To explain further, let us get the current definitions of second and
kilometer as per International System of Units (SI):
Second is measured by the duration of 9,192,631,770 periods of the
radiation corresponding to the transition between two hyperfine
levels of the ground state of caesium133 atom at rest at a
temperature of 0°K. This value of second, when defined in terms of
the period of revolution of the Earth around the Sun, is equal to
the fraction 1/31,556,925.9747 of the tropical year for 1900 January
0 at 12 hours ephemeris time.
Kilometer is the length of the path traveled by light in vacuum
during a time interval of 1/_{299 792 .458} of a second.
The velocity of light in vacuum is 299,792.458 kilometers/second.
There are two points to be noted:
1) Either the velocity of light was fixed at 299,792.458
kilometers/second by which the kilometer was measured or the
kilometer was fixed by the length of a specially designed rod by
which the velocity of light was measured.
(The International Bureau of Weights and Measures (BIPM) created a
prototype bar in 1889 at the first General Conference on Weights and
Measures (CGPM), establishing the International Prototype Meter as
the distance between two lines on a standard bar composed of an
alloy of 90% platinum and 10% iridium, measured at the melting point
of ice. The original international prototype of the meter is still
kept at the BIPM under the conditions specified in 1889).
2) While Second is measured in terms of revolution of the Earth
around the Sun as well as its rotation on its own axis (one frame),
Kilometer is either measured by an arbitrarily accepted value of the
velocity of light or by an arbitrary length of a specially designed
rod or 10^{3} of the length of the Second’s Pendulum or 10^{4} of the
Earth’s meridian along a quadrant (another frame).
Irrelevancy of Unit
In the Chapter III of the book, we have stated that the orbital
velocity of a cosmic body remains more or less uniform in short
periods.
In the Diagram, ABC is the orbital path traversed by the Earth
around the Sun. Let us divide the path into 31,556,925.9747 parts,
where 31556925 parts are exactly equal to one another and the
remaining part is equal to 0.9747 of any of the other parts.
When the Earth completes one revolution around the Sun, let's say,
the Earth takes a time whose value is 31,556,925.9747 parts of ABC
and covers a distance whose value is 31,556,925.9747 x 29.834379
parts of ABC.
This means that: Unit time = 1 part of ABC and Unit distance =
29.834379 parts of ABC. So, Unit Distance / Unit Time = 29.834379
which is the orbital velocity of the Earth around the Sun.
Since the Earth makes 365.2421987 rotations on its axis while
completing 1 revolution around the Sun, the time taken by it for 1
rotation = (31,556,925.9747 ÷ 365.24219870) = 86400 or (24 x 60 x
60) parts of ABC and the distance covered by it in 1 rotation along
its orbital path is (86400 x 29.834379) or 2577690 parts of ABC.
In other words, time is counted when the Earth moves from one point
to another in its orbital path. So we can identify the position of
the Earth in the orbit either by time or by distance. In fact, time
was zero at the Origin of the Cosmos. We began to count it when ppS6
began to move in its projectile path. Therefore, there is no
difference between the nature of time and that of distance when
these are measured in the same frame. If the system comprising of
the Earth, its orbital path and the Sun is treated as a big clock,
we can say that a certain amount of displacement of the Earth
creates a certain amount of time. Either of these amounts which are
numbers or parts of ABC, signifies the displacement of the Earth in
its orbit or the time taken for such displacement. In your
wristwatch too, you understand the time when the hands undergo
displacement from one set of points to another.
Since we have been taught through text books that values of distance
and time should be accompanied by their units viz. km and second, we
have developed a mindset that distance = 1 or time = 1 gives no
clear meaning. To our minds, distance = 1 may mean 1 km or 1 meter
or 1 centimeter. Similarly, time = 1 may mean 1 second or 1 minute
or 1 hour. But these divisions of distance and time have been
created by us to minimize the occurrence of large or
fraction/decimal values in calculations. We use nanometer as unit
for microscopic distances and lightyear or parsec as unit for
astronomical distances. We also use nanosecond as unit in cycle time
for radio frequencies and year as unit in revolution time of the
Earth around the Sun. While these units are used in specific fields
of science and in the micro or macro level of the Cosmos to avoid
large or decimal values in calculations, a single unit for distance
or time may be used at all levels of the Cosmos. In the book, we
have used km as unit of distance for calculations pertaining to
radius of a primary particle as well as distance between two cosmic
bodies. While the radius of ppS_{6} is of the order of 10^{26} km, the
present distance between S_{6} and S_{5} of our Universe is about 8 x 10^{23}
km. Similarly, we have used second as unit of time for calculations
of time for creation of ppS_{6}, which is 10^{52} second and also for the
lifespan of the Cosmos, which is 9.67458816 x 10^{21} seconds.
