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