De thi vat li quoc te IPHO nam 20054

Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (84.46 KB, 5 trang )

(1)<div class='page_container' data-page=1>

Th1 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 
⇒
=
= 0
2
0
2
0
0

0 mv

v

R
g
v
m
r
L T
0
2
0 v
R
g
m

L = T

⇒
−
=
−
=
−
= 2
0
2
0
0
2
2
0
0

2
0
0 2
1
2
1
2
1
mv
mv
r
m
R
g
mv
r
m
M
G
mv

E T T 0 02

2
1

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
0
2
2
2
2
0
4
2
2
2
2

0 1 r

v
R
g
m
R
g

v
R
g
m
m
M
G
L
l T
T
T
T
0
r
l=

The eccentricity value is

3
2
2
2
0
2 2
1
m
M
G
L
E

T
+
=
ε

where E is the new satellite mechanical energy

(

)

0
2
0
2
0
2
2
0 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
1
2

1 2 2

0
2

0
2
2

0 ⎟⎟= −

⎠
⎞
⎜
⎜
⎝
⎛
−

= ∆ mv β

v
v
mv
E

Combining both, one gets ε=β

</div>
(2)<div class='page_container' data-page=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

(

)

⇒

−
=
=
=

β
α

cos
1

0
0

r
r

r

π
α =

2.3

From the trajectory expression one immediately obtains that
the maximum and minimum values of r correspond to θ =0 and

θ = respectively (see Figure 4). Hence, they are given by

ε
−
=

l

rmax ε

+
=

l
rmin

that is

β
−
=

0
max

r

r and = +β

0
min

r
r

Forβ =1/4, one gets

m
10
38
3
m;

10
63

5 ⋅ 7 = ⋅ 7

= . 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

=
=

−
=

−

0
2
0

2
2
2

0 2

1
1
2

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
0

0
3
2

r
T
a
T

where a, the semi-major axis of the ellipse, is given by

2
0

2 = −β

= r r r

a max min

Therefore

( )

2 3/2

−

−
=T β

T

15⎞ 3/2

⎛ −

π
α=
min

r rmax

vr

∆

0
vr P

0
r

</div>
(3)<div class='page_container' data-page=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 = 1

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 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

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
0

2
1

1
2

∞
=
−

= mv mv

E β ⇒

( )

2 1/2

0 −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 0 = ∞

⇒
=

∞

v
v
r

b 0 0

( )

2
/
1
2

0 1

−

=r β

b

v

∆

0
v

∞

v

Asymptote

b

Figure 5

asym

0
r

</div>
(4)<div class='page_container' data-page=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 0

1 β θasym ⎟⎟

⎠
⎞
⎜⎜
⎝
⎛

= −

β
θasym cos 1 1

According to Figure 5

⇒
+

=π θasym

2 ⎟⎟⎠

⎞
⎜⎜
⎝
⎛
+

= −

β
π

φ cos 1

2
1

For

2
3
2

3
=

= βesc

</div>
(5)<div class='page_container' data-page=5>

Th 1 ANSWER SHEET

Question Basic formulas and 
ideas used

Analytical results Numerical results Marking 
guideline 
1.1

m
10
22
.

4 7

0= ⋅

r 0.3

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

=
=

0
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
1

r
r
r

0
2
0

v
mgR

L = T

2
0
0

2
1

mv
E =−

0.4

2.1 l=r0

β
ε=

0.4
0.5

2.2

Hint on the conical curves

α= 1.0

2.3

Results of 2.1, or
conservation of E and L

β
β

+
=

−
=

1
1

0
0

r
r

min

max

m
10
38
.
3

m
10
63
.
5

7
min

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 = ∞ or

rmax = ∞

1
=
esc

β 0.5

3.2 ε = 1 and results of 2.1

2
0
r

rmin′ = 1.0

4.1 Conservation of E v∞ =v0

( )

β2−11/2 1.0

4.2 Conservation of L b=r0

( )

β2−1−1/2 1.0

4.3 Hint on the conical curves ⎟⎟

⎠
⎞
⎜⎜
⎝

⎛
+

= −

β
π

φ cos 1

</div>

De thi vat li quoc te IPHO nam 20054

<b>Th1 AN ILL FATED SATELLITE </b>

<b>SOLUTION </b>

(

)

(

)

(

)

( )

( )

(

)

(

)

( )

( )

( )

( )

( )

Tài liệu liên quan

Tài liệu bạn tìm kiếm đã sẵn sàng tải về