Ipho 2013 Theory 1 Solution 6h353c
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The Maribo Meteorite
T1
Solutions Top view
B
G Frame 161
D
B
Turned sideview, not to scale
Top view: Triangle MBC: | , so ( ) | | Then | | ( ) Triangle DBC: | | so ( | | Then | | (
,
, and .
.
1.3 ,and
,
, and
,
.
) )
, .
)
.
) (
|
|
,
Triangle EBC: | | so ( | | Then | | ( |
E
E
M
Triangle EDC: Maribo:| |
M
D
N
1.1
195 km
F
Frame 155 C
C
(
Horizontal distance traveled by
) )
( ) Side view: Triangle CDF: | | | | ( ) Triangle CEG: | | | | Thus vertical distance travelled by Maribo: | | | | Total distance travelled by Maribo from frame 155 to 161: | | √| | (| | | |) .
.
The speed of Maribo is
Newton's second law: By integration
yields (
)
.
.
1.2a Alternative solution: The average force on the meteoroid when the speed decreases from 0.7 ( ) . Using that the acceleto can be estimated to ( ) ration is approximately constant, , results in .
Page 1 of 3
The Maribo Meteorite 1.2b
(
[ ] [ ] [ ] [ ] [ so [ ] [ ] [ ] 1.3a Thus ,
1.3b
(
)
]
[ ] [ [ ]
] [ [ ] .
] [
,
)
] ,
, and
(
) and ( )
√
( )
and Rb Sr: ( ) ( ) ( Thus ) 1.4b equation of a straight line: ( )
Slope: )
.
0.4 0.3
( ) ( ) [ ( ), and dividing by ( )
(
(
( ) ( )]. we obtain the
0.7
( )
)
.
( )
and (
0.6
̅
( )
1.4c
. .
1.6 mm
1.4a Rb-Sr decay scheme:
So
0.3
.
)
From which (
T1
. 0.4
)
.
( )
Kepler's 3rd law on comet Encke and Earth, with the orbital semi-major axis of Encke 1.5
given by
(
). Thus
( )
.
For Earth around its rotation axis: Angular velocity
.
Moment of inertia Angular momentum 1.6a Asteroid:
.
. and angular momentum . is perpendicular to , so by conservation angular momentum: ( ) = .The axis tilt (so the North Pole moves ). At vertical impact
1.6b in rotation period is 1.6c
At tangential impact thus (
0.6
so ( (
) . Thus we obtain )
is parallel to ) (
(
(
so )
Page 2 of 3
)
, and since . The change .
)(
0.7
0.7
) and .
0.7
The Maribo Meteorite
Maximum impact speed (I) The velocity
arises from three contributions:
of the body at distance
√
(Earth orbit radius) from the Sun,
.
. 1.7 (II) The orbital velocity of the Earth, (III) Gravitational attraction from the Earth and kinetic energy seen from the Earth: ( ) ( ) . In conclusion:
T1
√(
)
1.6
.
Total
9.0
Page 3 of 3
T1
Solutions Top view
B
G Frame 161
D
B
Turned sideview, not to scale
Top view: Triangle MBC: | , so ( ) | | Then | | ( ) Triangle DBC: | | so ( | | Then | | (
,
, and .
.
1.3 ,and
,
, and
,
.
) )
, .
)
.
) (
|
|
,
Triangle EBC: | | so ( | | Then | | ( |
E
E
M
Triangle EDC: Maribo:| |
M
D
N
1.1
195 km
F
Frame 155 C
C
(
Horizontal distance traveled by
) )
( ) Side view: Triangle CDF: | | | | ( ) Triangle CEG: | | | | Thus vertical distance travelled by Maribo: | | | | Total distance travelled by Maribo from frame 155 to 161: | | √| | (| | | |) .
.
The speed of Maribo is
Newton's second law: By integration
yields (
)
.
.
1.2a Alternative solution: The average force on the meteoroid when the speed decreases from 0.7 ( ) . Using that the acceleto can be estimated to ( ) ration is approximately constant, , results in .
Page 1 of 3
The Maribo Meteorite 1.2b
(
[ ] [ ] [ ] [ ] [ so [ ] [ ] [ ] 1.3a Thus ,
1.3b
(
)
]
[ ] [ [ ]
] [ [ ] .
] [
,
)
] ,
, and
(
) and ( )
√
( )
and Rb Sr: ( ) ( ) ( Thus ) 1.4b equation of a straight line: ( )
Slope: )
.
0.4 0.3
( ) ( ) [ ( ), and dividing by ( )
(
(
( ) ( )]. we obtain the
0.7
( )
)
.
( )
and (
0.6
̅
( )
1.4c
. .
1.6 mm
1.4a Rb-Sr decay scheme:
So
0.3
.
)
From which (
T1
. 0.4
)
.
( )
Kepler's 3rd law on comet Encke and Earth, with the orbital semi-major axis of Encke 1.5
given by
(
). Thus
( )
.
For Earth around its rotation axis: Angular velocity
.
Moment of inertia Angular momentum 1.6a Asteroid:
.
. and angular momentum . is perpendicular to , so by conservation angular momentum: ( ) = .The axis tilt (so the North Pole moves ). At vertical impact
1.6b in rotation period is 1.6c
At tangential impact thus (
0.6
so ( (
) . Thus we obtain )
is parallel to ) (
(
(
so )
Page 2 of 3
)
, and since . The change .
)(
0.7
0.7
) and .
0.7
The Maribo Meteorite
Maximum impact speed (I) The velocity
arises from three contributions:
of the body at distance
√
(Earth orbit radius) from the Sun,
.
. 1.7 (II) The orbital velocity of the Earth, (III) Gravitational attraction from the Earth and kinetic energy seen from the Earth: ( ) ( ) . In conclusion:
T1
√(
)
1.6
.
Total
9.0
Page 3 of 3
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