We have used units of distance and time in the book to avoid
confusion in the minds of the readers who believe that values of
distance and time have no meaning unless they are expressed with
units. But we have shown earlier that distance, time and velocity
can be expressed by numerical values only and that units are
irrelevant in any expression if distance and time are defined with
respect to the same frame of reference. In that case, time can be
expressed by square of the radius as has been done in the Equation
(10). But if, for the sake of convenience, we use km and second as
the units of distance and time in all expressions, they should carry
the same units irrespective of their relationship (as has been done
in the book). The relationship R_{6} = 1 / V_{5} in the Equation (8)
states that the value (or magnitude) of R_{6} is equal to the inverse
of V_{5} . Since the concept of units is unnecessary, irrelevant and
illogical, such relationship is understandable and true.

Prof. H. C. Bhatt, Dean (Academic) Indian Institute of
Astrophysics on 3.2.2010:
1) It may be useful to see if the proposed theory is consistent with
the basic conservation laws of physics like energy and momentum.
Answer:
This comment necessitates differentiation between two theories  the
Big Bang theory and the one mentioned in the book.
Big Bang theory:
It is observed that the distance between galaxy clusters has been
increasing. So it can be inferred that they were closer together in
the past. Following the laws of conservation of energy, this
inference has been considered back in time to extreme densities and
temperatures and has led to the assumption, known as Big Bang theory
that the Universe has expanded from a primordial hot and dense
initial condition around 13.3 to 13.9 billion years ago.
Such extreme conditions have not yet been created and probed in any
machine including particle accelerators. Since there is no evidence
associated with the earliest instant of the expansion, the Big Bang
theory is unable to provide any explanation for such an initial
condition and gives a general description of the evolution of the
Universe since that instant.
Theory as stated in the Book:
A tiny primary particle, termed as ppS_{6}, is created at the Origin.
Having enormous rotational velocity, it creates rings around its
base and an equivalent quantity of dark particles from nothing (no
particle). The rings are broken and the broken parts of the rings
are transformed into primary particles. As ppS_{6} traverses a
projectile path, it continues to create primary particles after
every 2 x 10^{52} second during its ascent. During descent of ppS_{6} to
the originating base, the primary particles around it, are
transformed into rings which together with an equivalent quantity of
dark particles, undergo complete dissolution.
Although the expansion and contraction of the Cosmos have been
described in the Chapter I to give the readers an idea about its
shape, this phenomenon or the characteristics of the Cosmos have
been supported by the Equations (10) and (16) derived in the Chapter
IV. So the Creation and Dissolution processes of the Cosmos are
true. Since the Cosmos appears and expands from zero state or no
particle and then undergoes complete dissolution, the laws of
conservation of energy do not hold good for such happenings.
In the Chapter IV page 19, the initial orbital velocity of a primary
particle has been defined as V^{2}D (= K, a constant), which is equal
to the rotational velocity of the parent primary particle. So, the
rotational velocity of a primary particle remains constant during
the life of the Cosmos although it constantly creates rings having
rotational velocities. Hence the laws of conservation of momentum
are not followed by a rotating primary particle.
The Wheels (atoms of elements), described in Chapter XIX page 85,
neither have parent primary particle at their common center of
rotation nor they create primary particles or rings. In a chain
reaction (described in page 8889), a Wheel (X) transfers a part of
its velocity to another Wheel (Y) with lesser velocity when X
strikes Y in the direction of Y's motion. Consequently, a Sound
Wheel may be transformed into a Heat wheel and a Heat wheel may be
transformed into a Light Wheel. Therefore, the laws of conservation
of momentum and energy hold good for such actions of atoms of
elements, which we experience on Earth.

