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satellite met > chapter 2: Orbits and Navigation > Flashcards

chapter 2: Orbits and Navigation Flashcards

1
Q

The gravitational and astronomical laws were originally formulated to explain

A

the motion of planets in the solar system and their orbits around the sun.

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2
Q

…………………………were originally formulated to explain
the motion of planets in the solar system and their orbits around the sun.

A

The gravitational and astronomical laws

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3
Q

These laws (Newtonian and Keplerian) are equally applicable to

A

the orbits of artificial satellites placed around the earth.

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4
Q

…………………….the orbits of artificial satellites placed around the earth.

A

These laws (Newtonian and Keplerian) are equally applicable to

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5
Q

Newton’s Law of Universal Gravitation

A

The force of attraction between two point masses m1 and m2 separated by a distance r is

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6
Q

where G is

A

is theUniversal gravitation constant(6.67259x10‐11 N m2 kg‐2).

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7
Q

the following is

A

Newton’s Law of Universal Gravitation

The force of attraction between two point masses m1 and m2 separated by a distanceris:

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8
Q

Consider the simple circular orbit shown in Figure. Assuming that the Earth is a ……………………, we can treat it as ………………….

A

sphere, we can treat it as a point mass

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9
Q

The centripetal force is required to

A

keep the satellite in a circular orbit is rnv^2/r, here v is the orbital velocity of the satellite.

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10
Q

………………………………….. that balances this centripital force is ……………………..

A

The force of gravity (F)

Gmem/r2

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11
Q

Theforce of gravity (F) that balances thiscentripetal forceis Gmem/r2
,where me is

and m is

A

the mass of the Earth (5.97370x1024kg)

the mass of the satellite

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12
Q

Theforce of gravity (F) that balances thiscentripetal forceis Gmem/r2
,whereme is the mass of the Earth (5.97370x1024kg) and mis the mass of the
satellite. Equating the two forces gives:

A
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13
Q

Division bym eliminates the mass of the satellite from the equation, which
means that

A

the orbit of a satellite is independent of its mass

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14
Q

Orbital Period (T):

A

The time taken by a satellite to travel around its orbit once is known as the
period.

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15
Q

The period of an orbit simply depends on

A

its altitude

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16
Q

The period of the satellite is

A

the orbit circumference divided by the velocity:
T=2πr/v

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17
Q

orbital period =

A
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18
Q

For any given height above the Earth’s surface, a satellite will

A

take a fixed time
to complete an orbit, regardless of the mass of the satellite.

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19
Q

for …………………………………………… a satellite will take a fixed time to complete an orbit, regardless of the mass of the satellite.

A

For any given height above the Earth’s surface

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20
Q

Kepler’s laws of motion:

A
  1. law of orbits
  2. law of areas
  3. law of periods
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21
Q

Kepler’s laws of motion state that:

1.

A

Law of orbits: All planets travel in elliptical paths with the sun at one focus.

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22
Q

Kepler’s laws of motion state that:

2.

A

Law of areas: The radius vector from the sun to a planet sweeps out equal areas in equal times

  • This empirical law discovered by Kepler arises from conservation of angular momentum.
  • When the planet is closer to the sun,it moves faster, sweeping through a longer path in a given time.
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23
Q

Kepler’s laws of motion state that:

3.

A

Law of Periods: The ratio of the square of the period of revolution of a planet to the cube of its semimajor axis (radius of orbit) is the same for all planets revolving around the sun.

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24
Q

Kepler’s laws of motion

These laws are also applicable to

A

the artificial satellites placed in elliptical orbits (also called Keplerian orbits) around the earth at one focus.

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25
Q

define perigee

A

The point where the satellite most closely approaches the Earth

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26
Q

perigee more generally the

A

perifocus

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27
Q

apogee is also known as

A

apofocus

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28
Q

apogee or apofocus:

A

The point where the satellite is furthest from the Earth

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29
Q

semimajor axis is denoted by the symbol

A

a

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30
Q

semimajor axis:

A

The distance from the center of the ellipse to the perigee (or apogee)

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31
Q

the eccentricity (E):

A

The distance (aE) from the center of the ellipse to one focus (to the center of the Earth) divided by the semimajor axis (a)

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32
Q

For an ellipse, the eccentricity is a number between

A

0 and 1 (0 < E < 1)

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33
Q

A circle is an ellipse with

A

zero eccentricity

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34
Q

The equation for the elliptical path that the satellite follows, in polar coordinates with the Earth as origin is given as:

A
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35
Q

the angle O- is

A

the true anomaly and is always measured counter clockwise (the
direction of satellite motion) from the perigee.

