The initial and final orbits cross at P, where the satellite engine fired instantaneously (see Figure 4)... Only if the satellite follows an open trajectory it can escape from the Earth [r]
(1)Th1 AN ILL FATED SATELLITE
SOLUTION
1.1 and 1.2
⎪ ⎪ ⎪ ⎩ ⎪⎪ ⎪ ⎨ ⎧ ⋅ = ⇒ = ⋅ = ⇒ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⇒ ⎪ ⎪ ⎪ ⎪ ⎭ ⎪⎪ ⎪ ⎪ ⎬ ⎫ = = = m/s 10 07 m 10 22 4 0 2 2 0 0 2 . v r g R v . r T R g r R GM g T r v r v m r m M G T / T T T T π π 1.3 ⇒ = = 0 2 0
0 mv
v R g v m r L T v R g m
L = T
⇒ − = − = − = 2 0 2 0 0 2 2 mv mv r m R g mv r m M G mv
E T T 02
2
mv E =−
2.1
The value of the semi-latus-rectuml is obtained taking into account that the orbital angular momentum is the same in both orbits That is
⇒ = = = = 0 2 2 2 2
0 r
v R g m R g v R g m m M G L l T T T T r l=
The eccentricity value is
3 2 2 m M G L E T + = ε
where E is the new satellite mechanical energy
(
)
0 2 2 2 2 mv v m E v m r m M G v v m
E= +∆ − T = ∆ + = ∆ −
that is
(
1)
1
1 2
0
0 2
0 ⎟⎟= −
⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −
= ∆ mv β
v v mv E
Combining both, one gets ε=β
(2)R.S.E.F 2.2
The initial and final orbits cross at P, where the satellite engine fired instantaneously (see Figure 4) At this point
(
)
⇒− = = =
α β α
θ
cos
0
r r
r
2
π α =
2.3
From the trajectory expression one immediately obtains that the maximum and minimum values of r correspond to θ =0 and
π
θ = respectively (see Figure 4) Hence, they are given by
ε − =
1
l
rmax ε
+ =
1
l rmin
that is
β − =
1
0 max
r
r and = +β
1
0
r r
Forβ =1/4, one gets
m 10 38 m;
10 63
5 ⋅ = ⋅
= . r .
rmax min
The distancesrmax andrmin can also be obtained from mechanical energy and angular momentum conservation,
taking into account that rr and vr are orthogonal at apogee and at perigee
( )
r v m v mgR L
r m gR mv mv
E
T
T
= =
− =
− =
0
2 2
2
0 2
1
1 β
What remains of them, after eliminating v, is a second-degree equation whose solutions are rmax and rmin
2.4
By the Third Kepler Law, the period T in the new orbit satisfies that
3
2
r T a T
=
where a, the semi-major axis of the ellipse, is given by
2
1
2 = −β
+
= r r r
a max min
Therefore
( )
2 3/201
−
− =T β
T
15⎞ 3/2
⎛ −
2
π α= min
r rmax
vr
∆
0 vr P
0 r
(3)Only if the satellite follows an open trajectory it can escape from the Earth gravity attraction Then, the orbit eccentricity has to be equal or larger than one The minimum boost corresponds to a parabolic trajectory, with ε =
β
ε= ⇒ βesc =1
This can also be obtained by using that the total satellite energy has to be zero to reach infinity (Ep = 0) without
residual velocity (Ek = 0)
(
1)
2
1 2
0 − =
= mv esc
E β ⇒ βesc =1
This also arises from T =∞ or from rmax =∞
3.2
Due to ε=βesc =1, the polar parabola equation is
θ cos 1−
= l
r
where the semi-latus-rectum continues to be l=r0 The minimum Earth - satellite distance corresponds to θ =π, where
2
0 r rmin′ =
This also arises from energy conservation (for E = 0) and from the equality between the angular momenta (L0) at the initial point P and at maximum approximation, where rr and vr are orthogonal
4.1
If the satellite escapes to infinity with residual velocityv∞, by energy conservation
(
)
2
2 1
1
∞ = −
= mv mv
E β ⇒
( )
2 1/20 −1
=
∞ v β
v
4.2
As ε=β >βesc =1 the satellite trajectory will be a hyperbola The satellite angular momentum is the same at P than at the point where its residual velocity is v∞ (Figure 5), thus
b v m r v
m 0 = ∞
So
⇒ =
∞
v v r
b 0
( )
2 /
0
−
− =r β
b
φ
v
∆
0 v
∞
v
Asymptote
Asymptote
b
Figure
asym
θ
asym
θ
asym
θ
0 r
(4)R.S.E.F 4.3
The angle between each asymptote and the hyperbola axis is that appearing in its polar equation in the limitr→∞ This is the angle for which the equation denominator vanishes
⇒ =
− cos
1 β θasym ⎟⎟
⎠ ⎞ ⎜⎜ ⎝ ⎛
= −
β θasym cos 1
According to Figure
⇒ +
=π θasym
φ
2 ⎟⎟⎠
⎞ ⎜⎜ ⎝ ⎛ +
= −
β π
φ cos
2
For
2
3 =
= βesc
(5)Th ANSWER SHEET
Question Basic formulas and ideas used
Analytical results Numerical results Marking guideline 1.1
m 10 22
4
0= ⋅
r 0.3
1.2
2 0
0 2
0
T T T
R GM g
T r v
r v m r
m M G
= =
=
π
0
r g R
v = T v0=3.07⋅103m/s 0.3 + 0.1
1.3
r Mm G mv E
v r m L
− =
× =
2
2
r r r
0
v mgR
L = T
2 0
2
mv E =−
0.4
0.4
2.1 l=r0
β ε=
0.4 0.5
2.2
Hint on the conical curves
2
π
α= 1.0
2.3
Results of 2.1, or conservation of E and L
β β
+ =
− =
1
0
r r
r r
min max
m 10 38
m 10 63
7
7 max
⋅ =
⋅ =
r r
1.0 + 0.2
2.4 Third Kepler's Law T =T0
( )
1−β2 −3/2 T =26.4h 0.5 + 0.2 3.1 ε = 1, E = 0, T = ∞ orrmax = ∞
1 = esc
β 0.5
3.2 ε = and results of 2.1
2 r
rmin′ = 1.0
4.1 Conservation of E v∞ =v0
( )
β2−11/2 1.04.2 Conservation of L b=r0
( )
β2−1−1/2 1.04.3 Hint on the conical curves ⎟⎟
⎠ ⎞ ⎜⎜ ⎝ ⎛ +
= −
β π
φ cos
2