\(\text {The gravitational field }=-\frac{G m}{r^{2}} \hat{r} \)
\(\frac{G m}{x^{2}}=\frac{G \times 4 m}{(r-x)^{2}} \)
\(\frac{m}{x^{2}}=\frac{4 m}{(r-x)^{2}} \)
\(\frac{1}{x^{2}}=\frac{4}{(r-x)^{2}} \)
\(\frac{1}{x}=\frac{2}{r-x}\)
r - x = 2x
\(\mathrm{r}=3 x \quad \therefore x=\frac{r}{3} \)
\(\text {Gravitational potential } \mathrm{V}=-\frac{G m}{r}=-\frac{9 \mathrm{Gm}}{r}\)

The weight of an object is defined as the downward force whose magnitude W is equal to the upward force that must be applied to the object to hold it at rest or at constant velocity relative to the Earth.

The gravitational potential at a distance r due to a mass is defined as the amount of work required to bring unit mass from infinity to the distance r and it is denoted as V(r).
\(\mathrm{V}(\mathrm{r})=-\frac{\mathrm{Gm}}{r}\)

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 force of attraction is given by
\(\vec { F } =-\frac { { Gm }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } \)
From the figure, r = 10 m.
First, we can calculate the magnitude of the force
\(F=-\frac { { Gm }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } =\frac { 6.67\times { 10 }^{ -11 }\times 1\times 2 }{ 100 } \)
= 13.34\(\times\)10^-13N.
It is to be noted that this force is very small. This is the reason we do not feel the gravitational force of attraction between each other. The small value of G plays a very crucial role in deciding the strength of the force.
The force of attraction \(\left( { \vec { F } }_{ 21 } \right) \) experienced by the mass m₂ due to m₁ is in the negative ‘y’ direction ie. \(\hat{r}=-\hat{j}.\) According to Newton’s third law, the mass m₂ also exerts equal and opposite force on m₁. So the force of attraction \(\left( { \vec { F } }_{ 12 } \right) \) experienced by m₁ due to m₂ is in the direction of positive ‘y' axis ie., \(\hat{r}=\hat{j}.\)
\(\overrightarrow { F } _{ 21 }=-13.34\times { 10 }^{ -13 }\hat { j } \)
\(\overrightarrow { F } _{ 12 }=13.34\times { 10 }^{ -13 }\hat { j } \)
The direction of the force is shown in the figure,

Gravitational force of attraction between m₁ and m₂
\({ \vec { F } }_{ 12 }=-{ \vec { F } }_{ 21 }\) which confirms Newton’s third law.

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

(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.
 

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

A satellite of mass M to move in a circular orbit, centripetal force must be acting on the satellite. This centripetal force is provided by the Earth's gravitational force.
\(\frac { { mv }^{ 2 } }{ \left( { R }_{ E }+h \right) } =\frac { { GMM }_{ E } }{ { \left( { R }_{ e }+h \right) }^{ 2 } } \)
\({ v }^{ 2 }=\frac { { GM }_{ E } }{ \left( { R }_{ E }+h \right) } \)
\(v=\sqrt { \frac { { GM }_{ E } }{ \left( { R }_{ E }+h \right) } } \)
As h increases, the speed of the satellite decreases.

(i) The magnitude of \(\overrightarrow { E } \) decreases as the distance r increases.
(ii) To calculate gravitational interaction. It carries energy and momentum in space.
(iii) The concept of field is inevitable in understanding the behaviour of charges.

(i) Contact forces are the forces applied where one object is in physical contact with the other. The movement of the object is caused by the physical force exerted through the contact between the object and the agent which exerts force
(ii) Consider the case of Earth orbiting around the Sun. Though the Sun and the Earth are not physically in contact with each other, there exists an interaction between them.
(iii) This is because of the fact that the Earth experiences the gravitational force of the Sun.
(iv) This gravitational force is a non-contact force.

Angular momentum \(\mathrm{L} =5 t^{2} \hat{i}-6 t \hat{j}+3 \hat{k} \)
\(\text {Torque }  \propto \frac{d L}{d t}  \)
\(=\frac{d}{d t}\left[5 t^{2} \hat{i}-6 t \hat{j}+3 \hat{k}\right]=10 t \hat{i}-6 \hat{j}\)

The value of semi major axis is directly proportional to twice the square of time period of a planet.
i.e, a \(\propto 2 \mathrm{~T}^{2}\)
It is given that for planet Kurinji,
\(\mathrm{T}_{1}=2, \quad \mathrm{a}_{1}=8=2 \times 2^{2} \Rightarrow 2 \mathrm{~T}_{1}{ }^{2}\)
For planet Mullai \(\quad \mathrm{T}_{2}=3, \quad \mathrm{a}_{2}=18=2 \times 3^{2} \Rightarrow 2 \mathrm{~T}_{2}{ }^{2}\)
For planet Marutham \(\quad \mathrm{T}_{3}=4, \quad \mathrm{a}_{3}=32=2 \times 4^{2} \Rightarrow 2 \mathrm{~T}_{3}{ }^{2}\)
For planet Neithal \(\quad \mathrm{T}_{4}=5, \quad \mathrm{a}_{4}=50=2 \times 5^{2} \Rightarrow 2 \mathrm{~T}_{4}{ }^{2}\)
For planet Paalai \(\quad \mathrm{T}_{5}=6, \quad \mathrm{a}_{5}=72=2 \times 6^{2} \Rightarrow 2 \mathrm{~T}_{5}^{2}\)
\(\therefore \alpha \propto 2 \mathrm{~T}^{2}\)

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 field at a point which is at a distance r from m, is defined as the gravitational force experienced by unit mass placed at that point.
It's unit is N/kg or ms^-2.

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)

(i) Newton verified his law of universal gravitation by comparing the acceleration of a terrestrial object to the acceleration of the moon.
(ii) He knew that the distance from the center of earth to the center of two spheres of known mass at either end of a light rod suspended by a then fiber from the center of the rod.
(iii) He had earlier found the small force that was needed to twist the fiber.
(iv) By bringing a third sphere close to one of the suspended spheres.
(v) He was able to measure the force of gravity between the spheres and hence gravitation.

 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.