2) All motions are not Periodic (orbital). There are chaotic
motions too.
Answer:
You will find in the Chapter III that the motions of all cosmic
bodies around their respective parent cosmic bodies are periodic.
All of them move in circular orbits like the bobs in conical
pendulums. The recent observation is that the Sun orbits around the
Milky Way galaxy in zigzag path. This is true for different reasons.
As described in the Chapter I and derived in subsequent chapters, in
the Milky Way galaxy (Diagram A), the Sun orbits around a star,
named by QS_{2} which orbits around another star called QS_{3} which too
orbits around a cosmic body (black cosmic body or black hole 
please refer to Chapter XIII, page 60) called QS_{4}. QS_{4} is the
central cosmic body of the Milky Way galaxy. So in time t, as QS_{3}
moves from position A to position B on its orbital path around QS_{4},
QS_{2} moves from position C to position D on its orbital path around
QS_{3} and the Sun completes several revolutions around QS_{2}. In other
words, if the galaxy is observed during time t from a distance (or
from the Earth), the Sun is seen to orbit around the center of the
galaxy by undergoing revolutions around QS_{2} and simultaneous
displacements along CD and AB. This gives us an impression that the
Sun orbits around the Milky Way galaxy in zigzag path. For similar
reasons, all S_{1}s and S_{2}s of a galaxy are seen to traverse zigzag or
chaotic paths around the center of the galaxy. On the other hand,
all S_{3}s of a galaxy are seen to traverse circular paths around the
center of the galaxy. So by observing the paths over a period of
time, we can identify whether a cosmic body belongs to S_{1}, S_{2} or S_{3}
group.
The rotational axes of all cosmic bodies are directed towards the
center of ppS_{6}. So the continuous changes in the directions of the
rotational axes of Earthlike bodies caused by their orbital motions
along the polar plane of the Cosmos result in continuous climatic
changes (refer to pages 106 – 108, Chapter XX) and fluctuation in
the population of living creatures in them.

Dr. Saradindu Sengupta, ExProfessor, IIT Kharagpur, India on
28.3.2010:
1) It is not clear why formation of the number of layers of rotating
galaxy is limited to 7.
Answer:
Please refer to the paragraphs under the title "Number of Subrings
in a set" in Chapter XIV. It has been derived that in the gap
between two primary rings, maximum number of subrings having
specific dimensions (limited by Equation 35) formed is seven.
Wherever there is a tendency for formation of more than seven
subrings, the breakage of some of the subrings takes place. Because
of that, in solar system, we find Asteroids and Kuiper Belt objects.
The primary particle formed from the subring of a primary ring
creates an S_{4} (the central cosmic body of a galaxy). Since seven
subrings are formed in the gap between two primary rings, seven S_{4}s
are formed. So we have seven galaxies in a Universe. All galaxies in
a Universe orbit around an S_{5}.

2) What is the basis of assuming that the plane of rotation forms
a cone?
Answer:
There is no room for assumption in Science. Every argument in the
book is supported by derivation(s). The planes of orbital paths of S_{5} form a cone. I suppose
you have asked the question by referring to Chapter XVII.
By Equation (16),
Time t ∝ D_{5}^{2}, where D_{5} is the distance between S_{5} and S_{6}.
This means that during the Creation process, S_{5} moves away from S_{6}
while traversing an orbital path around the latter.
Referring to Diagram 41, we find that the line joining the centers
of S_{5} and S_{6} forms an angle (∠AOD) with the equatorial axis of S_{6}.
Since by Equation (16), square of both polar and equatorial
distances of S_{5} from S_{6} are proportional to time, the ratio of these
distances is a constant. But this ratio is equal to tan ∠AOD.
Therefore, ∠AOD is a constant. This means that S_{5} moves away from S_{6}
along the line OA. Since S_{5} orbits around S_{6}, its orbital paths form
the shape of a cone (please refer to Diagram 5 in page 3).

3) Why is the shape of the Cosmos a hemisphere?
