Polar Orbit Change and Line of Sight Time

[orbitsnerd]

Homework Statement

A satellite is in a circular polar orbit 240 km altitude. When the satellite is over the South Pole the engine is fired to achieve a polar orbit that has apogee directly over the North Pole. After the impulsive burn an observer on the North Pole observes the satellite has a velocity of 9.183 km/s. As the satellite comes over the horizon at the North Pole it is determined to have an altitude of 3670.9 km. How long with the observer be able to communicate with the satellite assuming only line of sight communications with the satellite?

Homework Equations

If Hohmann transfer, those equations.
deltav1=v2-v1
deltav2=v4-v3

The Attempt at a Solution

rp=240km+6378km
ra=3670.9km+6378km
If it is necessary to find e, I can with rp and ra. e=.206
v3=9.183km/s
Possible use of swath angle with Hohmann transfer information?

 

[D H]

The question is not asking about Hohmann transfers. It is asking about line of sight communications: How long is the satellite in view? What are the relevant equations for this?

BTW, that velocity is not right for the Earth. This must be some other planet.

 

[orbitsnerd]

The question is not asking about Hohmann transfers. It is asking about line of sight communications: How long is the satellite in view? What are the relevant equations for this?

BTW, that velocity is not right for the Earth. This must be some other planet.

I have to go off of that it is Earth even though the velocity doesn't seem to make sense I guess. :/

As for the relevant equations for line of sight/viewing geometry I have:
central angle to the horizon=cos angle=re/re+h so at 240 km angle is 15.477 degrees and at 3670.9 km angle is 50.6 degrees.
I can also find the angle from the nadir to the spacecraft by saying cos central angle=sin rho.

Swath width=2re*central angle to the horizon

There are a handful of other equations that define Dh and other angles. I guess my biggest question is: why was I given velocity and the initial orbit altitude if it is only for line of sight?
Thanks!