g = \(\frac{GM_e}{R^2_e}\)
M_e = 2 M_e      R_e= 2 R_e then,
g' = \(\frac{G \times 2M_e}{(2R_e)^2}\) = \(2 \frac{Gm_e}{4R_e^2}\)
\(2 \frac{Gm_e}{4R_e^2}\) = \(\cfrac g2\)

Time period T = \(\frac{2\pi}{\sqrt GM_E} (R_E+ h)^\frac{3}{2}\)
\(\therefore\) It is independemt of mass

Acceleration to Newton's III law, the force with which the earth vice-versa towards the apple is equal to the force with which earth attracts the apple. However the mass of the earth is extremely large as compared to that of apple. So acceleration of the earth is very small & is not noticeable.

(i) if F be the gravitational force on a body of mass m then
F=\(\frac{GMm}{R^2}\)= mg  ஃ\(\frac{GM}{R^2}\)
(ii) F \(\alpha\) m but g does not depend on m. So all bodies fall with same speed if there is no air resistance.

\(\mathrm{g}_{\text {latitude'}} \ g^{\prime}=g-\omega^{2} R \cos ^{2} \lambda\)
Value of latitude of g at Chennai \(\simeq 13^{\circ}\)
\(\operatorname{Cos} 13^{\circ} =0.2268 \mathrm{rad}  \)
\(\omega =\frac{2 \pi}{T}=\frac{2 \pi}{24 \times 3600}  \)
\(=\frac{2 \pi}{86400}=\frac{2 \times 3.14}{86400}  \)
\(\therefore \omega^{2} R =\left(\frac{2 \times 3.14}{86400}\right)^{2} \times\left(6400 \times 10^{3}\right)  \)
\(=3.4 \times 10^{-2} \mathrm{~m} / \mathrm{s}^{2} \)
\(\mathrm{g}_{\text {Cbeanai }} =\mathrm{g}-\omega^{2} R \cos ^{2} \lambda  \)
\(=9.8-\left(3.4 \times 10^{-2}\right)^{2} \cos (0.2268)^{2}  \)
\(\mathrm{~g}_{\text {Chennai }} =9.7677 \mathrm{~m} / \mathrm{s}^{2}\)
Value of latitude at Kanyakumari
\(=8.088^{\circ} \mathrm{N}=8.08=8.1^{\circ} \mathrm{N}  \)
\(\omega =\frac{2 \pi}{T}=\frac{2 \times 3.14}{86400}  \)
\(\omega^{2} R =\left(\frac{2 \times 3.14}{86400}\right)^{2} \times\left(6400 \times 10^{3}\right)  \)
\(=3.4 \times 10^{2} \mathrm{~m} / \mathrm{s}^{2}\)
\(\mathrm{g}_{\text {Kanyakumari }} =g-\omega^{2} R \cos ^{2} \lambda \)
\(=9.8-\left(3.4 \times 10^{2}\right)^{2}\left[\cos \left(8.1^{\circ}\right)\right]^{2} \)
\(g_{\text {Kanyakumari }} =9.798 \mathrm{~ms}^{-2} \)
\(\Delta g =9.798-9.767=0.031 \mathrm{~ms}^{-2}\)

1. Law of orbits
Each planet moves around the Sun in an elliptical orbit with the Sun at one of the foci.
2. Law of area
The radial vector (line joining the Sun to a planet) sweeps equal areas in equal intervals of time.
3. Law of period
The square of the time period of revolution of a planet around the Sun in its elliptical orbit is directly proportional to the cube of the semi major axis of the ellipse. It can be written as :
\(T^{2} \propto a^{3}  \)
\(\frac{T^{2}}{a^{3}}=\text { constant. }\)

The gravitational force experienced by the apple due to Earth
\(F=\frac { { GM }_{ E }M_{ A } }{ { R }^{ 2 } } \)
Here M_A– Mass of the apple, M_E – Mass of the Earth and R – Radius of the Earth.
Equating the above equation with Newton’s second law
\(M_{ A }a_{ A }=\frac { { GM }_{ E }M_{ A } }{ { R }^{ 2 } } \)
Simplifying the above equation we get,
\(a_{ A }=\frac { { GM }_{ E } }{ { R }^{ 2 } } \)
Here a_A is the acceleration of apple that is equal to ‘g’.
Similarly the force experienced by Moon due to Earth is given by
\(F=-\frac { { GM }_{ E }M_{ m } }{ { R }_{ m }^{ 2 } } \)
Here R_m- distance of the Moon from the Earth, M_m – Mass of the Moon
The acceleration experienced by the Moon is given by
\(a_{ m }=\frac { { GM }_{ E } }{ { R }_{ m }^{ 2 } } \)
The ratio between the apple’s acceleration to Moon’s acceleration is given by
\(\frac { { a }_{ A } }{ { a }_{ m } } =\frac { { R }_{ m }^{ 2 } }{ { R }^{ 2 } } \)
From the Hipparchrus measurement, the distance to the Moon is 60 times that of Earth radius. R_m = 60R.
\({ a }_{ A }/{ a }_{ m }=\frac { { \left( 60R \right) }^{ 2 } }{ { R }^{ 2 } } =3600.\)
The apple’s acceleration is 3600 times the acceleration of the Moon.
The same result was obtained by Newton using his gravitational formula. The apple’s acceleration is measured easily and it is 9.8 ms^-2. Moon orbits the Earth once in 27.3 days and by using the centripetal acceleration formula, (Refer unit 3).
\(\frac { { a }_{ A } }{ { a }_{ m } } =\frac { 9.8 }{ { 0.00272 } } =3600\)
which is exactly what he got through his law of gravitation.

