Mécanique du point : forces Newtoniennes (PCSI)

−→
F=−K
r2
−→
ur
r=mC2
K
1
1+e cosθ C=r2˙
θ
e, K, m C2
Ep=−K
r=−K2 mC2(1 + e cosθ)
v2= ˙r2+r2˙ θ2=dr
dθ 2
˙
θ2+r2˙ θ2
= dr
dθ 2
+r2!C2
r4
=
 mC2
K
−e sinθ
(1 + e cosθ)2!2
+mC2
K
1
1 + e cosθ 2
C2
r4
= m2C4
K2
e2sin2θ
(1 + e cosθ)4+m2C4
K2
1
(1 + e cosθ)2!C2K4
m4C8(1 + e cosθ)4
=K2
m2C2e2sin2θ+K2 m2C2(1 + e cosθ)2
=K2
m2C2e2sin2θ+e2cos2θ+ 1 + 2 ecosθ
=K2
m2C2e2+ 1 + 2e cosθ
Ec=1
2mv2=K2 2mC2e2+ 1 + 2e cosθ
Ec+Ep=K2
2mC2e2−1

−→
F=−K
r2
−→
ur
r=p
1+e cosθ
−→
v0b
b
r˙ θ
N
N−→ uθ
vNN ON b v0
0 = r2
minv2
0+ 2GMTrmin −v2 0b2
bcb < bc
b>bc
−→
F=−G MTm
r2
−→
urT−→ r∧−→ F= 0
−→
r∧−→ v=−−→ cste −→ r∧−→ v
−→
r∧−→ v
−→
r∧−→ v= (r−→ ur)∧˙r−→ ur+r˙ θ−→ uθ=r2˙ θ−→ uz
r2˙ θ=cste
m−→ r∧−→ v=mr2˙
θ=m−→ r0∧−→ v0=m(−x0−→ ux−b−→ uy)∧(v0−→ ux) = mbv0−→ uzr2˙ θ=v0b
−→
v= ˙r−→ ur+r˙ θ−→ uθ˙r= 0 r−→ vN=rmin ˙ θ−→ uθ=v0b
rmin
−→
uθ
Em=1
2mv2
0=1
2mv2
N−GMTm
rmin =1 2mv2
0b2
r2
min
−GMTm
rmin 0 = r2
minv2
0+ 2GMTrmin −v2 0b2
rmin =RTv0bc=pR2
Tv2
0+ 2GMTRTbc=RTq1 + 2GMT
RTv2
0
=
RTr1 + v2
lib
v2
0

C m = 2,5 1015kg
dmax = 5.104 a a
T0
a a T0
v0
G MS
G Msv0a
dmdM
dma v0dMvMvmC
dma C
vM
dm'a
r=R+h h
v
T T r
λ W
W

M m
r1
r2
P
∆vP
A
∆vA
r(θ)
e
∆vP∆vA
r1r2
r=p
1+e cosθ , Em=K2 2mC2(e2−1) = K 2p(e2−1) a=1 2p 1−e+p 1+e=p 1−e2
Em=−K
2a
r2˙ θ=cste r =cste ˙ θ=cste −→ a=−v2
r
−→
ur
−mv2
r
−→
ur=−mMTG
r2
−→
urv2=MTG
r
Em=Ec+Ep=1
2mMTG r−MTGm r=−mMTG 2r
r=p
1+e cosθ
rmin =p
1+e=r1rmax =p 1−e=r21+e 1−e=r2 r1e=r2−rA r2+r1
Em=−GMTm
2a2a=r1+r2Em=−GMTm r1+r2
A−mMTG
2r1
−GMTm
r1+r2−GMTm r1+r2−−mMTG 2r1=GMTm1 2r1−1 r1+r2=
GMTm
2r1r2−r1 r1+r2∆v1∆Ec=1 2m(v1+ ∆v1)2−v2
1=
GMTm
2r1r2−r1 r1+r2∆v2
1+2v1∆v1−GMT
r1r2−r1 r1+r2= 0 v1=qMTG r1∆v1=−qMTG r1±
rMTG
r1+GMT r1r2−r1 r1+r2=qMTG r1r1 + r2−r1 r1+r2−1=qMTG r1q2r2 r2+r1−1

B A −GMTm
r1+r2
−mMTG
2r2−mMTG 2r1−−GMTm r1+r2=GMTm1 r1+r2−1 2r2=
GMTm
2r2r2−r1 r1+r2∆v2∆Ec=1 2mv2
2−(v2−∆v2)2=
GMTm
2r1r2−r1 r1+r2