Download HW6 class solution

AE 245 homework #6 solutions
Tim Smith
15 March 2000
1 Problem 1
The Hubble Space Telescope was placed in a circular orbit about the Earth at an altitude of 330 mi. Identify the
following orbital characteristics:
1. What is the circular velocity of the Hubble Telescope?
2. What is the orbital period of the Hubble Telescope?
3. What are the specific total energy and specific angular momentum of the Hubble Telescope?
1.1 Circular velocity
In a circular orbit, the gravitational acceleration is a centripetal acceleration; thus,
µ
r2
=
v2
r
(1)
where µ = GM is the Earth’s gravitational parameter. Rearranging Eqn. 1 gives the circular speed
v=
rµ
(2)
r
Since the Earth’s mean equatorial radius r = 3963:2 miles, the orbital radius r = (3963:2 + 330) mi = 2:2668 107
ft, so
s
v=
1:4076 1016 ft
2:2668 107 s
= 24; 919
ft
s
1.2 Orbital period
Since a circular path has a length 2πr, a steady speed of v gives an orbital period
T
=
2πr
v
1
(3)
Substituting Eqn. 2 into 3 and rearranging gives
T
=
2πr3=2
pµ
=
2π(2:2668 107)3=2
p
s = 5715:6 s = 1 hr 35 min 15:6 s
1:4076 1016
(4)
1.3 Specific total energy and angular momentum
The specific total energy for any orbit is
E
v2
2
, µr
(5)
Substituting Eqn. 2 for a circular orbit, the Hubble Space Telescope specific total energy
µ
E =, 2r
=
1016
, 21(24076
2668 107)
:
:
ft2
s2
2
= 3:1048
108 fts2
(6)
Likewise, the specific angular momentum for any orbit is
h ~r ~v
~
(7)
which, for our circular orbit, has the constant scalar value
h = rv = 2:2668 107(1:4076 1016) ft2 =s = 3:7555 1023 ft2 =s
(8)
2 Problem 2
Consider a satellite in circular orbit about the Earth.
1. Plot the gravitational acceleration of the satellite due to Earths gravity as a function of the altitude of the satellite
as measured from the surface of the Earth. Scale your plot so that the altitude goes from 100 km to 10,000 km.
2. On the same graph, plot the gravitational acceleration of the satellite due to Earths gravity and the (maximum)
gravitational acceleration of the satellite due to the Suns gravity as a function of the altitude of the satellite as
measured from the surface of the Earth. Scale your plot so that the altitude goes from 100 km to 10 million km.
3. Use your graphs in part (b) to decide (approximately) how large the radius of acircular orbit can be so that the
two body model of the Earth and satellite is accurate to within 2%.
2.1 Earth’s gravitational acceleration
The gravitational acceleration due to the Earth on the satellite is
g =
2
µ
r2
(9)
where r is the sum of the altitude z and the mean equatorial radius of the Earth r . Figure 1 is produced by the
following IDL code:
10
acceleration (m/s^2)
8
6
4
2
0
0
2000
4000
6000
altitude (km)
8000
10000
Figure 1: Gravitational acceleration of satellite due to the Earth as a function of altitude.
pro hw6_2a
; Define fundamental constants & conversions.
; Source: Bate, Mueller & White, _Fundamentals of Astrodynamics_, 1971.
mu_e
r_e
= 3.986012e14
= 6.378145e6
; Earth’s GM, mˆ3/sˆ2
; mean Earth radius, m
; Create altitude array.
h_min
h_max
bins
range
h
h
=
=
=
=
=
=
1.e5
1.e7
401
alog(h_max/h_min)
range*findgen(bins)/(bins-1)
h_min*exp(h)
; minimum altitude, m
; maximum altitude, m
; number of array bins
; altitude, m
; Calculate and plot acceleration due to Earth’s gravity.
g_e
xtitle
ytitle
file
scale
xsz
= mu_e/(r_e + h)ˆ2
=
=
=
=
=
’altitude (km)’
’acceleration (m/sˆ2)’
’fig6_2a.eps’
1.0
4.0*scale
; x-axis title
; y-axis title
; file name
3
ysz
= 3.0*scale
set_plot, ’ps’
device,/encapsulated,/preview,filename=file
device,/inches, xsize=xsz, ysize=ysz,
$
font_size = 7
plot, h/1.e3, g_e, xtitle=xtitle,
$
ytitle=ytitle, font=0 ;
device, /close
; set plot device to PostScript
; open file & save EPS
; close the file
end
2.2 Solar gravitational acceleration
The maximum solar gravitational acceleration for a given altitude from the Earth’s surface occurs when the satellite is
closest to the Sun. This happens when the satellite is on a line between the Earth and the Sun, so that the radius of the
satellite from the Sun’s center
r = a , r
where a is the distance between the Earth’s and Sun’s centers (1 AU, or 1:4960 1011 m). Thus, the solar gravitational
acceleration
g =
µ
(a
, r)2
Figure 2 is produced by the following IDL code:
pro hw6_2b
; Define fundamental constants & conversions.
