At first we solve by numeric method.(Analytically can't be solved. Maple and Mathematica can't solve)
Mathematica code:
G = 6.67408*10^-11;
M = 5.9722*10^24;
R = 6378.14*1000;
F = 68000;
m = 2340;
k = 3000;
r = 68;
t0 = k/r // N
when $t=k/r$ gives: 44.11764706
second.
$$h''(t)=\frac{F}{k+m-r t}-\frac{G M}{(h(t)+R)^2}$$
sol = NDSolve[{h''[t] == F/(m + k - r *t) - (G*M)/(R + h[t])^2,
h[0] == 0, h'[0] == 0}, h, {t, 0, t0}];
((h[t] /. sol) /. t -> t0)
putting numeric values to solution:
Rocket will lift at the height: 6192.405405
meters (6.2
km).
At second we will a approximation at point h=0
expanding with series:
$$\frac{G M}{(h(t)+R)^2}=\frac{G M}{R^2}-\frac{2 (G M) h(t)}{R^3}+O\left(h(t)^2\right)$$
putting to equation:
$$h''(t)=\frac{F}{k+m-r t}+\frac{2 G M h(t)}{R^3}-\frac{G M}{R^2}$$
With help CAS we can solve:
$h(t)=-\frac{R e^{-\frac{2 \sqrt{2} \sqrt{G} \sqrt{M} (k+m+r t)}{r R^{3/2}}} \left(\sqrt{2} F \sqrt{R} \text{Ei}\left(-\frac{\sqrt{2}
\sqrt{G} (k+m) \sqrt{M}}{r R^{3/2}}\right) e^{\frac{\sqrt{2} \sqrt{G} \sqrt{M} (3 k+3 m+r t)}{r R^{3/2}}}-\sqrt{2} F \sqrt{R}
\text{Ei}\left(\frac{\sqrt{2} \sqrt{G} (k+m) \sqrt{M}}{r R^{3/2}}\right) e^{\frac{\sqrt{2} \sqrt{G} \sqrt{M} (k+m+3 r t)}{r
R^{3/2}}}-\sqrt{2} F \sqrt{R} e^{\frac{\sqrt{2} \sqrt{G} \sqrt{M} (3 k+3 m+r t)}{r R^{3/2}}} \text{Ei}\left(-\frac{\sqrt{2} \sqrt{G}
\sqrt{M} (k+m-r t)}{r R^{3/2}}\right)+\sqrt{2} F \sqrt{R} e^{\frac{\sqrt{2} \sqrt{G} \sqrt{M} (k+m+3 r t)}{r R^{3/2}}}
\text{Ei}\left(\frac{\sqrt{2} \sqrt{G} \sqrt{M} (k+m-r t)}{r R^{3/2}}\right)-2 \sqrt{G} \sqrt{M} r e^{\frac{2 \sqrt{2} \sqrt{G} \sqrt{M}
(k+m+r t)}{r R^{3/2}}}+\sqrt{G} \sqrt{M} r e^{\frac{\sqrt{2} \sqrt{G} \sqrt{M} (2 k+2 m+r t)}{r R^{3/2}}}+\sqrt{G} \sqrt{M} r
e^{\frac{\sqrt{2} \sqrt{G} \sqrt{M} (2 k+2 m+3 r t)}{r R^{3/2}}}\right)}{4 \sqrt{G} \sqrt{M} r}$
sol2 = DSolve[{h''[t] ==
F/(m + k - r *t) - (G M)/R^2 + (2 (G M) h[t])/R^3, h[0] == 0,
h'[0] == 0}, h[t], t] // FullSimplify
h[t] /. sol2 /. t -> t0
(* {{h[t] -> -(1/(4 Sqrt[G] Sqrt[M] r))
E^(-((2 Sqrt[2] Sqrt[G] Sqrt[M] (k + m + r t))/(r R^(3/2))))
R (E^((Sqrt[2] Sqrt[G] Sqrt[M] (2 (k + m) + r t))/(
r R^(3/2))) (-1 + E^((Sqrt[2] Sqrt[G] Sqrt[M] t)/R^(
3/2)))^2 Sqrt[G] Sqrt[M] r +
Sqrt[2] F Sqrt[
R] (E^((Sqrt[2] Sqrt[G] Sqrt[M] (3 (k + m) + r t))/(
r R^(3/2))) (ExpIntegralEi[-((
Sqrt[2] Sqrt[G] (k + m) Sqrt[M])/(r R^(3/2)))] -
ExpIntegralEi[-((Sqrt[2] Sqrt[G] Sqrt[M] (k + m - r t))/(
r R^(3/2)))]) +
E^((Sqrt[2] Sqrt[G] Sqrt[M] (k + m + 3 r t))/(
r R^(3/2))) (-ExpIntegralEi[(
Sqrt[2] Sqrt[G] (k + m) Sqrt[M])/(r R^(3/2))] +
ExpIntegralEi[(Sqrt[2] Sqrt[G] Sqrt[M] (k + m - r t))/(
r R^(3/2))])))}}*)
putting numeric values to solution:
Rocket will lift at the height: 6192.406503
meters.
The difference form numerics and analytically is very small about 1
millimeter!
Another approximation.See-> Altitude:
$$h''(t)=\frac{F}{k+m-r t}-g$$
where $g$ is standard acceleration of free fall.
g = (G*M)/R^2
(* 9.798004977 *)
and solution:
$$h(t)=\frac{r t (2 F-g r t)+2 F (k+m-r t) (\log (k+m-r t)-\log (k+m))}{2 r^2}$$
sol3 = DSolve[{h''[t] == F/(m + k - r *t) - g, h[0] == 0, h'[0] == 0},
h[t], t] // FullSimplify
(h[t] /. sol3 /. t -> t0)
(* {{h[t] -> (r t (2 F - g r t) +
2 F (k + m - r t) (-Log[k + m] + Log[k + m - r t]))/(2 r^2)}}*)
Rocket will lift at the height: 6190.114097
meters.
The difference form numerics and analytically is about 2
meters!
F,m,k,r
? $\endgroup$