► The known type
**G M / r**^{2}
gives us the **acceleration** **g** where result from the force **F**g
of gravity at one body with mass **Μ,** that is to say
**g **=** G M / r**^{2} and in radius **r**.
Well. if we put the data of our planets we will find approximate the acceleration
that result the gravitational field in the radius **r**.

**g **=** G M / r**^{2} →
(6,6725 ×10^{-11} ) × (5,973 ×10^{24} ) / (6,3787
×10^{6} )^{2}
= 39,8548 ×10^{13} / 40,68781 ×10^{12} =
__9,795 m / sec__^{2}

► Now, we will find what acceleration
result from the same type g = G M / r^{2} when we put in the denominator the
middle radius of the Earth until the Sun
**r**ers
= 1,495978 × 10^{11}
m and with mass of Earth
**M****er**
= The mass of Earth:

**g **=** G M****er
/ r**ers^{2}
→
(6,6725 ×10^{-11} ) × (5,973 ×10^{24} )
/
(1,495978
× 10^{11}
)^{2} = **17,8086 ×10**^{-9} m/s^{2}

We find the same acceleration as from the formula Fers / Ms when we put mass of Sun in
the denominator
__and - attention- not mass of Earth__. That is to say

**g
**=** G M****er **/** r**ers^{2}
=
**F**ers
/ **M**s

**g **=** ****F**ers/
**M**s
→
35,4383 × 10^{21}
/ 1,99 ×10^{30}
= **17,808 ×10**^{-9} m/s^{2}

►
Now, with mass **Ms**
= mass of Sun :

**g **=** G M****s****
/ r**ers^{2}
→
(6,6725 ×10^{-11} ) × (1,99
×10^{30}
) /
(1,495978
× 10^{11}
)^{2} = **5,9332 ×10**^{-3} m/s^{2}

We find the same acceleration as from the formula Fers / Mer when we put mass of Earth in
the denominator
__and - attention- not mass of Sun__. That is to say

**g
**=** G M****s**
/** r**ers^{2}
=
**F**ers
/ **M**er

**g **= **F**ers / **M**er
→
35,4383 × 10^{21}
/ 5,973 ×10^{24} = **5,933 ×10**^{-3} m/s^{2}

We observe:
G M_{1} / r^{2} = F / M_{2}

► If the formula change in
**GM/r** its result in units is a __speed at square
____V__^{2}. With root on the result we find the clear speed **V**, ** V **
=**
√GM/r**. The **speed of rotation in orbit**
for
a satellite around of earth is given from this relation.

From the same formula V = √GM/r
we find **the speed of Earth orbit** around the Sun at the middle radius :
r = 1,495978 × 10^{11}
m

**V** = **√(G
M****sun
/ r ****)**
= √(
13,2782 ×10^{19} / 1,495978 ×10^{11} )=
√8,875982 ×10^{8} →

**V** = __2,979 ×10__^{4}
m/sec (This is the middle speed of Earth around the Sun).

This same
speed V=2,979 ×10^{4}
of the formula V = √G Msun
/ r we find from the formula √ **F**ers x **r**ers
/ **M**er
when we put the other mass (of Earth **Μ**er
and not of Sun). That is to say :

√**(G**
**M**sun
/ **r**ers
**)** = √**(F**ers x **r**ers
/ **M**er
**)** = V

(because G Mer
Ms / r = Fers
r = E )

► Also, we can
find the speed of the Earth
orbit from the simple type **V = r 2π / Τ** when we know the distance **r 2π** from
the central mass and time **T **of one complete rotation.

►
The formula
V = √ (2 G M / r )
give the
speed of a body that is
require so as to escape from the gravitational field of one body mass **Μ** and with
radius **r** and this speed called in physics "**Escape
velocity**". For example, from mass of Earth is result :

V = √(2 G 5,973 ×10^{24} / 6,3787 ×10^{6} ) = √(79,709685
×10^{13} / 6,3787 ×10^{6} )→

V =
√12,496227 ×10^{7} =
__11,178 × 10__^{3} m/sec

►
According to
Newtonian physics,
we found the force F between mass of
the Earth and Sun
and the speed of the Earth in the orbit (centripetal speed
V=2,979 ×10^{4>}
m/sec) around of the Sun. We know
their middle distance and the length of the orbit, where the Earth follow r × 2π = 9,4 × 10^{11}
m

The known formulas, allow us to find the period of one
complete orbit of the Earth around of the Sun, time until come one
rotation. This time, we find from the formula T = r
×
2π / Vκ

**T = r
×
2π / V****κ**
→ 9,4 × 10^{11}
/ 2,979 ×10^{4}
= 3,155421 × 10^{7}
sec

The result is in seconds, because we use a speed in
meters per second and the length of the orbit in meter, again. If we convert the seconds in
days, then we will see the known number of the days of one earthly year.

3,155421 × 10^{7}
sec / 60 / 60 / 24 = 365,21 days (in 24hours). The small divergence from the accurate result,
here do not engross us.

► **The force F** between the Sun and the Earth
is a centripetal force and this is expressed from the formula :

F = M V^{2}
/ r → F = 5,973
×10^{24}
(2,979 ×10^{4}
)^{2}
/ 1,495978 × 10^{11}
= 35,43303 ×10^{21}
N

We find __the same force in N__
as from the Newton's type with 2 masses. Observe that in type F = M V^{2}
/ r
we put mass of Earth M that finds in orbit of radius r with speed V and not the central mass
of the Sun.

The relation
**V**^{2}**
/ r **=** a **=**g**
as the **F / M**er

We can write
**F **=** M V**^{2}**
/ r **=** M g**

With force F that we find from the formula F = M V^{2}
/ r we can find the second mass M_{2} , where that is applied or would be applied
bilaterally the above force F. It will need the Newton's type :

**M**_{2} =** F r**^{2}**
/ G M**_{1} → 35,43303 ×10^{21}
× (1,495978 × 10^{11} )^{2}
/ G 5,973 ×10^{24} = __1,9896 ×10__^{30}__
kg__ (The mass of Sun)

We put the numbers that we know about mass M of the
Earth, speed V in the orbit and the distance r from the sun and with the last type, we found the
central mass of Sun. With the same formulas we can find any other second mass, when we imagine any
mass in orbit, with any speed and radius.