To solve this problem we will use Kepler's third law for which the period is defined as
[tex]T= 2\pi \sqrt{\frac{a^3}{GM}}[/tex]
The perigee altitude is the shortest distance between Earth's surface and Satellite.
The average distance of the perigee and apogee of a satellite can be defined as
[tex]a = R+\frac{r_1+r_2}{2}[/tex]
Here,
R = Radius of Earth
[tex]r_1[/tex]= Lower orbit
[tex]r_2[/tex]= Higher orbit
Replacing we have,
[tex]a = 6378.1+\frac{400+600}{2}= 6878.1km[/tex]
Time Taken to fly from perigee to apogee equals to half of orbital speed is
[tex]t = \frac{T}{2}[/tex]
[tex]t = \pi \sqrt{\frac{a^3}{GM}}[/tex]
Replacing,
[tex]t =\pi \sqrt{\frac{(6878.1*10^3)^3}{(6.67*10^{-11})(6*10^{24})}}[/tex]
[tex]t = 2832.79s[/tex]
[tex]t = \text{47min 12.8s}[/tex]
Therefore will take around to 47 min and 12.8s to coast from perigee to the apogee.
