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. Simple!

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)