andrei andryieuski, andrei lavrinenkoย ยท 2017. 3. 7.ย ยท maximum impact speed ๐ฃimpmax arises...
TRANSCRIPT
The Maribo Meteorite T1
Page 1 of 3
Solutions
1.1
Top view: Triangle MCB: |CM| = 195 km, โ MCB = 230ยฐ โ 180ยฐ = 50ยฐ, and โ MBC = 75ยฐ, so โ CMB = 180ยฐ โ 75ยฐ โ 50ยฐ = 55ยฐ. Then |CB| = |CM| sin(โ CMB)
sin(โ MBC) = 165.4 km. Triangle DCB: |CB| = 165.4 km, โ DCB = 215ยฐ โ 180ยฐ = 35ยฐ,and โ DBC = 75ยฐ, so โ CDB = 180ยฐ โ 75ยฐ โ 35ยฐ = 70ยฐ. Then |CD| = |CB| sin(โ DBC)
sin(โ CDB) = 170.0 km. Triangle ECB: |CB| = 165.4 km, โ ECB = 221ยฐ โ 180ยฐ = 41ยฐ, and โ EBC = 75ยฐ, so โ CEB = 180ยฐ โ 75ยฐ โ 41ยฐ = 64ยฐ. Then |CE| = |CB| sin(โ EBC)
sin(โ CEB) = 177.7 km. Triangle ECD:โ ECD = 41ยฐ โ 35ยฐ = 6ยฐ. Horisontal distance travelled by Maribo:|DE| = |DC| sin(โ ECD)
sin(โ CED) = 19.77 km Side view: Triangle CFD: |FD| = |CD| tan(โ FCD) = 59.20 km Triangle CGE: |GE| = |CE| tan(โ GCE) = 46.62 km Thus vertical distance travelled by Maribo: |FD| โ |GE| = 12.57 km. Total distance travelled by Maribo from frame 155 to 161: |FG| = ๏ฟฝ|DE|2 + (|FD| โ |GE|)2 = 23.43 km. The speed of Maribo is ๐ฃ = 23.43 km
2.28 sโ1.46 s= 28.6 km/s
1.2
1.2a Newton's second law: ๐M
d๐ฃd๐ก
= โ๐๐atm๐๐ M2 ๐ฃ2 yields 1๐ฃ2
d๐ฃ = โ๐๐atmฯ๐ M2
๐M d๐ก.
By integration ๐ก = ๐M๐๐atm๐๐ M
2 ๏ฟฝ 10.9โ 1๏ฟฝ 1
๐ฃM= 0.88 s.
0.7
1.2b ๐ธkin๐ธmelt
=12 ๐ฃM
2
๐sm(๐smโ๐0) + ๐ฟsm = 4.2ร108
2.1ร106= 2.1 ร 102 โซ 1. 0.3
The Maribo Meteorite T1
Page 2 of 3
1.3a
[๐ฅ] = [๐ก]๐ผ[๐sm]๐ฝ[๐sm]๐พ [๐sm]๐ฟ = [s]๐ผ[kg mโ3]๐ฝ[m2 sโ2Kโ1]๐พ [kg m sโ3Kโ1]๐ฟ, so [m] = [kg]๐ฝ+๐ฟ[m]โ3๐ฝ+2๐พ+๐ฟ[s]๐ผโ2๐พโ3๐ฟ[K]โ๐พโ๐ฟ. Thus ๐ฝ + ๐ฟ = 0, โ3๐ฝ + 2๐พ + ๐ฟ = 1, ๐ผ โ 2๐พ โ 3๐ฟ = 0, and โ๐พ โ ๐ฟ = 0.
From which (๐ผ,๐ฝ, ๐พ, ๐ฟ) = ๏ฟฝ+ 12
,โ12
,โ12
, + 12๏ฟฝ and ๐ฅ(๐ก) โ ๏ฟฝ ๐sm๐ก
๐sm๐sm.
0.6
1.3b ๐ฅ(5 s) = 1.6 mm ๐ฅ/๐ M = 1.6 mm/130 mm = 0.012. 0.4 1.4a Rb-Sr decay scheme: Rb37
87 โ Sr 3887 + eโ1
0 + ๏ฟฝฬ ๏ฟฝe 0.3
1.4b
๐87Rb(๐ก) = ๐87Rb(0)eโ๐๐กand RbโSr: ๐87Sr(๐ก) = ๐87Sr(0) + [๐87Rb(0) โ ๐87Rb(๐ก)]. Thus ๐87Sr(๐ก) = ๐87Sr(0) + (e๐๐ก โ 1)๐87Rb(๐ก), and dividing by ๐86Sr we obtain the equation of a straight line:
๐87Sr(๐ก)๐86Sr
= ๐87Sr(0)๐86Sr
+ (e๐๐ก โ 1) ๐87Rb(๐ก)๐86Sr
.
0.7
1.4c Slope: e๐๐ก โ 1 = ๐ = 0.712โ0.700
0.25= 0.050 and ๐ยฝ = ln(2)
๐= 4.9 ร 1010 year.
So ๐๐ = ln(1 + ๐) 1
๐= ln(1+๐)
ln(2) ๐ยฝ = 3.4 ร 109 year . 0.4
1.5 Kepler's 3rd law on comet Encke and Earth, with the orbital semi-major axis of Encke
given by ๐ = 12
(๐min + ๐max). Thus ๐กEncke = ๏ฟฝ ๐๐E๏ฟฝ32 ๐กE = 3.30 year = 1.04 ร 108 s.
0.6
1.6a
For Earth around its rotation axis: Angular velocity ๐E = 2๐24 h
= 7.27 ร 10โ5 sโ1.
Moment of inertia ๐ผE = 0.83 25๐E๐ E2 = 8.07 ร 1037 kg m2.
Angular momentum ๐ฟE = ๐ผE๐E = 5.87 ร 1033 kg m2sโ1. Astroid ๐ast = 4๐
3๐ ast3 ๐ast = 1.57 ร 1015 kg and angular momentum ๐ฟast =
๐ast๐ฃast๐ E = 2.51 ร 1026 kg m2sโ1. ๐ฟast is perpendicular to ๐ฟE, so by conservation angular momentum: tan (ฮ๐) = ๐ฟast/๐ฟE = 4.27 ร 10โ8. The axis tilt ฮ๐ = 4.27 ร 10โ8 rad (so the north pole move ๐ E ฮ๐ = 0.27 m).
