dr

And that should do it. Unless I made a mistake somewhere ...