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36
Q

The two basic types of meteorological satellite orbits are:

A
  1. Geostationary orbit – high altitude orbit over the equator
  2. Polar orbit – low altitude orbit from pole to pole
37
Q

Geostationary orbit:

A
  • A satellite in this orbit will remain stationary relative to the earth’s surface.
  • For any orbit to be geostationary, it must be geosynchronous and circular with zero inclination over the equator.
38
Q

A geosynchronous orbit is

A

any orbit with an orbital period equal to the earth’s rotational period.

39
Q

A geosynchronous satellite completes one orbit around the earth in the same
time

A

that the earth takes to make one rotation

40
Q

A geosynchronous satellite completes one orbit around the earth in the same time that the earth takes to make one rotation

This time period is known as …………………………. and is equivalent to ………………………….

A

onesidereal day and is equivalent to 23 h 56 m 04 s

41
Q

This time period is known as onesidereal dayand is equivalent to 23 h 56 m 04 s .It represents the

A

time taken by the earth to rotate on its axis relative to the stars, and is almost four minutes shorter than the solar day.

42
Q

To achieve this earth’sorbital period,the satellite should have

A

the same angular velocity as the earth.

43
Q

To achieve this earth’sorbital period,the satellite should have the sameangular
velocityas the earth. Based on the Kepler’s laws, to achieve this, the satellite
must be placed at a height of

A

35,786 km above the earth

44
Q

To achieve this earth’sorbital period,the satellite should have the sameangular
velocityas the earth. Based on the Kepler’s laws, to achieve this, the satellite
must be placed at a height of 35,786 km above the earth.

That is, to achieve the earth’s angular velocity of ……………………………………. the satellite altitude can be calculated from

A

ς = 7.292 x 10-5 s-1

45
Q

To ensure that a satellite remains over a particular point on the earth’s surface, the orbit must also be

A

circular and havezero inclination over the equator.

46
Q

However, a geosynchronous orbit has an inclination of

A

20 degrees

47
Q

However, a geosynchronous orbit has an inclination of 20 degrees, because of
which a geosynchronous satellite will move

A

north and south of the equator during its orbit while the geostationary satellite will not.

48
Q

Therefore, geostationary satellites are those which orbit with a period

A

equal to the earth’s rotational period and with zero eccentricity and zero inclination

49
Q

The theoretical coverage area of a geostationary satellite extends to about

A

40% of the Earth’s surface.

50
Q

………………………………..can provide full coverage of the Earth, except for ………………………….

A

A constellation of three equally spaced geostationary satellites

the polar regions

51
Q

As satellites in geostationary orbit continuously cover a large portion of the
Earth, this makes it an ideal orbit for

A

telecommunications or for monitoring continent‐wide weather patterns and environmental conditions

52
Q

As satellites in geostationary orbit continuously cover a large portion of the
Earth, this makes it an ideal orbit for telecommunications or for monitoring
continent‐wide weather patterns and environmental conditions.
It also ………….. cost as ……………………………………………………………

A

decreases

as ground stations do not need to track the satellite.

53
Q

Major disadvantage of this high altitude satellite is, of course,

A

a low resolution in the images (the typical resolution can be 1 to 4 km).

54
Q

Some of the operational geostationary satellites are:

A

GOES (US), METEOSAT (Europe) and INSAT (India)

55
Q

Polar orbit

A

The main objective of the satellites rotating in these low‐altitude, high inclination polar‐orbits is ‘global coverage’ with high spatial resolution.

56
Q

The polar orbiting satellites fly in ………….orbits

A

much lower orbits typically at around 850 km

57
Q

The polar orbiting satellites fly in much lower orbits typically at around 850 km, with the orbital plane at

A

an angle of about 80o to the equator.

58
Q

the period of these satellites is given by:

A
59
Q

Thus, these satellites circle the earth from

A

pole to pole once in less than 2 hours (T = 104 min) i.e., they circle the earth 14 times each day.

