36th International Physics Olympiad. Salamanca (España) 2005
R.S.E.F.
Th1
AN ILL FATED SATELLITE
SOLUTION
1.1 and 1.2
G
MT m
r02
m








  







v 02
r0
2 r0
T0
v0 
GM T
g
RT2
 g RT2 T02
r0  
 4 2





1/ 3
g
r0
v 0  RT
 r0  4.22  10 7 m
 v 0  3.07  10 3 m/s
1.3
L0  r0 m v0 
E0 
g RT2
v02
m v0

L0 
m g RT2
v0
M m 1
g RT2 m 1 2
1 2
mv 0  G T  mv 02 
 mv 0  mv 02
2
r0
2
r0
2

E0  
1
mv 02
2
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
l
L20
G MT m2

m 2 g 2 RT4
1
v02
g RT2 m 2

g RT2
v02
 r0

l  r0
The eccentricity value is
 2  1
2 E L20
G 2 M T2 m 3
where E is the new satellite mechanical energy


E
M m 1
1
1
2
2
2
2 1
2
m v0   v  G T  m v  E0  m v  mv 0
2
r0
2
2
2
E
  v2
 1
1
mv 02  2  1  mv 02  2  1
 v
 2
2
 0

that is
Combining both, one gets


 
This is an elliptical trajectory because     1 .
Th 1 Solution
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36th International Physics Olympiad. Salamanca (España) 2005
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
r      r0 
r0
1   cos 



2

v0
2.3
P
From the trajectory expression one immediately obtains that

the maximum and minimum values of r correspond to   0 and
r0
   respectively (see Figure 4). Hence, they are given by
rmax
l

1 
rmin
rmin
l

1 
v


2
rmax
that is
rmax 
r0
1 
and
rmin 
r0
1 
Figure 4
For   1/ 4 , one gets
rmax  5.63  107 m; rmin  3.38  107 m
The distances rmax and rmin can also be obtained from mechanical energy and angular momentum conservation,


taking into account that r and v are orthogonal at apogee and at perigee
E


2
gR m
1
1
mv 2  2  1  mv 2  T
2 0
2
r
L0 
2
mgRT
 mvr
v0
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
T2
a3

T02
r03
where a, the semi-major axis of the ellipse, is given by
a
rmax  rmin
r0

2
1  2
Therefore

T  T0 1   2
For  = 1/4

3 / 2
 15 
T  T0  
 16 
3 / 2
 26 .4 h
Th 1 Solution
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36th International Physics Olympiad. Salamanca (España) 2005
R.S.E.F.
3.1
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  = 1
 
 esc  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)
E


1
mv 2  2  1  0
2 0 esc
 esc  1

This also arises from T   or from rmax   .
3.2
Due to    esc  1 , the polar parabola equation is
r
l
1 cos 
where the semi-latus-rectum continues to be l  r0 . The minimum Earth - satellite distance corresponds to    , where
 
rmin
r0
2
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 r and v are orthogonal.
4.1
If the satellite escapes to infinity with residual velocity v  , by energy conservation
E


1
1
mv 02  2  1  mv 2
2
2

Asymptote


v  v0  2  1
v0
1/ 2
4.2
As      esc  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
P
v
 asym
r0

 asym
 asym
m v 0 r0  m v  b
b
So
b  r0
v0
v



b  r0  2  1
1 / 2
Asymptote
Figure 5
Th 1 Solution
v
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36th International Physics Olympiad. Salamanca (España) 2005
R.S.E.F.
4.3
The angle between each asymptote and the hyperbola axis is that appearing in its polar equation in the limit r   .
This is the angle for which the equation denominator vanishes
1   cos asym  0

1


 asym  cos 1 
According to Figure 5

For  

2
  asym

3
3
 esc  , one gets
2
2


1
 cos 1  
2

  138º  2.41rad
Th 1 Solution
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36th International Physics Olympiad. Salamanca (España) 2005
R.S.E.F.
Th 1
Question
1.1
Basic formulas and
ideas used
G
MT m
r02
v0 
1.2
1.3
g
m
ANSWER SHEET
Analytical results
v 02
r0
Marking
guideline
r0  4.22 10 7 m
0.3
v0  3.07 10 3 m/s
0.3 + 0.1
2 r0
T0
GM T
RT2

 
L  m r v
1
Mm
E  mv 2  G
2
r
2.1
Hint on the conical curves
g
r0
v 0  RT
L0 
mgRT2
v0
E0  
0.4
1
mv 02
2
0.4
l  r0
0.4
 
0.5
2.2

Results of 2.1, or
2.3
Numerical results
conservation of E and L
rmin
2.4
Third Kepler's Law
3.1
 = 1, E = 0, T =  or
r0
1 
r
 0
1 
rmax 
2
rmax  5.63 10 7 m

T  T0 1   2

3 / 2
rmax = 
 = 1 and results of 2.1
 
rmin
4.1
Conservation of E
v  v0  2  1
4.2
Conservation of L
b  r0  2  1
1.0 + 0.2
T  26.4 h
0.5 + 0.2
 esc  1
0.5
4.3
Hint on the conical curves

1.0



1.0
rmin  3.38 10 7 m
r0
2
3.2



1/ 2
1.0
1 / 2
1
 cos 1  
2

1.0
  138º  2.41rad
1.0 + 0.2
Th 1 Solution
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