Answer:
While S_{5}s orbit in planes which are parallel to the equatorial plane
of S_{6} (Diagram 15), the S_{4}s, S_{3}s, S_{2}s, S_{1}s, planets and satellites
orbit in planes which are not exactly parallel to the polar plane of
S_{6} but are slightly inclined towards the same ( Diagrams 18 and 19).
Since there are three nearly identical Universes in every layer of
Universeformations (please refer to page 15  'Number of
Universes'), the inner formation of the Cosmos looks like a
nearlyclosed flower (Diagram 1 in page 1).
As stated in pages 5 and 6, the formations of a primary particle and
its rings are accompanied by the formation of an equivalent amount
of dark particles around them. Since the inner formation of the
cosmos resembles to a nearlyclosed flower (or spheroid), the dark
particles form layers around the said formation. The formation of
dark particles, therefore, takes the shape of a spheroid. In other words, the dark particles,
after formation, move in all directions but remain attracted towards
S_{6} because of the latter's gravitational effect (please refer to
page 8  'Does the primary particle attract the dark particles?')
and therefore, the Cosmos, as a whole, looks like a spheroid (please
also refer to Chapter XVIII).
 Prof. Achyuta Kumar Chakraborty, formerly of City College, Kolkata,
India on 23.04.2010
From the present theories and experiments, we know that: 'g' of a body decreases below the Earth's surface as the body travels towards the center of the Earth. We further know that 'g' of a body decreases above the surface of the Earth as it moves away from its surface. This means that as per present theories, 'g' is maximum on the surface of the Earth and is zero at its center and also at an infinite point above its surface. In other words, 'g' of a body is zero at an infinite point above the surface of the Earth, increases as the body travels towards the Earth's surface, is maximum on the Earth's surface and decreases as the body travels from the Earth's surface towards its center.
It is stated in Chapter II page 8 of the book that the rotational velocity of the primary particle at the center of a cosmic body is responsible for the gravitational attraction towards all particles around it. It is further stated in Chapter IV page 19 that gD² = K, a constant (where D is the distance between the center of a body and that of the Earth) is true for the body situated below, on or above the surface of the Earth. By this relationship, g
∝ 1 / D², which means that 'g' of a body is zero at an infinite point above the Earth's surface but constantly increases as the body travels from that infinite point towards the center of the Earth. Accordingly, 'g' is maximum at the center of the Earth.
Which one is true?
Answer:
In the Chapter IV page 19, gD^{2} = K is deduced from the following three relationships that we presently know in respect of planetary motion:
1) D^{3} = kT^{2
}
2) T = 2πD / V
3) g = V^{2} / D
Where D is the distance between the center of a planet and that of its orbital path, k is a constant, T is the orbital time period of the planet, V is the orbital velocity of the planet and g is the gravitational acceleration or retardation on the planet.
If the above three relationships are true, gD^{2} = K is also true. In other words, g ∝ 1 / D^{2} is true.
Therefore, 'g' of a body is zero at an infinite point
above the surface of the solid part of the Earth but constantly
increases as the body travels from the infinite point towards the
center of the Earth. Accordingly, 'g' is maximum at the center of
the Earth.
'g' inside the solid part of the Earth
As stated in Chapter II, page 8 (Topic: Gravitation): The rotational
velocity of a primary particle creates a strong field of attraction
in the surrounding area. Any material or dark particle situated in
that area is attracted towards the primary particle by the shortest
distance.
Therefore, the large field of attraction is a unique characteristic
of a parent primary particle, which ultimately causes dissolution of
all material and dark particles situated within its field of
attraction, during the 'Dissolution process'. An atom, on the other
hand, is a bunch of rotating ringsystem having no primary particle
at the common center. Hence, it does not, under normal conditions,
create a field of attraction beyond its tiny 'cosmoslike' formation
(refer to Chapter XIX). In other words, beyond the surface of this
formation, the field of attraction of the unexcited atom does not
exist. Therefore, an unexcited atom does not attract another
unexcited atom situated beyond the surface of its 'cosmoslike'
formation. Besides, the fields of attraction of two atoms do not
combine to create a larger field of attraction because all unexcited
atoms exist as separate entities and their fields of attraction are
limited to the boundaries of their 'cosmoslike' formations. So a
body situated inside the solid part of the Earth is not attracted by
the numerous atoms above it. Nor is it attracted by the numerous
atoms below it. The gravitational attraction to the body is caused
by the rotating parent primary particle at the center of the Earth
by creating a large gravitational field. As discussed in the Chapter
VII (Topic: Role of Rotational Velocity) page 34, the magnitude of
gravitational attraction of a cosmic body depends on the magnitude
of the rotational velocity of the primary particle at its center.