60
Q

As the earth rotates to the …………….. beneath the satellite, each pass ………………………………………to the ………… of the previous pass

A

east

monitors an area to the west of the previous pass.

61
Q

Each polar‐orbiting satellite can typically observe

A

the entire planet twice in 24‐ hours, once during daylight and once at night

62
Q

Each polar‐orbiting satellite can typically observe the entire planet twice in 24‐
hours, once during daylight and once at night, with …………………. resolution than the ………………………………….

A

better resolution than the geostationary satellites.

63
Q

The orbital plane of these satellites is…………………………, which means ……………………………………….. . hence, this orbit is also known as

A

fixed relative to the Sun

that the plane of the orbit keeps a constant angle with the sun throughout the year.

sun‐synchronousorbit

64
Q

The sun‐synchronous orbit ensures that the satellite

A

passes over a given location on the earth at the same local time each day.

65
Q

Some of the operational polar orbit satellites are:

A

TIROS‐N / NOAA‐A (US), METEOR‐2 (Russia), SPOT (France), ERS‐1 (Europe)

66
Q

Comparison of the characteristics of Geostationary and Polar‐orbiting satellites

A
67
Q

To define the position of a point on the celestial sphere (……………………..) we use the …………………………….. known as

A

an imaginary sphere of infinite radius surrounding the earth

celestial equatorial coordinate system

known as the Right Ascension‐Declination Coordinate System.

68
Q

Like terrestrial coordinates (i.e., latitude and longitude), the two coordinates to define a point on the celestial sphere are:

A

the declination and the right ascension.

69
Q

declination

A

The coordinate indicating where an object is between the celestial poles

70
Q

Declinationis measured

A

from the celestial equator

71
Q

……………………is measured from the celestial equator

A

declination

72
Q

Declinationis measured from the celestial equator

it extends from

A

0° at the celestial equator to +90° at the north celestial pole and from 0° at celestial equator to‐90° at the south celestial pole.

73
Q

right ascension

A

the second coordinate in the celestial equatorial system

  • It is analogous to longitude of the terrestrial coordinates.
  • Like Greenwich which is the arbitrary zero point for longitude, right ascension also has a zero reference point atVernal Equinox.
  • Right ascension, consequently, is measured in (sidereal) hours, 0h to 24h east from the Vernal Equinox Point. That is, east is the direction of increasing right ascension.
74
Q

equinoxes

A

The intersections of theecliptic(apparent path of the satellite) with the celestial equator

75
Q

Consequently, the vernal equinox point:

the autumnal equinox point:

A

Consequently, the vernal equinox point (RA = 00h 00m 00s) and the autumnal equinox point (RA = 12h 00m 00s).

76
Q

The position of the satellite at any given time can be correctly identified, if

A

a set of parameters (orbital elements) fully describe a satellite orbit, are known.

77
Q

The sixorbital elements, which completely determine the motion of a satellite are:

A
78
Q

The semimajor axis and eccentricity define the

A

size and shape of the ellipse

79
Q

right ascension of ascending node, inclination angle andargument of perigee combine to

A

position the ellipse relative to the Earth.

80
Q

These five parameters define the

A

shape and location of the orbital path

81
Q

mean anomaly:

A

“how far round” the ellipse the satellite lies at any given time

82
Q

Thus the first two parameters give the

A

dimensions of the ellipse

83
Q

the final three define the

A

position of the orbit relative to the Earth

84
Q

the mean anomaly indicates

A

the location of the satellite along the orbital path.

85
Q

Keplerian orbit

A

For a satellite in a truly elliptical orbit

86
Q

For a satellite in a truly elliptical orbit (also known as aKeplerian orbit)allthe
elements

A

remain constant, with the exception of the mean anomaly which changes with the motion of the satellite, and so increases with time.

87
Q

Satellite Tracking

A

Knowing the position of the satellite in orbit

88
Q

navigation

A
  • In addition to knowing where a satellite is in its orbit, it is necessary to know the Earth coordinates (latitude, longitude) of the particular scene it is viewing and calculating the Earth coordinates is known as thenavigation.
  • Navigation– Calculating the location (latitude and longitude) of the spot being sensed by the satellite.

satellite met (6 decks)

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