; Source: Bate, Mueller & White, _Fundamentals of Astrodynamics_, 1971.
mu_e
r_e
mu_s
au
=
=
=
=
3.986012e14
6.378145e6
1.3271544e20
1.4959965e11
;
;
;
;
Earth’s GM, mˆ3/sˆ2
mean Earth radius, m
solar GM, mˆ3/sˆ2
Earth-Sun distance, m
; Create altitude array.
h_min
h_max
bins
range
h
h
=
=
=
=
=
=
1.e5
1.e10
401
alog(h_max/h_min)
range*findgen(bins)/(bins-1)
h_min*exp(h)
; minimum altitude, m
; maximum altitude, m
; number of array bins
; altitude, m
; Calculate and plot acceleration due to Earth’s & solar gravity.
g_e
= mu_e/(r_e + h)ˆ2
; Earth’s accel, m/sˆ2
4
(10)
102
acceleration (m/s^2)
100
10-2
10-4
10-6
102
103
104
105
altitude (km)
106
107
Figure 2: Terrestrial (solid line) and solar (dotted line) gravitational acceleration as a function of altitude.
g_s
xtitle
ytitle
file
scale
xsz
ysz
= mu_s/(au - r_e - h)ˆ2
=
=
=
=
=
=
; solar accel, m/sˆ2
’altitude (km)’
’acceleration (m/sˆ2)’
’fig6_2b.eps’
1.0
4.0*scale
3.0*scale
; x-axis title
; y-axis title
; file name
set_plot, ’ps’
device,/encapsulated,/preview,filename=file
device,/inches, xsize=xsz, ysize=ysz,
$
font_size = 7
plot, h/1.e3, g_e, xtitle=xtitle,
$
ytitle=ytitle, font=0, /xlog, /ylog
oplot,h/1.e3, g_s, linestyle=1
device, /close
; set plot device to PostScript
; open file & save EPS
; close the file
end
2.3 Validity of two-body model
The combined terrestrial and solar gravity model, g , g , is more accurate than the two-body model, g . The percent
error is
5
ε=
g , (g , g )
g , g
=
g
g , g
(11)
Letting ε = 0:02 in Eqn. 11 gives the requirement that
g = 51:0g
(12)
at our desired location. Substituting Eqn. 9 and 10 into 12 yields
µ
r2
=
51:0µ
, r)2
(13)
(a which can be rearranged to give a radius from Earth
r=
a
p
1 + 51 0µ
:
(14)
=µ
Since r = r + z, the altitude for 2% error is
0
1
11
1
4960
10
q 51 0µ , r = @ q 51 0 1 371510 , 6 3781 106A m = 29 326 km
a
z=
1+
:
:
µ
1+
:
(
:
3:9860105
11)
:
;
(15)
3 Problem 3
Assume that each of the planets Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune is in a circular orbit about the
Sun.
1. Make a table showing the mean orbital radius (in km), the mean circular velocity (in km/sec), and the mean
orbital period (in Earth years) for each of these planets. Check that the assumptions of a circular orbit for these
planets is justified.
2. It is not accurate to assume that the planets Mercury and Pluto are in circular orbit about the Sun. Show this
from data for their mean orbital radius (in km), mean circular velocity (in km/sec), and mean orbital period (in
Earth years) for Mercury and Pluto.
3.1 Circular planetary orbits
Table 1 compares tabulated data for the planets1 to values calculated by the circular orbit equations for circular velocity
vc =
rµ
r
and period
1 R.
R. Bate, D. D. Mueller, and J. E. White, Fundamentals of Astrodynamics, Dover Publications, New York, 1971, p. 361.
6
(16)
s
T
= 2π
r3
µ
(17)
where the Sun’s gravitational parameter µ GM = 1:3272 1011 km3 /s2 .
Planet
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Table 1: Mean circular orbits for the given planets.
From Bate, Mueller & White
Calculated
Mean orbital Mean circular Mean orbital Mean circular Mean orbital
radius (km)
speed (km/s)
period (yr)
speed (km/s)
period (yr)
1:081 108
35.04
0.615
35.04
0.6147
1:495 108
29.79
1.000
29.80
0.9997
2:278 108
24.14
1.881
24.14
1.880
7:780 108
13.06
11.86
13.06
11.87
1:426 109
9.65
29.46
9.647
29.45
2:868 109
6.80
84.01
6.803
84.00
4:494 109
5.49
164.8
5.434
164.8
The calculated values are all within 0.05% of the tabulated values, indicating that a circular orbit assumption works
pretty well for all the planets except Mercury and Pluto.
3.2 Validity of circular orbits
Let’s continue Table 1 for Mercury and Pluto:
Planet
Mercury
Pluto
Table 2: Mean circular orbits for Mercury and Pluto.
From Bate, Mueller & White
Calculated
Mean orbital Mean circular Mean orbital Mean circular Mean orbital
radius (km)
speed (km/s)
period (yr)
speed (km/s)
period (yr)
5:790 107
47.87
0.241
47.88
0.2410
5:896 109
4.740
247.7
4.745
247.6
At first glance, this seems to show that the circular orbit assumption is also valid for Mercury and Pluto! One reason
for this may be that the standard mean radii are calculated from the orbital period.
Mercury and Pluto don’t have very circular orbits because their orbital eccentricities are too far from zero. Table 3
lists the eccentricities for all the planets.
7
Planet
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
Table 3: Mean orbital parameters for the planets.
Mean orbital Mean circular Mean orbital Eccentricity
radius (km)
speed (km/s)
period (yr)
7
5:790 10
47.87
0.241
0.2056
1:081 108
35.04
0.615
0.0068
1:495 108
29.79
1.000
0.0167
24.14
1.881
0.0934
2:278 108
7:780 108
13.06
11.86
0.0482
1:426 109
9.65
29.46
0.0539
2:868 109
6.80
84.01
0.0514
5.49
164.8
0.0050
4:494 109
5:896 109
4.740
247.7
0.2583
8