Download Accuracy of the Astronomical Unit

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Interplanetary Space Travel
Accuracy of the Astronomical Unit
When guiding a spacecraft to another world within the solar system it is imperative
that we know the distances to those worlds as well as possible. To ensure that happens,
an accurate measurement of the standard unit of measurement for distances within the
solar system, the Astronomical Unit, is crucial. The questions below illustrate how
uncertainties in the measurement of the Astronomical Unit can have a major impact on
the success of an interplanetary spacecraft mission.
1. The typical measurement of the Astronomical Unit used in many introductory
astronomy courses is, 1 AU = 1.5 x 108 km. If the percent uncertainty of this
measurement is 0.1%, calculate the actual uncertainty in the measurement.
1 AU = ______________  ___________ km
2. The uncertainty in our measurement of the Astronomical Unit will factor into how
well we can predict the position of a planet. The lower our uncertainty, the better
we can guide a spacecraft to a rendezvous with a distant world. If we wanted to
send a spacecraft to a Mars while it was 1 AU from Earth, then the uncertainty in
the AU measurement compared to the size of the target can provide an idea of
how challenging the mission might be. Calculate how many times greater the
uncertainty is compared to the size of Mars (D = 6794 km).
3. To get an idea of the scale of the uncertainty, imagine that the circle in Figure 7.1
represents the diameter of Mars. Measure the diameter of the scaled down
version of Mars and use proportions to determine how large the uncertainty would
be on that scale.
4. Using a ruler and a drawing compass, draw a boundary around the scaled down
version of Mars that would represent how far away the uncertainty could cause
the spacecraft to miss the planet altogether.
5. Go online and find the most current accurate measurement of the Astronomical
Unit and its uncertainty. Record them both below and calculate the percent
uncertainty.
1 AU = ______________  ___________ km
Percent Uncertainty = ________
Figure 7.1: Scaled-down version of Mars
Planetary Positioning
How we get a spacecraft will differ slightly depending on whether our target planet
is an inferior or superior planet. An inferior planet is any planet that has an orbit that is
closer to the Sun than Earth’s orbit. Conversely, a superior planet is any planet that has
an orbit that is beyond Earth’s.
6. Categorize the planets of our solar system based on whether they are an inferior
planet or a superior planet.
Inferior Planets
Superior Planets
Because they continuously are in motion around the Sun, planets are essentially
moving targets. For the best chances of a successful rendezvous with a planet, we need
to track the position of a planet relative to Earth at all times. This information will allow
us to both direct the spacecraft to the planet on a proper course and provide us the ability
to communicate with the spacecraft once it arrives. Because it is a source of radio waves,
the Sun can create interference in communications between the control center on Earth
and a spacecraft that has rendezvoused with a planet.
Figure 7.2 illustrates the orbit Earth with an inferior planet (a) and the orbit of Earth
with a superior planet (b). Keeping in mind that an orbit represents every point where a
planet could be during its trip around the Sun, several important alignments are
highlighted.
Inferior Conjunction (A): occurs when an inferior planet lies at a point in its orbit
that places the planet between the Earth and the Sun.
Superior Conjunction (B): occurs when an inferior planet lies at a point in its orbit
that places the planet behind the Sun when viewed from Earth.
Greatest Elongation: these mark the points where an inferior planet will achieve the
greatest separation from the Sun that its orbit will allow. This occurs at two points.
As viewed from Earth, the point to the left of the Sun is referred to as the Greatest
Eastern Elongation (C) while the point to the right of the Sun is referred to as the
Greatest Western Elongation (D).
E
B
C
A
D
F
(a)
= Earth
= Sun
(b)
= Inferior Planet
= Superior Planet
Figure 7.2: Planetary positions for an inferior planet (a) and a superior planet (b).
Superior Conjunction (E): like an inferior planet, a superior planet can appear
behind the Sun, as viewed from Earth. (Note that an inferior conjunction is not
possible because a superior planet’s orbit is outside of Earth’s.)
Opposition (F): occurs when a superior planet is on the opposite side of Earth from
the position of the Sun.
7. What is the best planetary position for Mars when trying to control the movement
of the rover, Spirit, that is currently on its surface? Explain.