0.7
1.6b At vertical impact ฮ๐ฟE = 0 so ฮ(๐ผE๐E) = 0. Thus ฮ๐E = โ๐E(ฮ๐ผE)/๐ผE, and since ฮ๐ผE/๐ผE = ๐ast๐ E2/๐ผE = 7.9 ร 10โ10 we obtain ฮ๐E = โ5.76 ร 10โ14 sโ1. The change in rotation period is ฮ๐E = 2๐ ๏ฟฝ 1
๐E+ฮ๐Eโ 1
๐E๏ฟฝ โ โ2๐ ฮ๐E
๐E2 = 6.84 ร 10โ5 s.
0.7
1.6c At tangential impact ๐ฟast is parallel to ๐ฟE so ๐ฟE + ๐ฟast = ( ๐ผE + ฮ๐ผE)(๐E + ฮ๐E) and thus ฮ๐E = 2๐ ๏ฟฝ 1
๐E+ฮ๐Eโ 1
๐E๏ฟฝ = 2๐ ๏ฟฝ ๐ผE+ฮ๐ผE
๐ฟE+๐ฟastโ 1
๐E๏ฟฝ = โ3.62 ร 10โ3 s. 0.7
The Maribo Meteorite T1
Page 3 of 3
1.7a Minimum impact speed is the escape velocity from Earth: ๐ฃimpmin = ๏ฟฝ2๐บ๐๐ธ๐ ๐ธ
= 11.2 km/s 0.5
1.7b
Maximum impact speed ๐ฃimpmax arises from three contributions: (I) The velocity ๐ฃb of the body at distance ๐E (Earth orbit radius) from the Sun,
๐ฃb = ๏ฟฝ2๐บ๐๐๐๐ธ
= 42.1 km/s.
(II) The orbital velocity of the Earth, ๐ฃE = 2๐๐๐ธ1 year
= 29.8 km/s. (III) Gravitational attraction from the Earth and kinetic energy seen from the Earth: 1
2(๐ฃb + ๐ฃE)2 = โ๐บ๐๐ธ
๐ ๐ธ+ 1
2๏ฟฝ๐ฃimpmax๏ฟฝ
2.
In conclusion: ๐ฃimpmax = ๏ฟฝ(๐ฃb + ๐ฃE)2 + 2๐บ๐๐ธ๐ ๐ธ
= 72.8 km/s.
1.2
Total 9.0
The Maribo Meteorite T1
Marking scheme
1.1
Top view: Triangle MCB: |CM| = 195 km, โ MCB = 230ยฐ โ 180ยฐ = 50ยฐ, and โ MBC = 75ยฐ, so โ CMB = 180ยฐ โ 75ยฐ โ 50ยฐ = 55ยฐ. Then |CB| = |CM| sin(โ CMB)
sin(โ MBC)= 165.4 km.
Triangle DCB: |CB| = 165.4 km, โ DCB = 215ยฐ โ 180ยฐ = 35ยฐ,and โ DBC = 75ยฐ, so โ CDB = 180ยฐ โ 75ยฐ โ 35ยฐ = 70ยฐ. Then |CD| = |CB| sin(โ DBC)
sin(โ CDB)= 170.0 km.
Triangle ECB: |CB| = 165.4 km, โ ECB = 221ยฐ โ 180ยฐ = 41ยฐ, and โ EBC = 75ยฐ, so โ CEB = 180ยฐ โ 75ยฐ โ 41ยฐ = 64ยฐ. Then |CE| = |CB| sin(โ EBC)
sin(โ CEB)= 177.7 km.
Triangle ECD:โ ECD = 41ยฐ โ 35ยฐ = 6ยฐ. Horisontal distance travelled by Maribo:|DE| = |DC| sin(โ ECD)
sin(โ CED)= 19.77 km
Side view: Triangle CFD: |FD| = |CD| tan(โ FCD) = 59.20 km Triangle CGE: |GE| = |CE| tan(โ GCE) = 46.62 km Thus vertical distance travelled by Maribo: |FD| โ |GE| = 12.57 km. Total distance travelled by Maribo from frame 155 to 161: |FG| = ๏ฟฝ|DE|2 + (|FD| โ |GE|)2 = 23.43 km. The speed of Maribo is ๐ฃ = 23.43 km
2.28 sโ1.46 s= 28.6 km/s
Marking: State ๐ฃ = ฮ๐
ฮ๐ก : 0.2 points
Correct drawing of all three triangles +0.4 points, Calculation of |๐ซ๐ฌ| +0.4 point Calculation of |๐ ๐| โ |๐๐| and speed +0.3 points. (-0.2 points for trivial trigonometric errors -0.1 points for more than 0.5 km/s deviation from correct result)
1.3
Page 1 of 4
The Maribo Meteorite T1
1.2a
Newton's second law: ๐M
d๐ฃd๐ก
= โ๐๐atm๐๐ M2 ๐ฃ2 yields 1๐ฃ2
d๐ฃ = โ๐๐atmฯ๐ M2
๐M d๐ก.
Solving by integration ๐ก = ๐M๐๐atm๐๐ M
2 ๏ฟฝ 10.9โ 1๏ฟฝ 1
๐ฃM= 0.90 s.
Marking: Newton's second law: ๐M
d๐ฃd๐ก
= โ๐๐atm๐๐ M2 ๐ฃ2 (0.2 points)
Solving by integration ๐ก = ๐M๐๐atm๐๐ M
2 ๏ฟฝ 10.9โ 1๏ฟฝ 1
๐ฃM= 0.90 s.
(+0.3 points for correct equation for t, +0.2 points for correct value) Alternative solution: Marking: Newton's second law: ๐M
d๐ฃd๐ก
= โ๐๐atm๐๐ M2 ๐ฃ2 (0.2 points) Approximately constant acceleration from 1.46 s to 2.28 s gives ๐ก = 0.79 s. (+0.3 points for correct equation for t, +0.2 points for correct value)
0.7
1.2b
๐ธkin๐ธmelt
=12 ๐ฃM
2
๐sm(๐smโ๐0) + ๐ฟsm = 4.2ร108
2.1ร106= 2.1 ร 102 โซ 1.