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=\(\frac{M}{2}\)

From Kepler's III law
T² \(\alpha\)r³
\({ T }_{ 1 }^{ 2 }\alpha { r }_{ 1 }^{ 3 }\& { T }_{ 2 }^{ 2 }\alpha { r }_{ 2 }^{ 3 }\)
For F to be maximum \(\frac { dF }{ dm } =0\)
\(\frac { { T }_{ 2 }^{ 2 } }{ { T }_{ 1 }^{ 2 } } =\frac { { r }_{ 2 }^{ 3 } }{ { r }_{ 1 }^{ 3 } } =\frac { { \left( 3R \right) }^{ 3 } }{ { \left( 6R \right) }^{ 3 } } \)
\(\frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { 1 }{ 2\sqrt { 2 } } \)        \(\quad \left[ \because { T }_{ 1 }=12 \right] \)
\({ T }_{ 2 }=\frac { 12 }{ 2\sqrt { 2 } } =\frac { 6 }{ \sqrt { 2 } } \)

(i) Case 1: If r < r' Since gravitational force is attractive, m₂ is attracted by m₁. Then m₂ can move from r' to r without any external work. Here work is done by the system spending its internal energy and hence the work done is said to be negative. 
(ii) Case 2: If r' > r' Work has to be done against gravity to move the object from r' to r. Therefore work is done' on the body by external force and hence work done is positive

\(\overrightarrow { F } _{ m }m\overrightarrow { E } \)
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. 

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.

The gravitational potential energy U(r) of a system of two masses m₁ and m₂ separated by a distance r as the amount of work done to bring the mass m₂ from infinity to a distance r assuming m₁ to be fixed in its position and is written as
\(\mathrm{U}(\mathrm{r})=-\frac{G m_{1} m_{2}}{r}\)

The gravitational force exerted by Earth on the mass m near the surface of the Earth is given by
\(\overrightarrow { F } =-\frac { { GMM }_{ e } }{ { R }_{ e }^{ 2 } } \hat { r } \)
Now equating Gravitational force to Newton's second law,
\(ma=-\frac { { GMM }_{ e } }{ { R }_{ e }^{ 2 } } \hat { r } \)
Hence, acceleration is,
\(\overrightarrow { a } =-\frac { { GM }_{ e } }{ { R }_{ e }^{ 2 } } \hat { r } \)
The acceleration experienced by the object near the surface of the Earth due to its gravity is called acceleration due to gravity. It is denoted by the symbol g. The magnitude of acceleration due to gravity is
\(\left| g \right| =\frac { { GM }_{ e } }{ { R }_{ e }^{ 2 } } \hat { r } \)
It is to be noted that the acceleration experienced by any object is independent of its mass. The value of g depends only on the mass and radius of the Earth.

The gravitational potential V (r) at a point of height h from the surface of the Earth is given by,
V(r = R + h) =\(\frac { { GM }_{ e } }{ \left( { R }+h \right) } \)
The gravitational potential V(r) on the surface of Earth is given by,
V(r = R) = \(\frac { { GM }_{ e } }{ R}\)
Thus we see that
V(r = R) < V(r = R + h)
Gravitational potential energy near the surface of the Earth at height h is mgh. The gravitational potential at this point is simply V(h) = U(h)/m = gh. In fact, the gravitational potential on the surface of the Earth is zero since h is zero. So the apple falls from a region of a higher gravitational potential to a region of lower" gravitational potential.

Consider an object of mass M on the surface of the Earth. When it is thrown up with an initial speed Vi' the initial total energy of the object is
E_i = \(\frac { 1 }{ 2 } { Mv }_{ i }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)   ............(1) 
where, M_E is the mass of the Earth and R_E- the radius of the Earth. The term \(\frac { GMM_{ E } }{ R_{ E } } \) is the potential energy of the mass M.
When the object reaches a height far away from Earth and hence treated as approaching infinity, the gravitational potential energy becomes zero [U(∝) = 0] and the kinetic energy becomes zero as well. Therefore the final total energy of the object becomes zero. This is for minimum energy and for minimum speed to escape. Otherwise Kinetic energy can be non-zero.
E_f = 0
According to the law of energy conservation,
E_i = E_f  .............(2)
Substituting (1) in (2) we get,
\(\frac { 1 }{ 2 } { Mv }_{ i }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)   =0
\(\frac { 1 }{ 2 } { Mv }_{ e }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)  = 0  .............(3)
Consider the escape speed, the minimum speed required by an object to escape Earth's gravitational field, hence replace v_i with v_e. i.e.,
\(\frac { 1 }{ 2 } { Mv }_{ e }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } \)
\(v_{ e }^{ 2 }-\frac { GMM_{ E } }{ R_{ E } } .\frac { 2 }{ M } \)
\(v_{ e }^{ 2 }=\frac { 2G{ M }_{ E } }{ { R }_{ E } } \)  ..............(4)
Using g  = \(\frac { G{ M }_{ E } }{ { R }_{ e } } \)  ..............(5)
\(v_{ e }^{ 2 }=2g{ R }_{ E }\)
\({ v }_{ e }=\sqrt { 2g{ R }_{ E } } \)   .................(6)