8. What is the best planetary position for Mercury when trying to control the
instruments aboard the MESSENGER spacecraft? Explain.
Newton’s Law of Gravity
Isaac Newton (1642 - 1727) first proposed the idea that an apple will fall from a tree
because of the force of gravity. Gravity, Newton says, is an inherent property of any
massive object. As long as an object has a mass, it will pull on other objects with mass.
Gravity, like all other forces, is a vector quantity in that we need to know the magnitude
(i.e. amount) of the force as well as the force’s direction. Luckily, the direction of the
gravitational force is always the same: toward the center of the objects in question. The
magnitude of the force can be calculated by using the Universal Law of Gravitation:
FG =
Gm1m2
, where
r2
FG  magnitude of the force of gravity (N)
m1, m2 mass of object 1 and object 2 (kg)
r  separation between the two objects (m)
G  Universal Gravitational Constant
= 6.67 x 10 -11 (
N  m2
)
kg 2
Example:
What
is the magnitude of the force of gravity between the Earth and the Sun?
To calculate the force of gravity between the Sun and Earth we will need to know the
following information, which is available in a textbook or on an appropriate website:
m1  mass of the Sun = 1.99 x 10 30 kg
m2  mass of the Earth = 5.96 x 10 24 kg
r  1.5 x 1011 m

To calculate the magnitude of the force of gravity, simple substitute the values in the
appropriate places in the Universal Law of Gravitation and then follow through with the
calculation.
Gm1m2
FG =
=
r2
6.67 x 10 1.99 x 10 5.96 x 10 
1.5 x 10 
-11
30
11 2
24
=
7.91 x 10 44
2.25 x 10 22
FG = 3.52 x 10 22 N

According to Newton’s 3rd Law of Motion, the force of 3.52 x 1022 N is felt equally
by both the Sun and the Earth, but in opposite directions. The Sun is being pulled toward
 the center of the Earth with 3.52 x 1022 N while the Earth is being pulled toward the
center of the Sun with the same force. How each object reacts to the force is different
and can be understood by looking at the Law of Inertia. Since the Sun is much more
massive than the Earth, the Sun will resist this force more intently than the Earth. As
such, it is adequate to visualize the Earth being held by the force of gravity in a closed
orbit around a nearly motionless Sun.
9. Use the Universal Law of Gravitation to calculate the force of gravity between the
Sun and Jupiter.
10. Calculate how many times greater the force of gravity between the Sun and
Jupiter is than the force of gravity between the Sun and Earth.
According to the Law of Inertia, a planet that is moving at some velocity (v) will do
so in a straight path so long as an outside force doesn’t act on it. However, planets are
under the influence of the force of gravity brought on by the presence of the Sun’s mass.
This centralized force causes the planet to be pulled (i.e. accelerated) toward the Sun’s
center (Figure 7.3). This combination of the planet’s desire to wander a straight path
through space with its experiencing centripetal acceleration, the planet ends up being
trapped in a closed orbit around the Sun.
Velocity
Planet
Acceleration
Sun
Resulting Orbital Motion
Figure 7.3: Orbital motion of a planet around the Sun.
Orbital Properties
Getting a spacecraft from one planet to another in the solar system is a challenging
task given that the targets are continuously moving in orbits around the Sun. Luckily, the
properties of orbital motion have been well understood since the time Johannes Kepler
(1571 - 1630) first proposed his three laws of planetary motion nearly four centuries ago.
In this section we will investigate each of Kepler’s laws of planetary motion in order to
better understand how spacecraft are sent to other worlds in our solar system.
Kepler’s 1st Law: Shapes of Orbits
Kepler’s first law identifies the shape of the orbital path of a planet as it travels
around the Sun is that of an ellipse and not a circle. An ellipse differs from a circle in
that it has two focal points (foci) that are both the same distance from the center, whereas
a circle has only one focal point located on the exact center (see Figure 7.4). This means
that for an ellipse, there are axes of many different sizes that can be drawn through the
center. There are two important axes that are important references for an ellipse. The
minor axis is the shortest axis that can be drawn across the ellipse through the center
while the major axis is the longest.
Foci
Focus
Ellipse
Figure 7.4: Comparison of an ellipse to a circle.
