► The known type
G M / r^{2}
gives us the acceleration g where result from the force
Fg
of the 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 x10^{11} ) x (5,973 x10^{24} ) / (6,3787
x10^{6} )^{2}
= 39,8548 x10^{13} / 40,68781 x10^{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
rers
= 1,495978 x 10^{11}
m and with mass of Earth
Mer
= The mass of Earth:
g = G Mer
/ rers^{2}
→
(6,6725 x10^{11} ) x (5,973 x10^{24} )
/
(1,495978
x 10^{11}
)^{2} = 17,8086 x10^{9} m/s^{2}
We find the same acceleration as from the type
Fers
/ Ms
when we put the mass of Sun in
the denominator
and  attention not the mass of Earth. That is to say
g = G Mer
/ rers^{2}
=
Fers
/ Ms
g =
Fers
/ Ms
→
35,4383 x 10^{21}
/
1,99 x10^{30}
= 17,808 x10^{9} m/s^{2}
►
Now, with mass
Ms
= the mass of Sun :
g = G Ms
/ rers^{2}
→
(6,6725 x10^{11} ) x (1,99
x10^{30}
) /
(1,495978
x 10^{11}
)^{2} = 5,9332 x10^{3} m/s^{2}
We find the same acceleration as from the type
Fers
/ Mer
when we put the mass of Earth in
the denominator
and  attention not the mass of Sun. That is to say
g = G Ms
/ rers^{2}
=
Fers
/ Mer
g =
Fers
/ Mer
→
35,4383 x 10^{21}
/
5,973 x10^{24} =
5,933 x10^{3} m/s^{2}
We observe:
G M_{1} / r^{2} = F / M_{2}
► If the type 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 type V = √GM/r
we find the speed of Earth orbit around the Sun at the middle radius :
r =
1,495978 x 10^{11}
m
V = √(G
Msun
/ r )
= √(
13,2782 x10^{19} /
1,495978 x10^{11} )=
√8,875982 x10^{8} →
V
= 2,979 x10^{4}
m/sec (This is the middle speed of Earth around the Sun).
This same
speed V=2,979 x10^{4}
of the type V = √G Msun
/ r we find from the
type √
Fers
x
rers
/ Mer
when we put the other mass (of Earth Μer
and not of Sun). That is to say :
√(G Msun
/
rers
) = √(Fers
x
rers
/ Mer
) = 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 the time T of one complete rotation.
►
The type
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 the physics "Escape
velocity". For example, from mass of Earth is result :
V = √ (2 G
5,973 x10^{24}
/ 6,3787 x10^{6} ) = √(79,709685
x10^{13} / 6,3787 x10^{6} )→
V
=
√
12,496227 x10^{7} =
11,178 x 10^{3} m/sec
►
According to
Newtonian physics,
we found the force F between the mass of
the Earth and Sun
and the speed of the Earth in the orbit (centripetal speed
V=2,979 x10^{4}
m/sec) around of the Sun. We know
their middle distance and the length of the orbit, where the Earth follow r x
2π = 9,4 x 10^{11}
m
The known types, allow us to find the period of one
complete orbit of the Earth around of the Sun, the time until come one
rotation. This time, we find from the type T = r
x
2π / Vκ
T = r
x
2π / Vκ
→ 9,4 x 10^{11}
/ 2,979 x10^{4}
= 3,155421 x 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 x 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 type :
F = M V^{2}
/ r → F = 5,973
x10^{24}
(2,979 x10^{4}
)^{2}
/ 1,495978 x 10^{11}
= 35,43303 x10^{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 the 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 / Mer
We can write
F = M V^{2}
/ r = M g
With force F that we find from the type 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 x10^{21}
x (1,495978 x 10^{11} )^{2}
/ G 5,973 x10^{24} = 1,9896 x10^{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 types we can find any other second mass, when we imagine any
mass in orbit, with any speed and radius.
