Thread: GRONDY PLEASE READ THIS!!!
Talking to myself, too...
Hey, any of you guests who've been around here could help out if you wanted...please...
here's the monster:
A satellite of mass m is in an elliptical orbit around the earth, which has a mass Me and a radius Me. The orbit varies from closest apporach of a at point A to maximum distance of b from the center of the Earth at point B . At point A, the speed of the satellite is Vo. Assume that the gravitational potential energy Ug = 0 when masses are an infinite distance apart. Express your answers in terms of a, b, m, Me, Re, Vo, and G. As the satellite passes point A, a rocket engine on the satellite is fired so that its orbit is changed to a circular orbit of radius a about the center of the Earth. Determine the work done by the rocket engine to effect this change.
blah...have fun...I'm pretty sure there's a simple way to do this, I'm just not seeing it.
Lemme see what I can do.
Here's a rough guide & some formulas that I can remember:
Mass of satellite = m
Mass of Earth = Me
Elliptical orbit distance: Perihelion = a; Aphelion = b
Circular orbit distance, r = a
Elliptical orbit velocity of satellite at perihelion = Vo
Elliptical orbit velocity of satellite at aphelion = V1
Circular orbit velocity of satellite = V
Gravitational constant = G
1) Assuming that the gravitational potential energy Ug = 0 when masses are an infinite distance apart, then
U(r) = - (intergral of, from infinity to r} F.dr =
F.dr = -GMe.m/r^2
So, U(r) = - (GMe.m)/ r
2) Velocity of satellite (circular orbit) = V1 = square root of (GMe/r)
3) For elliptical orbit:
Solving the equation of (1) conservation of angular momentum (m.Vo.a = m.V1.b), and
(2) conservation of energy (Eo = E1), where E = (1/2 . m.V^2 - (G.Me.m)/ r
Velocity of satellite (elliptical orbit) at perihelion squared = Vo^2 = 2.G.Me.((b/(a.(a+b))
V^2 = Vo^2 . a . (a+b)/ (2b) ----- (i)
4) Energy for circular orbit = Ec = -(G.Me.m)/2a ----------- (ii)
5) Energy for elliptical orbit = Ee = - (G.Me.m)/(a+b) --------------( iii)
6) U = Gravitational potential energy
K = kinetic energy = (1/2). m.v^2
E = total energy = K + U ----------- (iv)
7) Change in potential energy between staellite & surface of Earth at a = delta U
delta U = G. Me. m. (a-Re)/(a.Re)
8) So the work performed equals change in energy
U(elliptical at perihelion) - U(circular) = equations (iii) -(ii)
See if it works...from here I can see that your results from (8) would yield m, Me, Vo, a, b & G. I haven't tried simplifying & subsituting equation (7) - maybe that's where you can get the term Re in your equation.
Good luck...hope this helps.
didn't quite make the due date, but whatever...at least I understand the problem now...I think I ended up making up some stuff that was pretty close to what you said...hopefully I'll get some credit for that
Of course when I took physics, Kennedy hadn't yet told NASA to shoot for the moon and my physics books hadn't the slightest hint about astrophysics being merely the tying together of the force of attraction between two or more bodies(gravity), the conservation of energy (momentum), and eliptical geometry in at least three dimensional space.
Yeah - but at least you saw the moon landing. I was only a few months old then...wish my parents had gotten down to business earlier.
but seriously AAAAAAAAAHHHHHHHHHHHHHH!!!!!!!!!