Marking: 0.1 point for each energy term; (-0.1 if ๐ฟ๐ ๐ dropped without argument)
0.3
1.3a
[๐ฅ] = [๐ก]๐ผ[๐sm]๐ฝ[๐sm]๐พ [๐sm]๐ฟ = [s]๐ผ[kg mโ3]๐ฝ[m2 sโ2Kโ1]๐พ [kg m sโ3Kโ1]๐ฟ, so [m] = [kg]๐ฝ+๐ฟ[m]โ3๐ฝ+2๐พ+๐ฟ[s]๐ผโ2๐พโ3๐ฟ[K]โ๐พโ๐ฟ. Thus ๐ฝ + ๐ฟ = 0, โ3๐ฝ + 2๐พ + ๐ฟ = 1, ๐ผ โ 2๐พ โ 3๐ฟ = 0, and โ๐พ โ ๐ฟ = 0.
From which (๐ผ,๐ฝ, ๐พ, ๐ฟ) = ๏ฟฝ+ 12
,โ12
,โ12
, + 12๏ฟฝ and ๐ฅ(๐ก) โ ๏ฟฝ ๐sm๐ก
๐sm๐sm.
Marking: 0.1 points for each dimensional equation +0.2 points for solving equations correctly.
0.6
1.3b ๐ฅ(5 s) = 1.6 mm ๐ฅ/๐ M = 1.6 mm/130 mm = 0.012. Marking: Calculation of x: 0.3 points, calculation of ๐ฅ/๐ M +0.1 points.
0.4
Page 2 of 4
The Maribo Meteorite T1
1.4a
Rb-Sr decay scheme: Rb3787 โ Sr 38
87 + eโ10 + ๏ฟฝฬ ๏ฟฝe
Marking: Correct scheme with both electron and antineutrino (neutrino also accepted) 0.3 points. (Missing electron or other error concerning electron: no points awarded for this question. Missing neutrino: -0.1 points)
0.3
1.4b
๐87Rb(๐ก) = ๐87Rb(0)eโ๐๐ก and RbโSr: ๐87Sr(๐ก) = ๐87Sr(0) + [๐87Rb(0) โ๐87Rb(๐ก)] Thus ๐87Sr(๐ก) = ๐87Sr(0) + (e๐๐ก โ 1)๐87Rb(๐ก), and dividing by ๐86Sr we obtain ๐87Sr(๐ก)๐86Sr
= ๐87Sr(0)๐86Sr
+ (e๐๐ก โ 1) ๐87Rb(๐ก)๐86Sr
Marking: ๐87Rb(๐ก) = ๐87Rb(0)eโ๐๐ก (0.2 points) RbโSr: ๐87Sr(๐ก) = ๐87Sr(0) + [๐87Rb(0) โ ๐87Rb(๐ก)] (+0.2 points) Obtain ๐87Sr(๐ก)๐86Sr
= ๐87Sr(0)๐86Sr
+ (e๐๐ก โ 1) ๐87Rb(๐ก)๐86Sr
(+0.3 points) (Wrong slope: -0.2 points, wrong intersection: -0.1)
0.7
1.4c
Slope: e๐๐ก โ 1 = ๐ = 0.712โ0.7000.25
= 0.050 and ๐ยฝ = ln(2)๐
= 4.9 ร 1010 year. So ๐๐ = ln(1 + ๐) 1
๐= ln(1+๐)
ln(2)๐ยฝ = 3.4 ร 109 year .
Marking: Slope (3.4 ยฑ ๐.๐ ร 109 year) 0.1 point, calculation of ๐๐ +0.3 points.
0.4
1.5
Kepler's 3rd law on comet Encke and Earth, with the orbital semi-major axis of Encke
given by ๐ = 12
(๐min + ๐max). Thus ๐กEncke = ๏ฟฝ ๐๐E๏ฟฝ32 ๐กE = 3.30 year = 1.04 ร 108 s.
Marking: Correct expression for a: 0.2 points, apply Keplers third law correctly: +0.2 points, calculation of orbital period +0.2 points.
0.6
1.6a
For Earth around its rotation axis: Angular velocity ๐E = 2๐24 h
= 7.27 ร 10โ5 sโ1.
Moment of inertia ๐ผE = 0.83 25๐E๐ E2 = 8.07 ร 1037 kg m2.
Angular momentum ๐ฟE = ๐ผE๐E = 5.87 ร 1033 kg m2sโ1. Astroid ๐ast = 4๐
3๐ ast3 ๐ast = 1.57 ร 1015 kg and angular momentum ๐ฟast =
๐ast๐ฃast๐ E = 2.51 ร 1026 kg m2sโ1. ๐ฟast is perpendicular to ๐ฟE, so by conservation angular momentum: tan (ฮ๐) = ๐ฟast/๐ฟE = 4.27 ร 10โ8. The axis tilt ฮ๐ = 4.27 ร 10โ8 rad (so the north pole move ๐ E ฮ๐ = 0.27 m). Marking: Conservation of angular momentum with๐ฟE = ๐ผE๐E and ๐ฟast =๐ast๐ฃast:0.3 point,. Realize geometry: +0.2 points, Calculation: +0.2 points.
0.7
Page 3 of 4
The Maribo Meteorite T1
1.6b
At vertical impact ฮ๐ฟE = 0 so ฮ(๐ผE๐E) = 0. Thus ฮ๐E = โ๐E(ฮ๐ผE)/๐ผE, and since ฮ๐ผE/๐ผE = ๐ast๐ E2/๐ผE = 7.9 ร 10โ10 we obtain ฮ๐E = โ5.76 ร 10โ14 sโ1. The change in rotation period is ฮ๐E = 2๐ ๏ฟฝ 1
๐E+ฮ๐Eโ 1
๐E๏ฟฝ โ โ2๐ ฮ๐E
๐E2 = 6.84 ร 10โ5 s.
Marking: Conservation of angular momentum ฮ(๐ผE๐E) = 0: 0.4 points, expression for ฮ๐E: +0.2 points, calculation: +0.1 points. (Low precision: -0.1 points, wrong ๐ผ๐ธ + ๐ผ๐๐ ๐ก: -0.2 points)
0.7
1.6c
At tangential impact ๐ฟast is parallel to ๐ฟE so ๐ฟE + ๐ฟast = ( ๐ผE + ฮ๐ผE)(๐E + ฮ๐E) and thus ฮ๐E = 2๐ ๏ฟฝ 1
๐E+ฮ๐Eโ 1
๐E๏ฟฝ = 2๐ ๏ฟฝ ๐ผE+ฮ๐ผE
๐ฟE+๐ฟastโ 1
๐E๏ฟฝ = โ3.62 ร 10โ3 s.
