One of the essential concepts to solve this problem is the utilization of the equations of centripetal and gravitational force.
From them it will be possible to find the speed of the body with which the estimated time can be calculated through the kinematic equations of motion. At the same time for the calculation of this speed it is necessary to clarify that this will remain twice the ship, because as we know by relativity, when moving in the same magnitude but in the opposite direction, with respect to the ship the debris will be double speed.
By equilibrium the centrifugal force and the gravitational force are equal therefore
[tex]F_c = F_g[/tex]
[tex]\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}[/tex]
Where
m = mass spacecraft
v = velocity
G = Gravitational Universal Constant
M = Mass of earth
[tex]r \rightarrow R+h \Rightarrow[/tex] Radius of earth and orbit
Re-arrange to find the velocity
[tex]\frac{mv^2_{orbit}}{r} = \frac{GMm}{r^2}[/tex]
[tex]\frac{v^2_{orbit}}{r} = \frac{GM}{r^2}[/tex]
[tex]v^2_{orbit}=\frac{GM}{r}[/tex]
[tex]v_{orbit} = \sqrt{\frac{GM}{r}}[/tex]
[tex]v_{orbit} = \sqrt{\frac{GM}{R+h}}[/tex]
Replacing with our values we have
[tex]v_{orbit} = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{6.37*10^6+0.4*10^6}}[/tex]
[tex]v_{orbit} = 7676m/s[/tex]
From the cinematic equations of motion we have to
[tex]t = \frac{d}{2v_{orbit}} \rightarrow[/tex] Remember that the speed is double for the counter-direction of the trajectories.
Replacing
[tex]t = \frac{29000m}{7676m/s}[/tex]
[tex]t = 3.778s[/tex]
Therefore the time required is 3.778s
