The effective value of acceleration due to gravity
g' = g cos \(\alpha\)
\(T=2\pi \sqrt { \frac { l }{ g' } } \)
\(=2\pi \sqrt { \frac { l }{ cos\alpha } } \)

Let g & g' be the acceleration due to gravity on the earths surface at a height R above the earth's surface.
\(g'=\frac { g }{ \left( 1+\frac { n }{ R } \right) } =\frac { g }{ \left( 1+{ \left( \frac { R }{ R } \right) }^{ 2 } \right) } =\frac { g }{ 4 } \)
Time period of simple pendulum \(T=2\pi \sqrt { \frac { l }{ g } } \)
\(\frac { { T }_{ 2 } }{ { T }_{ 1 } } =\sqrt { \frac { g }{ g' } } =\sqrt { \frac { 9 }{ \frac { g }{ 4 } } } =2\)

As the gravitational force varies inversely as the n^th power of the distance, so the gravitational force on the planet is given by
\(F=\frac { GMm }{ { r }^{ n } } \)
This force provides the centripetal force mrw² to the planet
\(\frac { GMm }{ { r }^{ n } } ={ mrw }^{ 2 }\)
\(=mr\left( \frac { 2\pi }{ T } \right) \)
\(or\quad { T }^{ 2 }=\frac { v\times 4\pi 2\times { v }^{ 2 } }{ GM } =\frac { 4{ \pi }^{ 2 }{ r }^{ (n+1) } }{ GM } \)
\(T=\frac { 2\pi }{ \sqrt { GM } } .{ r }^{ (n+1)/2 }\)
\(\therefore T\alpha { r }^{ (n+1)/2 }\)

(i) The value of g varies with height has
g a\(\frac { 1 }{ { \left( R+h \right) }^{ 2 } } \ or\ g\ a\ \frac { 1 }{ { r }^{ 2 } } \)
Thus the graph of g versus V is the parabolic curve AB

(ii) The value of g varies with depth d as
\(g=g\left( 1-\frac { d }{ R } \right) \)i.e g\(\alpha \)(R-d)
Thus the graph of g versus depth d is the straight line AB.
 

Let m₁ =m ;m₂ =M-m
\(R=\frac { G(M-m).m }{ { r }^{ 2 } } =\frac { G(Mm-{ m }^{ 2 }) }{ { r }^{ 2 } } \)
Differentiating with respect to m,
\(\frac { dF }{ dm } =\frac { G }{ { r }^{ 2 } } (M-2m)\)
For F is maximum \(\frac { dF }{ dm } \)=0
\(\frac { G }{ { r }^{ 2 } } (M-2m)=0\)
M = 2m Þm = \(\frac{M}{2}\)
m₁= m₂=\(\frac{M}{2}\)

Mass of the earth M = 6.0\(\times\)10²⁴ kg ; m = 67 kg G = 6.67\(\times\)10^-11 Nm² kg-2
GravitationaI potentia V = -\(\frac{GM}{R}\)
\(=\frac { 6.67\times { 10 }^{ -11 }\times 6\times { 10 }^{ 24 } }{ 6.6\times { 10 }^{ 10 } } \)
V = -6.1\(\times\)103 Jkg^-1

\(\overrightarrow { F } _{ m }m\overrightarrow { E } \)
Now we can equate this with Newton's second Law \(\overrightarrow { F } =m\overrightarrow { a } \)
\(m\overrightarrow { a } =m\overrightarrow { E } \)
\(\overrightarrow { a } =\overrightarrow { E } \)
In other words, equation implies that the gravitational field at a point is equivalent acceleration experienced by a particle at that point.  However, it is to be noted \(\overrightarrow { a } \) that and \(\overrightarrow { E } \) are separate physical quantities that have the same magnitude and direction. 

(i) The force experienced by a mass 'm' which is on the surface of the Earth is given by F= \(\frac { { GM }_{ 1 }m }{ { r }_{ E }^{ 2 } } \)
(ii) M_E-mass of the Earth, m - mass of the object, R_E- radius of the Earth.
Equating Newton's second law, F= mg, to equation we get,
mg= \(\frac { { GM }_{ 1 }m }{ { r }_{ E }^{ 2 } } \)
g=\(\frac { { GM }_{ E }m }{ { r }_{ E }^{ } } \)
(iii) Now the force experienced by some other object of mass M at a distance r from the center of the Earth is given by,
F= \(\frac { { GM }_{ E }m }{ { r }_{ }^{ 2 } } \)
Using the value of g in equation, the force F will be,
\(F=-gM\frac { { R }_{ E }^{ 2 } }{ { r }^{ 2 } } \)
From this it is clear that the force can be calculated simply by knowing the value of g. It is to. be. noted that in the above calculation G is not required.

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.