Marking: Conservation of angular momentum ๐ฟE + ๐ฟast = ( ๐ผE + ฮ๐ผE)(๐E + ฮ๐E) 0.4 points, solving for ฮ๐E and calculating value: 0.3 points. [Both impact directions will be accepted] (Low precision: -0.1 points).
0.7
1.7
Maximum impact speed ๐ฃimpmax arises from three contributions: (I) The velocity ๐ฃb of the body at distance ๐E (Earth orbit radius) from the Sun,
๐ฃb = ๏ฟฝ2๐บ๐๐๐๐ธ
= 42.1 km/s.
(II) The orbital velocity of the Earth, ๐ฃE = 2๐๐๐ธ1 year
= 29.8 km/s. (III) Gravitational attraction from the Earth and kinetic energy seen from the Earth: 1
2(๐ฃb + ๐ฃE)2 = โ๐บ๐๐ธ
๐ ๐ธ+ 1
2๏ฟฝ๐ฃimpmax๏ฟฝ2.
In conclusion: ๐ฃimpmax = ๏ฟฝ(๐ฃb + ๐ฃE)2 + 2๐บ๐๐๐๐ธ
= 72.8 km/s.
Marking: ๐ฃb = ๏ฟฝ2๐บ๐๐๐๐ธ
= 42.1 km/s: 0.6 points, ๐ฃE = 2๐๐๐ธ1 year
= 29.8 km/s : 0.4 points, 12
(๐ฃb + ๐ฃE)2 = โ๐บ๐๐ธ๐ ๐ธ
+ 12๏ฟฝ๐ฃimpmax๏ฟฝ2: 0.6 points.
(Alternatively using energy conservation of asteroid with static Sun and Earth, to find speed of asteroid at ๐๐ธ and then adding ๐ฃ๐ธ : 1.5 points).
1.6
Total 9.0
Page 4 of 4
Plasmonic Steam Generator T2
Page 1 of 2
Solutions
2.1
Volume: ๐ = 43๐๐ 3 = 4.19 ร 10โ24 m3. Mass of ions: ๐ = ๐ ๐Ag = 4.39 ร 10โ20 kg
no. of ions: ๐ = ๐๐ด๐๐Ag
= 2.45 ร 105. Charge density ๐ = ๐๐๐
= 9.38 ร 109 C mโ3
Electron concentration ๐ = ๐๐
= 5.85 ร 1028 mโ3, charge ๐ = ๐๐ = 3.93 ร 10โ14 C, and mass ๐0 = ๐๐๐ = 2.23 ร 10โ25 kg.
0.7
2.2
For a sphere with radius ๐ and constant charge density, Gauss's law yields directly 4๐๐2๐0๐ฌ+ = 4
3๐๐3๐ ๐๐ , for ๐ < ๐ with ๐๐ being the unit radial vector pointing away
from the center of the sphere. Thus, ๐ฌ+ = ๐3๐0
๐. Likewise, inside a small sphere of radi-
us ๐ 1 and charge density โ๐ centered at ๐p the field is ๐ฌโ = โ๐3๐0
(๐ โ ๐p). Adding the two charge configurations gives the setup we want and inside the charge-free region ๏ฟฝ๐ โ ๐p๏ฟฝ < ๐ 1 the field is ๐ฌ = ๐ฌ+ + ๐ฌโ = ๐
3๐0๐p with the pre-factor ๐ด = 1
3.
1.0
2.3
With ๐p = ๐ฅp ๐๐ฅ and ๐ฅp โช ๐ we have from above ๐ฌind = ๐3๐0
๐ฅp ๐๐ฅ inside the particle. The number of electrons that produced ๐ฌind is negligibly smaller than the number of electrons inside the particle, so ๐ญ = ๐๐ฌind = (โ๐๐) ๐
3๐0๐ฅp ๐๐ฅ = โ 4๐
9๐0๐ 3๐2๐2 ๐๐.
The work done by ๐ญext on the electrons is ๐el = โโซ ๐น(๐ฅโฒ) d๐ฅโฒ =๐ฅp0
12
๏ฟฝ4๐9๐0
๐ 3๐2๐2๏ฟฝ ๐ฅp2.
0.9
2.4 The E-field is zero inside the particle, so ๐ฌ0 + ๐ฌind = 0, so ๐ฅp = 3๐0
๐๐ธ0 = 3๐0
๐๐๐ธ0.
Charge displaced through the ๐ฆ๐ง-plane: โฮ๐ = โ๐ ๐๐ 2๐ฅp = โ๐๐ 2 ๐๐ ๐ฅp. 0.5
2.5a The electric energy ๐el of a capacitor with capacitance ๐ถ holding a charge ยฑฮ๐ is ๐el = ฮ๐2
2๐ถ , so from (3c) and (4b) we obtain ๐ถ = 9
4๐0๐๐ = 6.26 ร 10โ19 F.
0.5
2.5b Combining (4b) and (5a) leads ๐0 = ฮ๐๐ถ
= 43๐ ๐ธ0. 0.3
2.6a ๐kin = 12๐๐๐ฃ2๐ = 1
2๐๐๐ฃ2 ๏ฟฝ
43๐๐ 3 ๐๏ฟฝ. The area ๐๐ 2 leads to ๐ผ = โ๐ ๐๐ฃ ๐๐ 2. 0.5
2.6b From (6a) and ๐kin = 12๐ฟ ๐ผ2 we obtain ๐ฟ = 4 ๐๐
3๐๐ ๐๐2= 2.57 ร 10โ14 H. 0.5
2.7a ๐p2 = (๐ฟ๐ถ)โ1 = ๐๐2(3๐0๐๐)โ1. 0.5
2.7b ๐p = 7.88 ร 1015 rad/s, ๐p = 2๐๐/๐p = 239 nm. 0.3
Plasmonic Steam Generator T2
Page 2 of 2
2.8a โจ๐heatโฉ = 1๐๐kin = 1
2๐๐๐โจ๐ฃ2โฉ ๏ฟฝ
43๐๐ 3 ๐๏ฟฝ. โจ๐ผ2โฉ = (๐๐ ๐๐ 2)2 โจ๐ฃ2โฉ = ๏ฟฝ3๐
4๐ ๏ฟฝ2โจ๐ฃ2โฉ . 1.0
2.8b ๐ heat = โจ๐heatโฉโจ๐ผ2โฉ
leads to ๐ heat = ๐kin๐๐ผ2
= 2๐๐3๐๐๐2๐ ๐
= 2.46 ฮฉ. 0.8
2.9a [๐scat] = J sโ1, [๐ ๐ฅ0] = C m, ๏ฟฝ๐p๏ฟฝ = sโ1, [๐] = m sโ1, [๐0] = C2(J m)โ1. Note that
๏ฟฝ๐2๐ฅ02
๐0๏ฟฝ = J m3 so ๏ฟฝ๐
2๐ฅ02
๐0
๐p4
๐3๏ฟฝ = J sโ1. Thus (๐ผ,๐ฝ, ๐พ, ๐ฟ) = (2,4,-3,-1) and ๐scat = ๐2๐ฅ02๐p
4
12๐๐0๐3.