Circle
Kepler also noted that the Sun is located at one of the focal points of a planet’s
elliptical orbit. This means that the distance between the Sun and a planet is always
changing as the planet moves around its orbit. As a standard way of comparing orbit
sizes, the average Sun-planet distance is always referenced. Below are several important
properties to know when discussing planetary orbits:
Semi-major axis (a): half the length of the major axis; this distance is equivalent to
the average Sun-planet distance.
Eccentricity (e): defines the “shape” of the orbit by taking the ratio of the Sun
center distance (c) to the semi-major axis (a){e = c/a}. See Figure 7.5.
Periapsis: the point in an orbit that is closest to the body being orbited.
Apoapsis: the point in an orbit that is farthest from the body being orbited.
Periapsis and apoapsis are general terms used when discussing orbits around an
unspecified body. More specific terms are used when the body being orbited is known.
For example, the terms perihelion and aphelion are used to refer to the point in a planet’s
orbit that is closest and furthest from the Sun, respectively.
c
a
Sun
c = 1.6
a = 3.5
e = c/a = 1.6/3.5 = 0.46
Figure 7.5: Determining the eccentricity (e) of a planetary orbit.
Kepler’s 2nd Law: Equal Area in Equal Time
Kepler’s Second Law of Planetary Motion states that an imaginary line connecting
the Sun and planet will “sweep out” equal areas within the orbital plane within equal
units of time. Though this is strictly an observation of planetary behavior, looking at this
discovery in terms of Newton’s laws reveals a deeper significance.
As a planet moves along its elliptical orbit, the distance separating the planet and the
Sun grows and shrinks over time. This change in the separation means that the force of
gravity between the two also changes. The result is that a planet will end up moving
faster in its orbit when it is closer to the Sun and slower when it’s further away.
Kepler’s 3rd Law: P2 = a3
Kepler’s Third Law of Planetary Motion draws a direct correlation between the
orbital period of a planet (measured in years) to its average distance from the Sun
(measured in AU). Though the accuracy of this law is improved upon using Newton’s
laws, Kepler’s third law in its stated form will be useful enough for our investigations.
Example: The orbital period of Venus is measured to be 0.6152 years. What is the
average distance between the Sun and Venus?
Solution:
2
Porb
= a3
a 3 = (0.6152) 2 = (0.3784)
a = 3 (0.3784)
a = 0.7233 AU

Use the orbit of a fictitious planet orbiting a generic star in Diagram 7.1 to answer
Questions 11 – 15.
Star
Planet
Diagram 7.1: Orbit of a fictitious planet orbiting a star.
11. Draw in the major axis of the planet’s orbit.
12. Mark the center of the major axis with an ‘x’.
13. Determine the length of the semi-major axis and the Sun-center distance.
Semi-major axis (a) = ________ cm
Sun-center distance (c) = ________ cm
14. Calculate the eccentricity (e) of the orbit.
e = ________
15. Label the point of periapsis and apoapsis on the planet’s orbit.
16. If the scale of Diagram 7.1 is, 1 cm = 3 AU, determine the planet’s true average
distance from the star it is orbiting.
True average distance = ________ AU
17. Between which two planets would this planet be located if it were orbiting within
our solar system?
18. Calculate the orbital period for the planet.
Porb = ________ yrs
19. Where will the planet be when it is moving with its greatest velocity?
Getting a Spacecraft to Mars
Before we can get a spacecraft to another planet, like Mars, we must first get it off
the surface of the Earth. To understand how this can be done we will first need to
understand how Earth’s gravity influences our ability to get a rocket into space.
Starting out with a simple example, imagine a person throwing a baseball at 50 mph
(22.2 m/s). The baseball will travel about 75 feet (~25 meters) before it falls to the
ground because of Earth’s gravity. Experience tells us that throwing the baseball faster
will cause it to go a further distance before coming back down to the ground. Since there
is a limit to how fast a person can throw a ball (~100 mph; 44.4 m/s), there will be a limit
to how far the baseball will travel before falling to the ground (175 ft; 57 meters).
