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]
Trang 1Th1 AN ILL FATED SATELLITE
SOLUTION
1.1 and 1.2
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
⋅
=
⇒
=
⋅
=
⇒
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⇒
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
=
=
=
m/s 10 07 3
m 10 22 4 4
2
3 0
0 0
7 0
3 1
2
2 0 2
0
2 0
0 0
0
2 0 2
0
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
⇒
=
2 0
2 0 0
v
R g v m r
0
2
R g m
L = T
⇒
−
=
−
=
−
0
2 0 0
2 2 0 0
2 0
1 2
1 2
1
mv mv r
m R g mv r
m M G mv
2
1
mv
E =−
2.1
The value of the semi-latus-rectum l is obtained taking into account that the orbital angular momentum is the same
in both orbits That is
⇒
=
=
=
2 0
2 2 2 2
0
4 2 2 2
2
r v
R g m R g v
R g m m M G
L
T
T T
0
r
l=
The eccentricity value is
3 2 2
2 0
1
m M G
L E
T
+
=
ε
where E is the new satellite mechanical energy
0
2 0
2 0
2 2
1 2
1 2
1 2
1
mv v
m E v m r
m M G v v m
E= +∆ − T = ∆ + = ∆ −
that is
( 1)
2
1 1 2
0 2
0
2 2
⎠
⎞
⎜
⎜
⎝
⎛
−
v
v mv E
Combining both, one gets ε=β
This is an elliptical trajectory becauseε=β <1
Trang 22.2
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 1
0 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
l
r min
that is
β
−
= 1
0 max
r
0 min
r r
Forβ =1/4, one gets
m 10 38 3 m;
10 63
r max min
The distancesr max andr min 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 0
2 2 2
2
1 1 2
What remains of them, after eliminating v, is a second-degree equation whose solutions are r max and r min
2.4
By the Third Kepler Law, the period T in the new orbit satisfies that
3 0
2 0 3
2
r
T a
T
=
where a, the semi-major axis of the ellipse, is given by
2 0
1
+
Therefore
−
2
3/
=
⎟
⎞
⎜
⎛
=
−
2
π
α=
min
vr
∆
0
vr
P
0
r
Figure 4
Trang 3Only 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 = 1
β
This can also be obtained by using that the total satellite energy has to be zero to reach infinity (E p = 0) without
residual velocity (E k = 0)
2
This also arises from T =∞ or from r max =∞
3.2
Due to ε=βesc =1, the polar parabola equation is
θ cos 1−
r
where the semi-latus-rectum continues to be l= The minimum Earth - satellite distance corresponds to r0 θ = , where π
2
0
r
r min′ =
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 0
2
1 1 2
1
∞
=
−
=
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
So
⇒
=
∞
v
v r
−
=r β
b
φ
v
∆
0
v
∞
v
Asymptote
Asymptote
b
Figure 5
asym
θ
asym
θ
asym
θ
0
r
P
Trang 44.3
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
⇒
=
⎠
⎞
⎜⎜
⎝
⎛
β
θasym cos 1 1 According to Figure 5
⇒ +
=π θasym φ
⎠
⎞
⎜⎜
⎝
⎛ +
β
π
2
1
For
2
3 2
3
=
= βesc
β , one gets φ=138º=2.41rad
Trang 5Th 1 ANSWER SHEET
Question Basic formulas and
ideas used
guideline
1.1
m 10 22
1.2
2 0
0 0
0
2 0 2
0
2
T T T
R
GM g
T
r v
r
v m r
m M G
=
=
=
π
0 0
r
g R
1.3
r
Mm G mv E
v r m L
−
=
×
=
2
2 1
r r r
0
2
0
v
mgR
L = T
2 0 0
2
1
mv
E =−
0.4
0.4
β
ε=
0.4
0.5
2.2
Hint on the conical curves
2
π
2.3
Results of 2.1, or
conservation of E and L
β
β
+
=
−
=
1
1
0
0
r r
r r
min
max
m 10 38 3
m 10 63 5
7 min
7 max
⋅
=
⋅
=
r
r
1.0 + 0.2
−
=T β
3.1 ε = 1, E = 0, T = ∞ or
3.2 ε = 1 and results of 2.1
2
0
r
=
−
=r β
⎠
⎞
⎜⎜
⎝
⎛ +
β
π
2