0.7
2.9b ๐ scat = โจ๐scatโฉโจ๐ผ2โฉ and โจ๐ฃ2โฉ = 1
2๐p2๐ฅ02 yields ๐ scat = ๐2๐ฅ02๐p
4
12๐๐0๐316๐ 2
9๐2โจ๐ฃ2โฉ= 8๐0
2๐ 2
27๐๐0๐3= 2.45 ฮฉ. 0.9
2.10a
At resonance the reactance of an LCR-circuit is zero: ๐๐ฟ + ๐๐ถ = 0. The voltage-current relation then becomes: โจ๐2โฉ = ๐๐ 2โจ๐ผ2โฉ = (๐ heat + ๐ scat)2โจ๐ผ2โฉ. From (2.5b) we have โจ๐2โฉ = 1
2๐02 = 8
9๐ 2๐ธ02, such that โจ๐ผ2โฉ = 8๐ 2๐ธ02
9(๐ heat+๐ scat)2.
This leads to โจ๐heatโฉ = ๐ heatโจ๐ผ2โฉ = 8๐ heat๐ 2
9(๐ heat+๐ scat)2๐ธ02 and โจ๐scatโฉ = ๐ scat
๐ heatโจ๐heatโฉ.
1.2
2.10b ๐ธ0 = ๏ฟฝ2๐/(๐0๐) = 27.4 kV/m. โจ๐heatโฉ = 6.82 nW. โจ๐scatโฉ = 6.81 nW. 0.3
2.11 Time per oscillation ๐กp = 2๐/๐p. Emitted energy per time = โจ๐scatโฉ. Energy per photon
๐ธp = โ๐p. Numbers of oscillation per photon ๐osc = โ๐p
๐กpโจ๐scatโฉ= โ๐p
2
2๐โจ๐scatโฉ= 1.53 ร 105. 0.6
2.12a
Total number of nanoparticles: ๐np = โ2๐ ๐np = 7.3 ร 1011. Total power into steam from Joule heating: ๐st = ๐np๐heat = 4.98 kW. In terms of heating up a mass of steam per second ๐st: ๐st = ๐st๐ฟtot, with ๐ฟtot = ๐wa(๐100 โ ๐wa) + ๐ฟwa + ๐st(๐st โ ๐100) = 2.62 ร 106 J kgโ1. Thus ๐st = ๐st
๐ฟtot= 1.90 ร 10โ3 kg sโ1.
0.6
2.12b ๐tot = โ2๐ = 0.01m2 ร 1 MW mโ2 = 10.0 kW, and thus ๐ = ๐st๐tot
= 4.98 kW10.0 kW
= 0.498. 0.2
Total 12.0
Plasmonic Steam Generator T2
Marking scheme
General issues
โข Number of significant digits: 3, no punishment if 2 or 4, punishment -0.05 if 1 or >4. โข Rounding: no punishment (for example, students result 2.34, correct result 2.47): no punishment if
there is small (+/-1) difference in the second from the left significant digit, provided that the formula is correct.
โข Wrong sign: depends on the assignment (no punishment for charge or current, but punishment for the wrong energy: -0.1 pt)
โข Cumulative mistakes: o we punish only one time, except for the case when the result is physically wrong (for example,
efficiency > 100%) or the dimensions are wrong (m=s^2 ~30-50%), โข Calculations: o If the pre-factor in the formula is wrong, but the value according to this formula is calculated
correct, the student receives points for the value, o if the formula is incorrect (dimensions), no points for the value. โข Forgotten units: o no punishment, if there are no units on the answer sheet, but there are units on the draft papers, o punishment deduct half the points for this point, if there are no units anywhere. โข Wrong fomula: o wrong dimensions (m=s^2): deduct 50% of the value. o wrong pre-factor: -0.1 โข We round up the points for each section to higher value in steps of 0.1 (eg. 12.05 -> 12.1).
2.1
0.1
0.1
0.1
0.1
0.1
No punishment for the sign
0.1
0.1 โข Correct 0.1, โข punishment for wrong value or no units -0.05 โข round up to the higher value
Page 1 of 4
Plasmonic Steam Generator T2
2.2
Check derivation in the papers Electric field in side a homogeneous sphere: 0.5 (just Gauss law 0.2) Principle of superposition:0.4 Vector calculations:0.2 Correct A: 0.1
1.2
2.3
F=Q Eind: 0.4 pt If F = ยฝ Q Eind: 0.2 pt Correct calculation: +0.1 pt
0.5
If integral: 0.4 If maxforce times distance: 0.2 Calculation of expression +0.1
0.5
punishment for the wrong sign 0.1 only for the negative work, not force 2.4 ๐ฅ๐ = 3๐0๐ธ0
๐๐
Mentioned that electric field in equilibrium is 0: 0.1 pt 0.3
No punishment for sign
0.3
2.5 a
Formula for the energy of the capacitor W=q^2/2C: 0.2 Try to estimate as a flat capacitor: 0.3 Just applying the formula for the capacitance of a single sphere: 0 Dimensional analysis: 0
0.6
a
0.1
b
Estimation of the voltage as 2ER: 0.2 pt
0.4
Page 2 of 4
Plasmonic Steam Generator T2
2.6 a
Punishment for forgetting N: -0.2 pt
0.4
a
No punishment for the sign 0.3
b
Energy of the inductor: 0.2 Dimensional analysis: 0
0.4
b
0.1
2.7 a
LC-circuit frequency: 0.3 Dimensional analysis 0.2 Alternative ways
0.5
b
No punishment for Hz 0.1
b
Correct formula 0.2 No punishment for m
0.3
2.8 a
โจ๐โ๐๐๐กโฉ =
โจ๐๐๐๐โฉ๐
=1
2๐๐๐ โฆ.
Formula for power as mean kinetic energy divided over tau: 0.4 If calculated in the assumption of constant acceleration: -0.2
0.5
a
0.5
b
Formula P=R<I^2>: 0.3
0.8
b
0.2
Page 3 of 4
Plasmonic Steam Generator T2
2.9
If they made <v^2> correct 0.4 pt If <v^2>=x0^2 omega^2: -0.2 P=I^2Rsc gives 0.3 pt
0.8
0.2
2.10 a
0.9 (0.3)
a
If at least one correct 0.9 If bulk expression with L and C: deduce 0.1 pt Vector diagram or differential equation: 0.3 Drawing equivalent RLC circuit: 0.2 If the student connect resistances in parallel: 0
0.3 (0.9)
b
0.1
b 0.1
b
0.1
2.11
Correct formula 0.4
0.6
If effficiency is calculated through the powers in single particle: 0
0.2
Page 4 of 4
The Greenlandic Ice Sheet T3
Page 1 of 4
Solutions
3.1 The pressure is given by the hydrostatic pressure ๐(๐ฅ, ๐ง) = ๐ice๐(๐ป(๐ฅ) โ ๐ง), which is zero at the surface. 0.3
3.2a
The outward force on a vertical slice at a distance ๐ฅ from the middle and of a given width โ๐ฆ is obtained by integrating up the pressure times the area:
๐น(๐ฅ) = โ๐ฆ๏ฟฝ ๐ice ๐ (๐ป(๐ฅ) โ ๐ง) d๐ง ๐ป(๐ฅ)
0=
12
โ๐ฆ ๐ice ๐ ๐ป(๐ฅ)2
which implies that โ๐น = ๐น(๐ฅ) โ ๐น(๐ฅ + โ๐ฅ) = โ d๐นd๐ฅโ๐ฅ = โโ๐ฆ ๐ice ๐ ๐ป(๐ฅ) d๐ป
d๐ฅโ๐ฅ.
This finally shows that
๐b =๐ฅ๐นโ๐ฅโ๐ฆ
= โ ๐ice ๐ ๐ป(๐ฅ)d๐ปd๐ฅ
Notice the sign, which must be like this, since ๐๐ was defined as positive and ๐ป(๐ฅ) is a decreasing function of ๐ฅ.
0.9
3.2b
To find the height profile, we solve the differential equation for ๐ป(๐ฅ):
โ๐b
๐ice ๐= ๐ป(๐ฅ)
d๐ปd๐ฅ
=12
dd๐ฅ
๐ป(๐ฅ)2
with the boundary condition that ๐ป(๐ฟ) = 0. This gives the solution:
๐ป(๐ฅ) = ๏ฟฝ2๐๐๐ฟ๐ice ๐๏ฟฝ
1 โ ๐ฅ/๐ฟ
Which gives the maximum height ๐ปm = ๏ฟฝ 2๐๐๐ฟ๐ice ๐
.
Alternatively, dimensional analysis could be used in the following manner. First notice that โ = [๐ปm] = ๏ฟฝ๐ice๐ผ ๐๐ฝ๐b
๐พ๐ฟ๐ฟ ๏ฟฝ. Using that ๏ฟฝ๐๐ice๏ฟฝ = โณโโ3, [๐] = โ ๐ฏโ2, [๐๐] =โณโโ1 ๐ฏโ2, demands that โ = [๐ปm] = ๏ฟฝ๐๐๐ผ๐๐ฝ๐๐๐พ๐ฟ๐ฟ๏ฟฝ = โณ๐ผ+๐พโโ3๐ผ+๐ฝโ๐พ+๐ฟ ๐ฏโ2๐ฝโ2๐พ, which again implies ๐ผ + ๐พ = 0, โ3๐ผ + ๐ฝ โ ๐พ + ๐ฟ = 1, 2๐ฝ + 2๐พ = 0. These three equations are solved to give ๐ผ = ๐ฝ = โ๐พ = ๐ฟ โ 1, which shows that
๐ปm โ ๏ฟฝ๐b
๐๐ice๐๏ฟฝ๐พ
๐ฟ1โ๐พ
Since we were informed that ๐ปm โ โ๐ฟ , it follows that ๐พ = 1/2. With the boundary condition ๐ป(๐ฟ) = 0, the solution then take the form
๐ป(๐ฅ) โ ๏ฟฝ๐b
๐ice ๐๏ฟฝ1/2
โ๐ฟ โ ๐ฅ
The proportionality constant of โ2 cannot be determined in this approach.
0.8
The Greenlandic Ice Sheet T3
Page 2 of 4
3.2c
For the rectangular Greenland model, the area is equal to ๐ด = 10๐ฟ2 and the volume is found by integrating up the height profile found in problem 3.2b:
๐G,ice = (5๐ฟ)2โซ ๐ป(๐ฅ) d๐ฅ๐ฟ0 = 10๐ฟ โซ ๏ฟฝ ๐b ๐ฟ
๐ice ๐๏ฟฝ1/2
๏ฟฝ1 โ ๐ฅ/๐ฟ d๐ฅ๐ฟ0 = 10๐ปm๐ฟ2 โซ โ1 โ ๐ฅ๏ฟฝ d๐ฅ๏ฟฝ1
0
= 10๐ปm๐ฟ2 ๏ฟฝโ23
(1 โ ๐ฅ๏ฟฝ)3/2๏ฟฝ0
1 = 20
3๐ปm๐ฟ2 โ ๐ฟ5/2,
where the last line follows from the fact that ๐ปm โ โ๐ฟ. Note that the integral need not be carried out to find the scaling with ๐ฟ. This implies that ๐G,ice โ ๐ด๐บ5/4 and the wanted exponent is ๐พ = 5/4.
0.5
3.3
According to the assumption of constant accumulation c the total mass accumulation rate from an area of width โ๐ฆ between the ice divide at ๐ฅ = 0 and some point at ๐ฅ > 0 must equal the total mass flux through the corresponding vertical cross section at ๐ฅ. That is: ๐๐๐ฅโ๐ฆ = ๐โ๐ฆ๐ปm๐ฃ๐ฅ(๐ฅ), from which the velocity is isolated:
๐ฃ๐ฅ(๐ฅ) =๐๐ฅ๐ปm
0.6
3.4
From the given relation of incompressibility it follows that d๐ฃ๐งd๐ง
= โd๐ฃ๐ฅd๐ฅ
= โ๐๐ปm
Solving this differential equation with the initial condition ๐ฃ๐ง(0) = 0, shows that: ๐ฃ๐ง(๐ง) = โ
๐๐ง๐ปm
0.6
3.5
Solving the two differential equations d๐งd๐ก
= โ๐๐ง๐ปm
and d๐ฅd๐ก
=๐๐ฅ๐ปm
with the initial conditions that ๐ง(0) = ๐ปm, and ๐ฅ(0) = ๐ฅ๐ gives ๐ง(๐ก) = ๐ปm eโ๐๐ก/๐ปm and ๐ฅ(๐ก) = ๐ฅ๐ e๐๐ก/๐ปm
This shows that ๐ง = ๐ปm ๐ฅ๐ /๐ฅ, meaning that flow lines are hyperbolas in the ๐ฅ๐ง-plane. Rather than solving the differential equations, one can also use them to show that
dd๐ก
(๐ฅ๐ง) =d๐ฅd๐ก๐ง + ๐ฅ
d๐งd๐ก
=๐๐ฅ๐ปm
๐ง โ ๐ฅ๐๐ง๐ปm
= 0
which again implies that ๐ฅ๐ง = const. Fixing the constant by the initial conditions, again leads to the result that ๐ง = ๐ปm๐ฅ๐/๐ฅ.
0.9
3.6 At the ice divide, ๐ฅ = 0, the flow will be completely vertical, and the ๐ก-dependence of ๐ง found in 3.5 can be inverted to find ๐(๐ง). One finds that ๐(๐ง) = ๐ปm
๐ln ๏ฟฝ๐ปm
๐ง๏ฟฝ. 1.0
The Greenlandic Ice Sheet T3
Page 3 of 4
3.7a
The present interglacial period extends to a depth of 1492 m, corresponding to 11,700 year. Using the formula for ๐(๐ง)from problem 3.6, one finds the following accumulation rate for the interglacial:
๐ig =๐ปm
11,700 yearsln ๏ฟฝ
๐ปm๐ปm โ 1492 m
๏ฟฝ = 0.17 m/year.
The beginning of the ice age 120,000 years ago is identified as the drop in ๐ฟ O 18 in figure 3.2b at a depth of 3040 m. Using the vertical flow velocity found in problem 3.4, on has d๐ง
๐ง= โ ๐
๐ปmd๐ก, which can be integrated down to a depth of 3040 m, using a
stepwise constant accumulation rate:
๐ปm ln ๏ฟฝ๐ปm
๐ปm โ 3040 m๏ฟฝ = โ๐ปm๏ฟฝ
1๐ง
๐ปmโ3040 m
๐ปmd๐ง
= ๏ฟฝ ๐ia d๐ก120,000 year
11,700 year+ ๏ฟฝ ๐ig d๐ก
11,700 year
0
= ๐ia(120,000 year-11,700 year)+๐ig11,700 year
Isolating form this equation leads to ๐ia = 0.12, i.e. far less precipitation than now.
0.8
3.7b Reading off from figure 3.2b: ๐ฟ O
18 changes from โ43,5 โฐ to โ34,5 โฐ. Reading off from figure 3.2a, ๐ then changes from โ40 โ to โ28 โ. This gives โ๐ โ 12 โ.
0.2
3.8a
From the area ๐ดG one finds that ๐ฟ = ๏ฟฝ๐ดG/10 = 4.14 ร 105 m. Inserting numbers in the volume formula found in 3.2c, one finds that:
๐G,ice =203๐ฟ5/2๏ฟฝ
2๐b๐ice๐
= 3.46 ร 1015 m3
This ice volume must be converted to liquid water volume, by equating the total masses, i.e. ๐G,wa = ๐G,ice
๐ice๐wa
= 3.17 ร 1015 m3, which is finally converted to a sea level rise,
as โG,rise = ๐G,wa๐ดo
= 8.78 m.
0.6
3.8b
From the area and aspect ratio of Antarctica and the volume to area scaling law found in problem 3.2c, it follows that
โA,rise
โG,rise=๐A,wa
๐G,wa=
25๏ฟฝ๐ฟA๐ฟG๏ฟฝ5/2
=25๏ฟฝ
52๐ดA๐ดG๏ฟฝ5/4
= ๏ฟฝ52๏ฟฝ1/4
๏ฟฝ๐ดA๐ดG๏ฟฝ5/4
which gives a sea level rise of โA,rise = 127 m.
0.2
3.9
From solving problem 3.8a, the total volume of the ice is known. From this number, we first deduce a radius of the spherical model of the Greenlandic ice sheet:
๐ ice = ๏ฟฝ3 ๐G,ice
4๐๏ฟฝ1/3
= ๏ฟฝ3 ร 3.46 ร 1015
4๐๏ฟฝ1/3
m = 93.8 km
This is an order of magnitude larger than the average sea depth of roughly 3 km, and therefore it doesnโt matter much if we locate the sphere at the mean Earth radius, rather
1.6
The Greenlandic Ice Sheet T3
Page 4 of 4
Figure 3.S1 Geometry of the ice ball (white circle) with a test mass ๐ (small gray circle).
than sea level. The total mass of the ice is
๐ice = ๐G,ice ๐ice = 3.17 ร 1018 kg = 5.31 ร 10โ7๐E
The total gravitational potential felt by a test mass ๐ at a certain height โ above the surface of the Earth, and at a polar angle ๐ (cf. figure 3.S1), with respect to a rotated polar axis going straight through the ice sphere is found by adding that from the Earth with that from the ice:
๐tot = โ๐บ๐E๐๐ E + โ
โ๐บ๐ice๐
๐= โ๐๐๐ ๐ธ ๏ฟฝ
11 + โ/๐ ๐ธ
+๐๐๐๐/๐๐ธ
๐/๐ ๐ธ๏ฟฝ
where ๐ = ๐บ๐๐ธ/๐ ๐ธ2. Since โ/๐ E โช 1 one may use the approximation given in the problem, (1 + x)โ1 โ 1 โ ๐ฅ, |๐ฅ| โช 1, to approximate this by
๐tot โ โ๐๐๐ ๐ธ ๏ฟฝ1 โโ๐ ๐ธ
+๐๐๐๐/๐๐ธ
๐/๐ ๐ธ๏ฟฝ.
Isolating โ now shows that โ = โ0 + ๐๐๐๐/๐๐ธ๐/๐ ๐ธ
๐ ๐ธ, where โ0 = ๐ ๐ธ + ๐tot/(๐๐). Using again that โ/๐ E โช 1, trigonometry shows that ๐ โ 2๐ E|sin(๐/2)|, and one has:
โ(๐) โ โ0 โ๐ice/๐E
2|sin(๐/2)|๐ ๐ธ โ1.69 m
|sin(๐/2)|.
To find the magnitude of the effect in Copenhagen, the distance of 3500 km along the surface is used to find the angle ๐CPH = (3.5 ร 106 m)/๐ ๐ธ โ 0.55, corresponding to โCPH โ โ0 โ 6.2 m. Directly opposite to Greenland corresponds to ๐ = ๐, which gives โOPP โ โ0 โ 1.7 m. From the radius of the ice ball, one finds that ๐GRL = ๐ ๐๐๐/๐ ๐ธ โ0.0147, which corresponds to โGRL โ โ0 โ 229.8 m. Finally, the differences become โGRL โ โCPH โ 223.5 m, and โCPH โ โOPP โ 4.5 m, where โ0 has dropped out.
Total 9.0
The Greenlandic Ice Sheet T3
Page 1 of 4
Marking scheme
3.1
๏ฟฝ(๏ฟฝ, ๏ฟฝ) = ๏ฟฝ๏ฟฝ๏ฟฝ( (๏ฟฝ) โ ๏ฟฝ)
Getting hydrostatic pressure right gives 0.3.
Deduct 0.2 for writing (๏ฟฝ) or ๏ฟฝ instead of (๏ฟฝ) โ ๏ฟฝ (wrong boundary condition)
Deduct 0.1 for writing ๏ฟฝ instead of (๏ฟฝ). 0.3
3.2a
๏ฟฝ = โ๏ฟฝ๏ฟฝ๏ฟฝ
0.5 points are given for solving the problem by dimensional analysis, which misses an
overall pre-factor.
No deduction for the sign, since one could in principle have looked at negative ๏ฟฝ.
0.9
3.2b
(๏ฟฝ) = ๏ฟฝ2๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ๏ฟฝ ๏ฟฝ1 โ ๏ฟฝ/๏ฟฝ
Setting up and solving the differential equation (integrating) gives 0.5.
Implementing the correct boundary conditions gives 0.3
No deductions /additions for absolute value around x.
Deduct 0.1 for missing factor of โ2 due to simple arithmetic error.
Deduct 0.2 for solving problem only by dimensional analysis (+boundary condition),
which also misses the factor of โ2 and leaves the pre-factor unknown.
Deduct 0.4 for applying wrong boundary condition at ๏ฟฝ = ๏ฟฝ.
Deduct 0.2 for expressing solution in terms of ๏ฟฝ, but not actually finding expression
for ๏ฟฝ in terms of the given parameters.
0.8
3.2c
๏ฟฝ = 5/4
Give 0.3 for getting the correct idea, involving an integral over the height profile times
the ground-area along y.
Give 0.2 for getting the scaling of ๏ฟฝ with ๏ฟฝ correct.
Deduct 0.1 for misunderstanding the definition of ๏ฟฝ, as in e.g. finding the scaling with ๏ฟฝ
rather than with .
0.5
The Greenlandic Ice Sheet T3
Page 2 of 4
3.3
!"(๏ฟฝ) = #๏ฟฝ $
Setting up mass balance and solving the problem gives 0.6.
Dimensional analysis cannot be used here since ๏ฟฝ/ ๏ฟฝ is dimensionless.
Deduct 0.2 for getting the sign and/or pre-factor wrong.
0.6
3.4
!%(๏ฟฝ) = โ #๏ฟฝ $
Setting up the correct differential equation and solve it gives 0.6.
No deduction for using !"(๏ฟฝ) without having solved for this first (i.e. without solving
problem 3.3)
Deduct 0.2 for getting the sign and/or pre-factor wrong.
Deduct 0.2 for wrong implementation of the boundary condition, like keeping an
undetermined integration constant.
0.6
3.5
๏ฟฝ = $๏ฟฝ&/๏ฟฝ
Finding that ๏ฟฝ โ 1/๏ฟฝ gives 0.6. If done by solving differential equation, then give 0.4
for setting up the equation and 0.2 for integrating it.
Implementing the initial condition gives 0.3.
No deduction if they happen to write (๏ฟฝ) instead of ๏ฟฝ.
0.9
3.6
((๏ฟฝ) = $# ln+ $๏ฟฝ ,
Setting up the right equation, including the integral gives 0.4.
Integrating (solving) the equation gives 0.3.
Implying the correct physical boundary conditions gives 0.3.
1.0
The Greenlandic Ice Sheet T3
Page 3 of 4
3.7a
#- = 0.17m/year c6 = 0.12m/year
Getting the correct value for #- gives 0.3.
Getting the correct value for #6 gives 0.5.
We will accept a range of significant digits between 2 and 4 without deduction.
If #6 is calculated by exactly the same method as #- (i.e. neglecting the influence of the
previous interglacial age) deduct 0.3 (i.e. then the nearly correct value (c6 =0.13m/year ) obtained this way will give a total of 0.2).
0.8
3.7b
โ9 โ 12โ
Given that the data are very noisy, give all 0.2 points for finding a number of โ9 = 12โ ยฑ 4โ. Give only 0.1 for โ9 = 12โ ยฑ 8โ, (and outside โ9 = 12โ ยฑ4โ), unless it is supplemented by a qualified and valid discussion, which can then
allow giving 0.2.
No deduction for wrong sign on โ9.
Deduct 0.1 for missing units on โ9.
0.2
The Greenlandic Ice Sheet T3
Page 4 of 4
3.8
โ?,@A๏ฟฝ = 8.78m
We subdivide the problem into four basic elements which are bound to enter the
solution in some way.
Give 0.1 for finding the correct value of ๏ฟฝ.
Give 0.2 for finding the correct values of ๏ฟฝ&BC.
Give 0.2 for converting ice volume to water volume, using the two different given
densities.
Give 0.1 for converting the water volume to a sea-level rise.
0.6
3.9
โDEF โ โGEE = 4.5 m
In grading this problem, we are searching for correct elements in three areas, and one
can earn up to 0.6 points from each of the following three categories:
1) Stating the relevant physical principles (fx. Involving gravitational potentials from
two bodies, sea level as equipotential surface, etc.) and putting them together in a
meaningful way.
2) Writing up the correct potentials, with the relevant lengths, masses, etc., and
expressing the ideas from pt.1 in correct mathematical form.
3) Actually solving the equations, making the appropriate expansions/approximations,
carrying out geometry/trigonometry.
1.8
Total 9.0