Newton relied on a similar analogy using canon ball trajectories to figure out how an
object can overcome the influence of Earth’s gravity. To start, Newton thought that if a
canon were placed at the top of a mountain, packed with gunpowder, and then fired, the
canon ball would be given such a velocity that the canon ball would fly a much greater
distance than a ball being thrown by a person. Packing more gunpowder would generate
greater velocity for the cannon ball, which, in turn means it can fly greater distances
along the Earth’s surface. Newton theorized that given enough gun powder, the canon
ball could be fired at such a high speed that it’s trajectory would take it around the Earth
completely before it ever fell down to the surface. The speed at which an object needs to
travel in order to make it around the Earth without hitting the ground is called the orbital
velocity. The orbital velocity for any world is given by the general equation:
vorb =
GM
, where
r
v orb  orbital velocity (m/s)
M  mass of world in question (kg)
 the world' s center (m)
r  distance away from
G  Universal Gravitational Constant

Example: What is the orbital velocity of a satellite orbiting 2000 km above the Earth’s
surface?
Solution:
In this example, the mass of the world is equal to the Earth’s mass (M = 5.96 x 1024
kg) and the distance away from the Earth’s center is equal to the sum of the Earth’s
radius and the orbital height (r = 6378 km + 2000 km = 8378 km = 8.378 x 106 m). The
satellite’s orbital velocity will be:
v orb =
GM
=
r
(6.67 x 10 -11)  (5.96 x 10 24 )
=
8.378 x 10 6
3.98 x 1014
=
8.378 x 10 6
4.74 x 10 7
v orb = 6.89 x 10 3 m/s
To get a satellite into orbit, a rocket carrying the satellite needs to be powered for a
long
enough time so that it can get to the desired altitude (in this example, 2000 km)

while maintaining the desired orbital velocity (6,890 m/s).
20. The Hubble Space Telescope travels around the Earth in a near-circular orbit at an
altitude of 560 km. What is its orbital velocity?
21. How long does it take the HST to complete one orbit around the Earth?
After a satellite has been placed into an orbit around the Earth, the next step is to get
it out of its Earth-bound orbit and guide it through the solar system to the desired target.
Being by gravity to the Earth, the satellite follows the Earth in its orbit around the Sun.
Therefore, the satellite’s orbital properties as it moves around the Sun are the same as
Earth’s. The trick to sending a satellite to another world is to change the satellite’s orbit
to one that will take it to the desired world. Such an orbit is referred to as the Hohmann
Transfer Orbit (HTO). Figure 7.6 illustrates such an orbit for a satellite that is being sent
to a superior planet.
Earth’s Orbit
Hohman Transfer Orbit
Superior Planet’s Orbit
Figure 7.6: Diagram of the Hohmann Transfer Orbit (HTO).
To get a spacecraft onto the desired Hohmann Transfer Orbit toward a superior
planet, the satellite’s energy needs to be increased. This is accomplished by installing
miniature thrusters onto the satellite. When the timing is right, firing the thrusters in the
direction of the satellite’s motion around the Sun will increase the energy. This will
create an HTO that will take it toward the superior planets.
The HTO has an elliptical shape with the Sun located at one of the foci. From the
geometry of the orbits involved, the following properties of the HTO are deduced:
- the HTO perihelion distance  the Earth-Sun distance
- the HTO aphelion distance  the superior planet-Sun distance
- the HTO semi-major axis  the average of the perihelion and aphelion distances
22. A satellite is sent on the Hohmann Transfer Orbit illustrated in Figure 7.6 to study
the atmosphere of Jupiter. Determine the following properties of the HTO.
a. What would the HTO perihelion distance be equal to?
b. What would the HTO aphelion distance be equal to?
c. What would the HTO semi-major axis be?
23. How long will it take the satellite to reach Jupiter? (Hint: the satellite needs to
travel half of its HTO to reach Jupiter).
24. In addition to finding the correct Hohmann Transfer Orbit, what considerations
need to be made in terms of the position of Jupiter in its orbit?
25. In the case where scientists want to send a satellite to study Venus, determine
what would be different about the following aspects of the satellite’s journey:
a. What would the HTO perihelion distance be equal to?
b. What would the HTO aphelion distance be equal to?
c. What would the HTO semi-major axis be?
d. How would the energy of the satellite need to be changed in order to get it on an
HTO toward an inferior planet, like Venus?