a rescue mission to a space station at 300 nmi circular orbi

a rescue mission to a space station at 300 nmi circular orbit starts with a launch with perigee at the surface of the earth and crosses the space station orbit with V= 9000 m/s and an elevation angle of 15 degree a) describe the trajectory b) what is the TOF to the space station orbit? how does it compare with the time of a Hohmann transfer? c) what is the required launch velocity?

Solution

Here in the question we are given with:-

r= 300nmi = 555.6km, V = 9000m/s, theta=15.

First of all we need to calculate the trajactory (e).

we have the formula:-

e= Rp*Vp^2/GM - 1 (1)

to calculate Rp->

Rp =(6378+555.6) * 1000 = 6.9*10^6 m (in meters)

radius of eartgh is 6378 km

Putting in eqn 1 we get the following form:-

e = 6.9*10^6*(9000)^2 / 3.98*10^14 - 1

(here GM of earth is 3.98*10^14)

e = 0.45

Which means trajectory is circular for this case

everything is known to us except the E

To calculate E

we use:-

E= 1.55 radians

Now to calculate the time of flight putting in egn 2 we get the following form: