CBB039 - Numerical Methods: Solar System Data Analysis Report

This document presents a student's solution to a numerical methods assignment focused on analyzing data related to the solar system. The assignment covers various aspects, including calculating mean densities and surface areas for Earth and Pluto, determining speeds of rotation and revolution, computing gravitational field strengths and escape velocities, and analyzing solar constants. It also involves scaling and representing planetary distances. The solution further includes analyzing atmospheric pressure at different heights and performing a logarithmic analysis of satellite orbital periods and radii. The final part involves calculating acceleration due to gravity, distance fallen, and speed after a given time on Pluto's surface, along with assessing the reliability of a given model. Desklib provides access to this and many other solved assignments to aid students in their learning journey.

Mathematics 1
Name:
Institution:
Date:

Mathematics 2
PART A
Question one
a) Mean density for earth and Pluto
- Mean density for Earth.
Mass = 5.978 x 1024 kg
Volume =? but volume= 4
3 π r3
radius of earth=6.376 × 106 metres
¿ 4
3 ×3.142 ×(6.376 ×106 ×6.376 ×106 × 6.376× 106 )
volume=1 . 0859 ×1021
density = mass
volume = 5.978× 1024
1.0859 ×1021 =5.505× 103
Mean density of earth=5.505 × 103 kg/ m3
- Mean density for Pluto.
Mass = 5x 1023 kg
Volume =? but volume= 4
3 π r3
radius of Pluto=3 ×106
¿ 4
3 ×3.142 ×(3× 106 ×3 ×106 ×3 ×106)
volume=1. 13 ×1020
density = mass
volume = 5 ×1023
1.13 ×1020 =4.42 ×103
Mean density of Pluto=4.42× 103 kg/m3

Mathematics 3
b) Surface area for earth and Pluto
- Surface area of earth
Since the earth is spherical;
surface area of a sphere=4 π r2
radius of earth=6.376 × 106 metres
¿ 4 ×3.142 ×( 6.376× 106 ×6.376 ×106 )
5.1 ×1014 m2
- Surface area of Pluto
surface area of a sphere=4 π r2
radius of Pluto=3 ×106 metres
¿ 4 ×3.142 ×(3 × 106 × 3 ×106 )
1 .13 ×1014 m2
c) Speed of rotation at the equator for earth and Pluto
- Speed of rotation of the earth
Speed= Distancetravelled
time taken
Distance at equator = 2 πr
where r=6.376× 106 m
¿ 2 ×3.142× ( 6.376 ×106 )
¿ 4.0 × 107 m
time∈hours for one rotation=23 .93 hrs
time∈s econds=23.93× 3600=86148 seconds
Therefore; Speed= 4.0× 107 m
8.6148× 104 s

Mathematics 4
speed=4.643× 102 m/ s
- Speed of rotation of Pluto
Speed= Distancetravelled
time taken
Distance at equator = 2 πr
where r=3× 106 m
¿ 2 ×3.142× ( 3 ×106 )
¿ 1.885 ×107 m
time ∈hours for one rotation=23.93 hrs
time∈seconds=23.93 × 3600=86148 seconds
Therefore; Speed= 1.885 × 107 m
8.6148× 104 s
speed=2.188 ×102 m/ s
d) Mean speed of travel around the sun
- Speed of earth round the sun
Speed= Distancetravelled
time taken
Distance round the sun = 2 πr
where r=1.496 × 1011 m
¿ 2 ×3.142× ( 1.496 ×1011 )
¿ 9.4 × 1011 m
time∈days for one rotation=365.3 days
time∈seconds=365.3 × 86400=3.156 ×107 seconds
Therefore; Speed= 9.4 ×1011 m
3.156× 107 s

Mathematics 5
speed=2.97 8 ×104 m/s
- Speed of Pluto round the sun
Sp eed= Distance travelled
time taken
Distance round the sun = 2 πr
where r=5 . 9 ×1012 m
¿ 2 ×3.142× ( 5.9 ×1012 )
¿ 3.708 ×1013 m
time∈days for one rotation=248 ×365.256 days
time∈s econds=248× 365.256 ×86400=7.826× 109 seconds
Therefore; Speed= 3.708× 1013 m
7.826 ×109 s
speed=4.738× 103 m/s
e) Gravitational field strength at the planet’s surface
- Gravitational field strength on earth’s surface
gravitational field strengt h , g= GM
R2
where G=6.673 x 10−11 N m2 kg−2
M =mass of the planet ( earth )=5.978 ×1024 kg
R=Radius of planet ( earth ) =6.378× 106 m
Theref ore gravitational field strength, g=( 6.673 x 10−11)(5.978 ×1024)
( 6.378× 106)2
¿ 3.989 ×1014
(6.378 ×106)2
¿ 3.989× 1014
4.068 ×1013

Mathematics 6
g=9.81 N /kg
- Gravitational field strength on Pluto’s surface
gravitational field strength , g= GM
R2
where G=6.673 x 10−11 N m2 kg−2
M =mass of the planet ( pluto ) =5× 1023 kg
R=Radius of planet ( pluto )=3 × 106 m
Ther efore gravitational field strength , g=(6.673 x 10−11)(5 ×1023 )
(3 ×106)2
¿ 3.337× 1013
9 ×1012
g=3.71 N /kg
f) Escape velocity at the surface
- Escape velocity on the surface of the earth
Escape v elocity= √ ( 2 GM
R )
where G=6.673 x 10−11 N m2 kg−2
M =mass of the planet ( earth )=5.978 ×1024 kg
R=Radius of planet ( earth ) =6.378× 106 m
2 GM=2 (6.673 x 10−11 × 5.978× 1024 )
2 ( 3.989 ×1014 ) =7.98× 1014
¿ 7.98 ×1014
6.378× 106 =1.25 ×108
√ 1.25× 108
1.12 ×104 Nm/ kg

Mathematics 7
- Escape velocity on the surface of Pluto
Escape velocity= √ ( 2GM
R )
where G=6.673 x 10−11 N m2 kg−2
M =mass of the planet ( pluto ) =5× 1023 kg
R=Radius of planet ( pluto )=3 × 106 m
2 GM=2 (6.673 x 10−11 × 5× 1023)
2 ( 3.337 ×1013 )=6.674 ×1013
¿ 6.674 ×1013
3 ×106 =2.224 ×107
√ 2.224 ×107
4.716 × 103 Nm /kg
g) Solar constant at the planet’s surface
- Solar constant at the earth’s surface
intensity of radiation at the surface= 1361W
m2
If the radiation intensity is inversely proportional to the square of distance from the sun,
therefore;
1361=k . 1
d2 ; where k is solar constant
d=1.496× 1011
Solar constant on earth surface therefore , k=136 1 d2
k =1361¿

Mathematics 8
k =1361(2.238 ×1022)
k =3.05 ×1025
- Solar constant at the Pluto’s surface
intensity of radiation at the surface= 1361W
m2
If the radiation intensity is inversely proportional to the square of distance from the sun,
therefore;
1361=k . 1
d2 ; where k is solar constant
d=5.9× 1012
Solar constant on Pluto surface therefore ,k =1361 d2
k =1361¿
k =1361(3.481× 1025)
k =4.737 × 1028
h) Radius at the pole
- Radius of the earth at the pole
radius at the poler =Rcos 900
radius at the poler =( 6.378× 106) cos 900
¿(6.378× 106 )× 0.44
r =2.81× 106 m
- Radius of the Pluto at the pole
radius at the poler =Rcos 900
radius at the poler =(3 × 106)cos 900

Mathematics 9
¿(3× 106 )× 0.44
r =1.32× 106 m
Question two
The distance between the earth and Pluto is;
( 5.9 ×1012 )−(1.496 ×1011 )
¿ 5.75 ×1012 meters
If 1 unit on paper represents 1012 on the ground then;
5.75× 1012
1012 =5.75 metres
The two planets will be 5.75 meters apart on paper.
Question three
24 meters to represent 5.9 x 1012 meters since this is the furthest distance from the sun.
scale ratio= 5.9 ×1012
24 =2.458× 1011
On the corridor the planets will be as below;
mercury= 5.791×1010
2.458× 1011 =0.236 meters
venus=1.082 ×1011
2.458 ×1011 =0. 44 meters
earth= 1.496 ×1011
2.458 ×1011 =0.6 09 meters
Earth
Pluto
5.75 m

Mathematics
10
mars= 2.279× 1011
2.458× 1011 =0. 927 meters
jupiter= 7.783 ×1011
2.458 ×1011 =3 .17 meters
saturn= 1.427× 1012
2.458× 1011 =5.8 meters
uranus= 2.869× 1012
2.458× 1011 =11.67 meters
neptune= 4.498× 1012
2.458 × 1011 =18.3 meters
pluto= 5.9 ×1012
2.458 ×1011 =24.0 meters
Question four
Comparing the planet earth and Pluto we find that the gravitational strength on the surface of
Pluto is less (3.71N/kg) compared to that on the surface of the earth which is 9.81N/kg. The solar
intensity on the planet Pluto compared to planet earth is much higher considering solar constant
on Pluto is 4.737 × 1028 and that on the surface of the earth is 3.05 ×1025 .
PART B
Height, h / km Pressure, Ph / Pa
0 101.3
10 31.24
20 9.63
30 2.97
40 0.96
50 0.28
60 0.087

Mathematics
11
ph = po e−( h
H )
atmospheric pressure ; ph= p0 e−( h
H )
Find the height at which the pressure is 20 kPa.
20=101.3 e−( h
8.5 )
0.1974=e−( h
8.5 )
log 0.1974=−h
8.5 log e1
log 0.1974=−h
8.5 log2.7182
log 2.7182
log 0.1974 = h
8.5
−h= log 0.1974
log 2.7182 × 8.5
h=13.79 km
PART C
a)
Satellite Period of orbit, T
/ days
Radius of orbit, R
/ x106 m
Log T Log R
Io 1.77 422 0.248 2.625
Europa 3.55 671 0.55 2.828
Ganymede 7.15 1070 0.854 3.03
Callisto 16.69 1883 1.22 3.27

Mathematics
12
b)
2.5 2.6 2.7 2.8 2.9 3 3.1 3.2 3.3 3.4
0
0.2
0.4
0.6
0.8
1
1.2
1.4
f(x) = 1.507129297824 x − 3.71032265933136
R² = 0.999967167593469
Graph of log T against Log R
Log R
Log T
The equation y=1.507 x−3.7103
comparing the above ¿ logT =bLog R+ log a
c) The value of b from the below compared equations is;
The equation y=1.507 x−3.7103
comparing the above ¿ logT =bLog R+ log a
b=1.507
PART D
a) Acceleration due to gravity is calculated as below;