Variation of g with depth:
Consider a particle of mass m which is in a deep mine on the Earth. (Example: coal mines -in Neyveli). Assume the depth of the mine as d. To calculate g' at a depth d, consider the following points.

The part of the Earth which is above the radius (R_e - d) do not contribute to the acceleration. The e· result is proved earlier and is given as
g' = \(\frac { GM' }{ ({ R }_{ e }-d)^{ 2 } } \)
Here M' is the mass of the Earth of radius (R_e - d)
Assuming the density of Earth p to be constant
  \(\rho =\frac { M' }{ V' } \)
where M is the mass of the Earth and V its volume, Thus
  \(\rho =\frac { M' }{ V' } \)
\(\frac { M' }{ V' } =\frac { M }{ V } \) and M' = \(\frac { M }{ V } V'\)
M' = \(\left( \frac { M }{ \frac { 4 }{ 3 } \pi { R }_{ e }^{ 3 } } \right) \left( \frac { 4 }{ 3 } \pi ({ R }_{ e }-d)^{ 3 } \right) \)
M'=\(\frac { M }{ { R }_{ e }^{ 3 } } \)(R_e - d)³
g' = G\(\frac { M }{ { R }_{ e }^{ 3 } } \)(R_e - d)³.\(\frac { 1 }{ ({ R }_{ e }-d)^{ 2 } } \)
g' = GM \(\frac { R_{ e }\left( 1-\frac { d }{ { R }_{ e } } \right) }{ { R }_{ e }^{ 3 } } \)
g' = GM \(\frac { \left( 1-\frac { d }{ { R }_{ e } } \right) }{ { R }_{ e }^{ 2 } } \)
Thus
g' = g \(\left( 1-\frac { d }{ { R }_{ e } } \right) \)
Here also g' < g. As depth increases, g' decreases. It is very interesting to know that acceleration due to gravity is maximum on the surface of the Earth but decreases when we go either upward or downward.

Whenever we analyze the motion of objects in rotating frames we must take into account the centrifugal force. Even though we treat the Earth as an inertial frame, it is not exactly correct because the Earth spins about its own axis. So when an object is on the surface of the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.

This centrifugal force is given by mω²R'.
 R' =  R cos λ       ....(1)
where λ  is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
 a_c = ω²R' cos λ = ω²R cos² λ
since R' = R cos λ
Therefore,
g' = g -ω²R cos² λ     ...(2)
From the expression (2), we can infer that at equator, λ  = 0; g' = g - ω²R. The acceleration due to gravity is minimum. At poles λ  = 90; g' = g, it is maximum. At the equator, g' is minimum.

By finding the apparent radii of the Earth's umbra shadow and the Moon, the ratio of the these radii can be calculated. This is shown in Figure.

The apparent radius of Earth's umbra shadow
\(=\mathrm{R}_{\mathrm{s}}=13.2 \mathrm{~cm}\)
The apparent radius of the Moon
\(=\mathrm{R}_{\mathrm{m}}=5.15 \mathrm{~cm}\)
The ratio
\(\frac{R_{S}}{R_{m}} \approx 2.56\)
The radius of the Earth's umbra shadow is
\(\mathrm{R}_{\mathrm{s}}=2.56 \times \mathrm{Rm}\)
The radius of Moon \(\mathrm{R}_{\mathrm{m}}=1737 \mathrm{~km}\)
The radius of the Earth's umbra shadow is
\(\mathrm{R}_{\mathrm{s}}=2.56 \times 1737 \mathrm{~km} \cong 4446 \mathrm{~km} \text {. }\)
The correct radius is 4610 km.
The percentage of error in the calculation

According to this model, the Sun is at the center of the solar system and all planets orbited the Sun. The retrograde motion of planets with respect to Earth is because of the relative motion of the planet with respect to Earth. The retrograde motion from the heliocentric point of view is shown in Figure.

Figure shows that the Earth orbits around the Sun faster than Mars. Because of the relative motion between Mars and Earth, Mars appears to move backward from July to October. In the same way the retrograde motion of all other planets was explained successfully by the Copernicus model. It was because of its simplicity the heliocentric model slowly replaced the geocentric model. Historically, if any natural phenomenon has one or more explanations, the simplest one is usually accepted. Though this was not the only reason to disqualify the geocentric model, a detailed discussion on correctness of the Copernicus model over to Ptolemy's model can be found in astronomy books.

The total energy of a satellite orbiting the Earth at a distance h from the surface of Earth is calculated as follows; The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,
U = \(\frac { { GN }_{ s }{ M }_{ E } }{ ({ R }_{ E }+h) } \) ......(1)
Here M_s - mass of the satellite, M_E - mass of the Earth, R_E - radius of the Earth 
The Kinetic energy of the satellite is
K.E = \(\frac { 1 }{ 2 } \) M_sv²,  ......(2)
Here v is the orbital speed of the satellite and is equal to
v = \(\sqrt { \frac { { GM }_{ E } }{ (R_{ E }+h) } } \)   ......(3)
Substituting the value of v in (2) the kinetic energy of the satellite becomes,
K.E = \(\frac { 1 }{ 2 } \frac { { GM }_{ E }{ M }_{ s } }{ 2({ R }_{ E }+h) } \)
Therefore the total energy of the satellite is
E = \(\frac { 1 }{ 2 } \frac { { GM }_{ E }{ M }_{ s } }{ ({ R }_{ E }+h) } -\frac { { GM }_{ 5 }{ M }_{ E } }{ ({ R }_{ E }+h) } \)
E = \(\frac { { GM }_{ E }{ M }_{ s } }{ 2({ R }_{ E }+h) } \)   .........(4)
The negative sign in the total energy implies that the satellite is bound to the Earth and it cannot escape from the Earth.
As h approaches infinity the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth's gravity and is not bound to Earth at large distances.