Therefore, as a body approaches the center of the Earth, its g
increases.
Measurement of g
If K is known, we can measure the value of acceleration due to gravity at any point on, above and below the
surface of the solid part of the Earth. For example:
Moon's distance (D) from the Earth = 384400 km and the orbital velocity (V) of moon = 1.02 km/sec. Therefore, Moon's K = V^{2}D = 4 x 10^{5} km^{3}/sec^{2}. In other words, K of a body situated within the field of gravitational attraction of the primary particle of the Earth is 4 x 10^{5} km^{3}/sec^{2}.
If g_{1} is the acceleration due to gravity on the
surface of the solid part of the Earth and R is the radius of the solid part of the Earth, g_{1}R^{2} = K
K = 4 x 10^{5} km^{3}/sec^{2}
R = Radius of the solid part of the Earth = 6378 km.
Therefore, g_{1} = K / R^{2} = (4 x 10^{5}) / 6378^{2} km/sec^{2} = 983.3111449 cm/sec^{2}.
Or, the value of 'g' on the surface of the solid part of the Earth is around 983 cm/sec^{2}.
(The surface of the solid part of the Earth is visible to us. So we call this
surface of the solid part as the surface of the Earth. But above this visible
surface of the solid part, there is a large invisible airspace. The air in the airspace and the solid and liquid substances are all material particles and together form the Earth. Therefore, the outer surface of the exosphere or a thin layer of airspace beyond that should ideally be called the surface of the Earth).
If g_{2} is the acceleration due to gravity at point X which is 1 kilometer below the
surface of the solid part of the Earth and R_{1} is the distance between X and the center of the Earth, g_{2}R_{1}^{2} = K
K = 4 x 10^{5} km^{3}/sec^{2}
R_{1} = (R  1) km = (6378 1) km = 6377 km.
Therefore, g_{2} = K / R_{1}^{2} = (4 x 10^{5}) / 6377^{2} km/sec^{2} = 983.6195621 cm/sec^{2}.
So, the difference between 'g' at a point which is 1 km below the
surface of the solid part of the Earth and that on the solid surface of the Earth = (983.6195621  983.3111449) cm/sec^{2} = 3 mm/sec^{2} which is very small. Such small variation in the value of 'g' can not be noticed if an experiment is conducted at a point which is one km or less below the
surface of the solid part of the Earth. Besides perfect laboratory conditions cannot be secured at such positions.
 Prof. Achyuta Kumar Chakraborty, formerly of City College,
Kolkata on 25.07.2010:
1) In the Chapter III, page 17 of the
book, you made a statement: "All cosmic bodies of the Cosmos orbit
in circular paths like conical pendulums to remain at their
respective positions at a given time without falling to or escaping
from the gravitational attraction of S_{6}". But we know
that the artificial satellites can orbit in both circular and
elliptical paths around the planets. Therefore, a cosmic body can
orbit around its parent cosmic body in an elliptical path. This has
been proved through observation of the solar system by the
astronomers. Besides it is the Sun's gravity that keeps all solar
planets in their respective orbits.
Answer
Please refer to the time of creation of ppS_{6} and its
ringsystems at the Origin of the Cosmos (Chapter II). At that time,
all primary rings and subrings had been orbiting around ppS_{6}
under the gravitational pull of the latter. When the primary
particles of all cosmic bodies were formed from the breakage of the
primary rings and subrings, they must have been still under the
gravitational pull of ppS_{6}. It will be illogical to say
that after the creation of the primary particles, this gravitational
pull by ppS_{6} towards other primary particles came to
naught. We know that the S_{5}s are still orbiting around
the S_{6} under the latter's gravitational pull because the
primary particles of the S_{5}s, after formation, were still
under the gravitational pull of ppS_{6}. Similarly, the primary
particles created from the subrings too were still under the
gravitational pull of ppS_{6}.
