Mécanique du point : problèmes à 2 corps (PCSI)

M1M2m1m2
G
R∗−→ i , −→ j , −→ k
−→
r1=−−−→ GM1−→ r2=−−−→ GM2−→
r=−−−−→ M1M2
−→
p1−→ p2
−→
v=−→ v2−−→ v1µ
G
M
µ−→ r G M1M2
M1M2M
M M1M2
P M R
B1B2m r G
d P G −−−→ f2→1=f2→1−→ ur−→ ur=
−−→
OG
OG B2B1
ωr
d
R0
B1B2
d=D
(1 + x)α= 1 + αx +O(x2)

−−−−→
FP→B1=−GMm 1
(d−r)2−→
ur−−−−→ FP→B2=−GMm 1
(d+r)2−→
ur
2md−→
vG
dt =−2mv2 d−→
ur=−GMm 1
(d−r)2+1 (d+r)2!−→
ur
=−GMm
d21+2r d+1−2r d−→
ur
=−2GMm
d2−→
ur
v2=d2ω2=GM
dω=qGM d3
B1−−−−→ FP→B1−−−−−→ Fgrav
B2→B1−−−→
f2→1−→ Fie
−GMm 1
(d−r)2+Gm21 (2r)2+f2→1+mω2(d−r)=0
−GMm 1
(d+r)2−Gm21 (2r)2−f2→1+mω2(d+r)=0
f2→1= 0
−GMm 1
(D−r)2+Gm21 (2r)2+mω2(D−r)=0
−GMm
D21+2r D+Gm21 4r2+mGM D21−r D= 0
−M−2Mr
D+mD2 4r2+M−Mr D= 0
−3Mr
D+mD2 4r2= 0
D=12M
m1/3r

P MPS MSMP
D M m MPMS
P M P
RGR(S, −→ uz)
ΩR RG
R
(x, y, z)R
z= 0
L1L2
x
d=D−x
•d < 0
P
•d > 0
L3
x
d=D+x
L4L5
P S
d=SM =P M
•d=D
L4L5
(1 + x)α= 1 + αx +O(x2)

T2
D3=4π2 GMS˙
θ=cste Ω =
2π
T=qGMS D3
•S M −−→ FSM =−GMSm
SM 3−−→
SM =−GMSm
(x2+y2+z2)3/2
x
y
z
•T M −−→ FT M =−GMTm
T M3−−→
T M =−GMTm
((x−D)2+y2+z2)3/2
x−d
y
z
•−→ Fie =−m−→ ae=−m−→ Ω∧−→ Ω∧−→ r=GMSm
D3
x
y
0
⇒
0 = −GMS
(x2+y2+z2)3/2x−GMT
((x−D)2+y2+z2)3/2(x−D) + GMS
D3x
0 = −GMS
(x2+y2+z2)3/2y−GMT
((x−D)2+y2+z2)3/2y+GMS
D3y
0 = −GMS
(x2+y2+z2)3/2z−GMT
((x−D)2+y2+z2)3/2z
z= 0
L1L2
y= 0 x d D
d
d3=±1 d2d
0 = −GMS
(D+d)3(D+d)−GMT d3d+GMS D3(D+d)
=−GMS
(D+d)2±GMT d2+GMS D3(D+d)
=−GMS
D21−2d D±GMT d2+GMS D21 + d D
⇒3d3=±MT
MSD3
L3
y= 0 x d D d −D < 0
d−2D
0 = −GMS
(d−D)3(d−D)−GMT (d−2D)3(d−2D) + GMS
D3(d−D)
= + GMS
(d−D)2+GMT (d−2D)2+GMS
D3(d−D)
=GMS
D21+2d D+GMT 4D21+2 d 2D+GMS D2d D−1
=−GMT
4D2−3dGMS+MT D3
⇒d=MT
12MSD
L4L5
x2+y2= (x−D)2+y2=d2y
0 = −GMS
d3−GMT d3+GMS D3y
⇒d=D
L4L5D
x y
x=D/2y=√3D/2
L4L5

L1, L2L3
L4
L5
L4L5