Thanks JayS, SSSji and others. Of course, JayS (and others who gave the same answers) is correct. SSS / JayS's posts are clear enough but let me add my comments as I said I would.
The problem is actually very simple -- (yet illustrate basic physics).
Since the sat (after the delta-V is given) encounters no more outside force (except gravity - and we are ignoring things like air drag, gravitational waves, solar wind, collision with space trash ityadi
in EACH case the orbit will be
, obviously,... Ellipse! (Obviously delta-V is too small to reach escape velocity so, as SSS pointed out, no hyperbolic orbit - also remember circle is a special case of ellipse and any ellipse whose perigee is less than radius of earth will crash on earth)
Also KE (and hence total Energy) is increased in all cases except C ==> the new orbit will be "higher" in all cases except C. C is the only case where it will be lower.
If you want to calculate/estimate "how much" it too is VERY easy if one just uses the most basic concepts/formulas.
(see below for details)
Note: Please note: By higher orbit, we mean average of perigee and apogee
(also called semi-major axis
of ellipse) is greater. The time period of the orbit is *longer* here. Time period, BTW depends only on semi-major axis and thus for any elliptical orbit it is same as equivalent circular orbit whose radius is equal to semi-major axis)
If you want to visualize the shape or calculate the numbers, just look at SriKumar's one of the very first post where the discussion was started in the first place a few months ago.
Easiest to work with is the formula:
(Assume M is mass of earth, G is gravitational constant, a=semi-major axis (=average of perigee and apogee), r = distance from center of earth, v= velocity of sat and assume the mass of sat=1 unit) Energy = KE+PE = (1/2)v^2 - GM/r = -GM/2a
Rest is simple, first for circular orbit - substitute r=a=radius_of_earth+500 in above to get orbital velocity v (it will come around 8km/sec). Let us call this value = u=8000 m/s.
Now substitute v=(u+100) for case B, v=u-100 for case C, and sqrt(u^2+100^2) for case A and D, and you will get the new 'a'.
(This will be about 170 Km more, for case B, about 170 Km less for case C and for A and D it is about 1Km more that before. So the answer in clear terms -
Case A - Slightly elliptical orbit, slightly higher (average of perigee and apogee height from ground ~ 501 Km, perigee about 470km) . Time period is increased by about 1 second. (JayS is right about shape and orientation - see his post).
Case B - Same perigee, higher apogee (about 840Km), average about 670Km. Time period increased a by about 3 minutes.
Case C - Same apogee, lower perigee (about 160 Km above earth's surface -- still will avoid hitting earth ) Time period decreased by about 3 minutes.
Case D - SAME shape of orbit as case A - (Just sort of mirror image but read JayS's description) .
(For some, it may be some what counter intuitive that a higher orbit is achieved, and it does not matter if you shoot up or down
(All the above values I calculated just now without a calculator (or even a scratch pad - I did the calculations while typing) so results may be a little off ( within few %) but it should not change the basic concepts)-- perhaps someone can feed the data in computer and draw.
Case B and C are actually used when one wants to transfer from one orbit to other.
(Edited later: Some minor typo's corrected)