Since a primary particle has the unique property of extending its
gravitational field beyond its surface, the primary particles of the
solar planets also came under the gravitational attraction of the
primary particle of their parent, the Sun, and began orbiting around
it. But the planes of their orbital paths faced the center of ppS_{6}
because they were still under the latter's gravitational pull. Had a
primary particle created from the subring of a primary ring not
been under the gravitational pull of ppS_{6}, it would have
been orbiting the primary particle created from the primary ring in
an orbital path which was parallel to that of the primary particle
created from the primary ring. The orbital paths of the primary
particles created from the subrings face the center of ppS_{6}
so that the gravitational pull of ppS_{6} towards them
remains uniform at all positions of their orbital paths. Therefore,
all cosmic bodies orbit around their respective parent cosmic bodies
like conical pendulums due to the gravitational attraction of S_{6}
 the attraction that is stretched out to all parts of the Cosmos.
At the same time, they are also attracted towards the centers of
their respective parent bodies as well as those of all other cosmic
bodies in their vicinities.
By referring to the Diagram (23A), page 17, Chapter III, we find
that the orbital paths of the solar planets are not in the same
plane. Therefore, the Sun is not exactly situated at the center of
their orbital paths. If that is so, the distances of the Sun from
different positions of a solar planet in its orbital path should be
different. So the question is whether the astronomers have drawn the
orbital path of the Earth by determining its distances from the Sun
for all adjacent positions in its orbital path or they have drawn it
by determining its distances from the Sun for the four positions
being four ends (on the Earth's orbital path) of two mutually
perpendicular lines through the center of the Sun.
However, since the derivations in the book are principally based on
the three experimentally proved equations viz. (1) D^{3} =
kT^{2} (2) T = 2πD / V and (3) g = V^{2} / D (refer
to page 19, Chapter IV) which are true for both circular and
elliptical orbits of the cosmic bodies, the derivations too are true
for both circular and elliptical orbits of the cosmic bodies.
2) In page 31, Chapter VII, you have mentioned that the
present distance of the Jupiter from the center of its orbital path
= 806513641 km and that the present orbital velocity of the Jupiter
= 12.90755952 km/sec. These values are higher than those determined
by the astronomers. Most of your subsequent deductions are based on
these two values. What are the reasons for these variations? How do
they affect your deductions?
Answer
As you have found at the end of the answer to your query (1) dated
25.07.2010, I raised a question about the position of the Sun around
the center of the orbital path of the Earth or any other solar
planet. These orbital paths are required to be drawn by determining
the distances of the solar planets from the Sun on a daily basis for
a year. Besides I am not convinced about the accuracy of the
measurement of such distance considering the fact that the light
rays from the distant planets and the Sun are deviated by the
different layers of dark particles and atmospheres and that the
orbital paths of the planets are in different planes, howsoever
close. Therefore, the distance of the Jupiter from the center of its
orbital path, being equal to 806513461 km, is not same as the
semimajor axis of its orbital path, being equal to 778547200 km as
determined by the astronomers. Interestingly, the value of the
present distance of the Earth from the center of its orbital path,
being 149841401 km (as mentioned in page 72, Chapter XVI) is very
close to that of the semimajor axis of its orbital path, being
149,598,261 km as determined by the astronomers (despite the fact
that the distancetime scale has been slightly altered in the book
by taking the value of the velocity of light as 299298.38 km/sec).
The reasons for the nearness of these values: the Earth is closer to
the Sun than the Jupiter, its distance from the Sun is directly
measured from a position on it and its plane of orbital path is
nearer to the Sun than the plane of the orbital path of the Jupiter.
The value of 806513461 km, being the distance of the Jupiter from
the center of its orbital path, works well and accurately for the
determination of time of commencement of the dissolution of life in
the Cosmos (page 99100, Chapter XX). As you may find  this value
has a direct link with the value of the radius of the Sun. It does
not affect any other deduction / calculation.
3) While answering to my query dated 23.04.2010, you
determined the Earth's surface gravity as 983.31 cm/sec^{2}
whereas the actual equatorial surface gravity as determined by the
scientists is 978.0327 cm/sec^{2}. What does this difference
signify?
