If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so during new Moon we can observe solar eclipse. But Moon's orbit is tilted 5^o with respect to Earth's orbit. Due to this 5^o tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment.

Newton's law of gravitation states that a particle of mass M₁ attracts any other particle of mass M₂ in the universe with an attractive force. The strength of this force of attraction was found to be directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.
\(\vec{F}=-\frac{G M_{1} M_{2}}{r^{2}} \hat{r}\)

Consider the Earth and mass system, with r, the distance between the mass m and the Earth's centre. Then the gravitational potential energy,
\(\mathrm{U}=-\frac{G M_{e} m}{r}\) ......(1)
Here r = R_e + h, where R_e is the radius of the Earth. h is the height above the Earth's surface.
\(\mathrm{U}_{\mathrm{c}}=-G \frac{M_{e} m}{\left(R_{e}+h\right)}\) .......(2)
If h << R_e, equation can be modified as
\(\mathrm{U} =-G \frac{M_{e} m}{\mathrm{R}_{e}\left(1+h / R_{e}\right)}  \)
\(\mathrm{U} =-G \frac{M_{e} m}{\mathrm{R}_{e}}\left(1+h / R_{e}\right)^{-1}\)  .......(3)
By using Binomial expansion and neglecting the higher order terms, we get
\(\mathrm{U}=-G \frac{M_{e} m}{\mathrm{R}_{e}}\left(1-\frac{h}{R_{e}}\right)\)   ........(4)
We know that, for a mass m on the Earth's surface,
\(G \frac{M_{e} m}{\mathrm{R}_{e}}=m g \mathrm{R}_{e}\)    ........(5)
Substituting equation (5) in (4) we get
\(\mathrm{U}=-\mathrm{mg} \mathrm{R}_{e}+\mathrm{mgh}\)   .......(6)
It is clear that the first term in the above expression is independent of the height h. For example, if the object is taken from height h₁ to h₂ then the potential energy at h₁ is
\(\mathrm{U}\left(\mathrm{h}_{1}\right)=-\mathrm{mg} \mathrm{} \mathrm{R}_{\mathrm{e}}+\mathrm{mgh}_{1}\)  .......(7)
and the potential energy at h₂ is
\(\mathrm{U}\left(\mathrm{h}_{2}\right)=-\mathrm{mg} \mathrm{} \mathrm{R}_{\mathrm{e}}+\mathrm{mgh}_{2}\)  .....(8)
The potential energy difference between h₁ and h₂ is
\(U\left(h_{2}\right)-U\left(h_{1}\right)=m g\left(h_{1}-h_{2}\right)\) .......(9)

 As the distance between two masses increases, the strength of the force tends to decrease because of inverse dependence on r². Physically it implies that the planet Uranus experiences less gravitational force from the Sun than the Earth since Uranus is at larger distance from the Sun compared to the Earth.

The gravitational forces between two particles always constitute an action reaction pair. It implies that the gravitational force exerted by the Sun on the Earth is always towards the Sun. The reaction-force is exerted by the Earth on the Sun. The direction of this reaction force is towards Earth.
The torque experienced by the Earth due to the gravitational force of the Sum is zero given by
\(\vec { \tau } =\vec { r } \times \vec { F } =\vec { r } \times \left( -\frac { { GM }_{ s }{ M }_{ E } }{ { r }^{ 2 } } \hat { r } \right) =0\)
Since \(\vec { r } =r\hat { r } ,(\hat { r } \times \hat { r } )=0\)
So, \(\hat { \tau } =\frac { d\vec { L } }{ dt } =0\)
It implies that angular momentum \(\vec{L}\) is a constant vector. The angular momentum of the Earth about the Sun is constant throughout the motion. It is true for all the planets. In fact, this constancy of angular momentum leads to the Kepler's second law.
The expression \(\vec{F}=-\frac{G M_{1} M_{2}}{r^{2}} \hat{r}\) has one inherent assumption that both M₁ and M₂ are treated as point masses. When it is said that Earth orbits around the Sun due to Sun's gravitational force, we assumed Earth and Sun to be point masses. This assumption is a good approximation because the distance between the two bodies is very much larger than their diameters. For some irregular and extended objects separated by a small distance, we cannot directly use the equation. Instead, we have to invoke separate mathematical treatment which will be brought forth in higher classes.
However, this assumption about point masses holds even for small distance for one special case. To calculate force of attraction between a hollow sphere of mass M with uniform density and point mass m kept outside the hollow sphere, we can replace the hollow sphere of mass M as equivalent to a point mass M located at the center of the hollow sphere. The force of attraction between the hollow sphere of mass M and point mass m can be calculated by treating the hollow sphere also as another point the center of the hollow sphere. It is shown in the Figure.
There is also another interesting result. Consider a hollow sphere of mass M. If we place another object of mass 'm' inside this hollow sphere as in Figure, the force experienced by this mass 'm' will be zero.
The triumph of the law of gravitation is that it concludes that the mango that is falling down and the Moon orbiting the Earth are due to the same gravitational force.

From the shadows it is revealed that Earth has spherical shape with bulging along equator and flat at poles.

The answer is wrong.
Actually, the seasons in the Earth arise due to the rotation of Earth around the sun with 23.5^o tilt.

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.