Answer
As experimentally determined by the scientists, the Earth's
equatorial surface gravity = 978.0327 cm/sec^{2} and its
equatorial radius = 6378 km.
So, K of the satellite (Moon) of the Earth
= gD^{2 }
= 0.009780327 x 6378^{2}
= 397853 km^{3}/sec^{2}
Moon's orbital velocity V = 1.022 km/sec.
So, the distance of the Moon from the center of its orbital path
= K / V^{2} (Since V^{2}D = K)
= 397853 / 1.022^{2}
= 380908 km.
But as determined by the scientists, the length of the semimajor
axis of the Moon's orbital path = 384399 km.
Therefore, the center of the orbital path of the Moon as determined
from the formula V^{2}D = gD^{2} = K is not the same
as the one determined by the scientists. This further confirms my
doubt regarding the presently established orbital paths of the
cosmic bodies as raised in my answers to your query (1) and (2).
Besides, while answering to your query dated 23.04.2010, I
determined the Earth's surface gravity as 983.31 cm/sec^{2}
because the distance of the Moon from the center of its orbital path
was taken as 384400 km (length of the semimajor axis of the Moon's
orbital path) whereas the actual distance as calculated above =
380908 km.
4) What should the exact orbital velocity of an artificial
satellite be around the Mercury if the satellite is designed to
orbit at a distance of 200 km from the surface of the solid part of
the Mercury?
Answer
An artificial satellite's orbital velocity around a solar planet at
a given height can be exactly determined by knowing the surface
gravity and radius of the solid part of the planet. In other words,
if the K of a satellite of the planet is known, the orbital velocity
of an artificial satellite can be determined at a given height.
Since the Earth, Mars, Jupiter, Saturn and Uranus have natural
satellites around them, we can determine the surface gravities of
these planets and Ks of their satellites. Therefore, we can estimate
the orbital velocity required by an artificial satellite of each of
these planets at a given height. The Mercury and the Venus do not
have natural satellites around them. So we can not measure their
surface gravities. The only option is to put an artificial satellite
in a circular path around Mercury / Venus by employing maneuvers and
then measure its orbital velocity and distance from the center of
the planet to determine K.
The Equation (31), page 32, Chapter VII describes relationship among
the Ks of the satellites of the solar planets. It works well for the
outer planets Jupiter, Saturn and Uranus. Successive Venus probes
by various Space Agencies estimated the surface gravity of the Venus
to be 887 cm/sec^{2}. With this value, K of the satellites
of the Venus = gD^{2} = 0.00887 x 6051.8^{2} km^{3}/sec^{2}
= 3.24857 x 10^{5} km^{3}/sec^{2}, which is
close to the value (4.0310784 x 10^{5} km^{3}/sec^{2})
drawn from the Equation (31). It is to be noted that anomaly in the
value of the Venus's gravity was reported during these probes. This
might have led to continuous adjustments of the orbital paths of
these probes through maneuvers to keep them around the Venus. The
surface gravity of the Venus, as per the Equation (31) = K / D^{2}
= 4.0310784 x 10^{5} / 6051.8^{2} or 1100 cm/sec^{2}
 more than that of the Earth. So, as per this value, ESA's Venus
Express could have performed better had it orbited the Venus with an
average orbital velocity of 3.08356 km/sec at an Apoapsis altitude
of 72300 km and a Periapsis altitude of around 400 km for a 24 hour
orbit. If the Equation (31) holds good for the Venus, it is true for
the Mars as well.
I am looking forward to the Messenger's entry into an orbit around
the Mercury in 2011. We shall then be able to determine K of the
satellite of the Mercury by knowing the orbital velocity of the
Messenger and its distance from the center of its orbital path. We
shall also be able to find out the surface gravity of the Mercury.
Using the Equation (31), the Mercury's surface gravity is seen to be
around 67700 cm/sec^{2}  about 69 times that of the Earth.
It is unbelievable!
Finding relationship among Ks in a set of cosmic bodies is very
important to determine the parameters of all cosmic bodies of our
Universe. The Messenger probe can prove or disprove the Equation
(31).
