\(\mathrm{g}_{\text {latitude'}} \ g^{\prime}=g-\omega^{2} R \cos ^{2} \lambda\)
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
\(=8.088^{\circ} \mathrm{N}=8.08=8.1^{\circ} \mathrm{N}  \)
\(\omega =\frac{2 \pi}{T}=\frac{2 \times 3.14}{86400}  \)
\(\omega^{2} R =\left(\frac{2 \times 3.14}{86400}\right)^{2} \times\left(6400 \times 10^{3}\right)  \)
\(=3.4 \times 10^{2} \mathrm{~m} / \mathrm{s}^{2}\)
\(\mathrm{g}_{\text {Kanyakumari }} =g-\omega^{2} R \cos ^{2} \lambda \)
\(=9.8-\left(3.4 \times 10^{2}\right)^{2}\left[\cos \left(8.1^{\circ}\right)\right]^{2} \)
\(g_{\text {Kanyakumari }} =9.798 \mathrm{~ms}^{-2} \)
\(\Delta g =9.798-9.767=0.031 \mathrm{~ms}^{-2}\)

From the figure, the distance of m₁ from the origin = a
From the figure, the distance of m₂ from the origin = a
Gravitational field \(\mathrm{E}=\frac{G M}{r^{2}} \hat{r}\)
At the origin (Point O) the change in gravitational field is
\(\vec{E}=\frac{G M}{a^{2}}\left[\left(m_{1}-m_{2}\right) \hat{i}+m_{3} \hat{j}\right]\)
It is given that
\(\mathrm{m}_{1} =\mathrm{m}_{2} \)
\(\therefore \vec{E} =\frac{G M}{a^{2}}\left[m_{3} \hat{j}\right]\)

Mass of the earth M = 6.024 5 10²⁴ kg
Radius of the earth R = 6.4\(\times\)10⁶ m
Gravitational constant G' = 6.67\(\times\)10¹¹
Gravitational constant G = 6.67\(\times\)10¹¹
Acceleration due to gravity g' =?
g' = 
g' = 9.8\(\times\)10²² m/s²
g = 9.8 m/s²
∴ g' = g\(\times\)10²² (or) 10²²
g = g' m/s²
Weight W = mg
W' = mg'
 = 10²² .W
W = 10²²

The gravitational potential energy
\(\mathrm{V} =-\frac{G M^{2}}{R}\left[1+\frac{4}{\sqrt{2}}\right]  \)
\(\mathrm{V} =-\frac{G M^{2}}{R}[1+2 \sqrt{2}]  \)
\(\text {Gravitational potential } \mathrm{V}_{\mathrm{o}}(\mathrm{r}) =-\frac{4 G M}{R}\)

Centripetal acceleration \(a=\frac{V^{2}}{R}\)
Centripetal force \(=\frac{M V^{2}}{R}\)
\(\therefore\) Speed \(V=\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)

Time period, T = 1.769 days
Orbital radius, r = 421700 x 10³ m
\(\mathrm{T} =2 \pi \sqrt{\frac{R^{3}}{G M}}  \)
\(\mathrm{T}^{2} =\frac{4 \pi^{2} \times R^{3}}{G M}  \)
\(\therefore \mathrm{M} =4 \pi^{2} \times G \times \frac{R^{3}}{T^{2}}  \)
\(\mathrm{~T}^{2} \propto \mathrm{R}^{3}  \)
\(\frac{R^{3}}{T^{2}} =\left[\frac{421700 \times 10^{3}}{1.769}\right]^{3 / 2}=\frac{\left(421.7 \times 10^{6}\right)^{3}}{(1.769)^{2}}  \)
\(=\frac{74991.3 \times 10^{3} \times 10^{18}}{3.1293}=\frac{7499.14 \times 10^{22}}{3.1293}  \)
\(=2396.4 \times 10^{22}\)
\(\mathrm{M} =4 \times 3.14 \times 6.67 \times 10^{-11} \times 2396.4 \times 10^{22}  \)
\(=1.898 \times 10^{27} \mathrm{~kg}\)

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

Gravitational force F = \(\frac { { GM }_{ 1 }{ M }_{ 2 } }{ { r }^{ 2 } } \hat{r}\)
Since mass and mutual distances are double there is no change in the gravitational force between them.

Let the distance of the Earth = R_E
The distance of unknown planet = R_P = 2 R_E
Let the time period of the Earth be T₁
The time period of the unknown planet be T₂
\(\text {Time period } \mathrm{T} =2 \pi \sqrt{\frac{R_{E}}{g}}  \)
\(\mathrm{~T}  \propto \sqrt{R_{E}}  \)
\(\therefore \frac{T_{1}}{T_{2}} =\sqrt{\frac{R_{E}}{R_{P}}}=\sqrt{\frac{R_{E}}{2 R_{E}}}  \)
\(\therefore \frac{T_{1}}{T_{2}} =\frac{1}{\sqrt{2}}  \)
\(\mathrm{~T}_{2} =\sqrt{2} T_{1}\)

If the Earth has us tilt then there would not be seasons of the Earth.

Moon will start to move in spiral path towards the earth and it may hit on the surface of the Earth.

Kinetic energy of the moon will vary but its total energy remains constant.

Mass of the sphere, m = 20 kg
Radius r = 0.25 m
Angular velocity 0 = 5 rad s^-1
Angular momentum \(L=I\omega =\frac{2}{5} mr^{2}\omega\)
\(=\frac{2}{5}\times 20 \times (0.25)^{2}\times 5 = 40 \times (0.0625)=2.5\)
L = 2.5 kg m² s^-1.

No change in speed of system as no external force is working.

Due to conservation of angular momentum.

No

Zero.

Angular momentum of planet.

Zero.

As \(\vec L\)= \(\vec r\) \(\times\) \(\vec mv\) i.e., \(\vec L\) magnitude decreases but direction remains constant.

\(K=\frac { { L }^{ 2 } }{ 2I } \Rightarrow K_{ A }>K_{ A }\)

Sphere of smaller density will have larger moment of inertia.

No, it changes with the position of axis of rotation.

P.E. is minimum.

Radial Component.

No. It can produce only rotatory motion.

\(\frac{1}{2}\)MR²

\(\sum { \bar { \tau } } =0\)

It will be about an axis passing through the centre of the cube and connecting the opposite corners. 

A cat is able to land on its feet after a fall. This is based on law of conservation of angular momentum. When the cat is about to fall, it curls its body to decrease the moment of inertia and increase its angular velocity. When it lands it stretches out its limbs. By which it increases its moment of inertia and intum it decreases its angular velocity. Hence the cat lands safety.

No, moment of inertia and radius of gyration depends on axis of rotation and also on the distribution of mass of the body about its axis.

(i) As we know L = I\(\omega \) = constant if the boy sits on the edge of revolving disc, its I will be increased in turn it reduces angular velocity.
(ii) If the boy starts moving towards the center of the disc, its I will decrease in turn that increases its angular velocity.

A rigid body is said to be in mechanical equilibrium when both its linear momentum and angular momentum remain constant.

If the straight path of the particle passes through the origin, then the angular momentum is zero, which is also a constant.

The torque about the axis will rotate the object about it and the torque perpendicular to the axis will turn the axis of rotation when both exist simultaneously on a rigid body the body will have a precession.

A rigid body is the one which maintains its definite and fixed shape even when an external force acts on it.

If ice of the pole melts then it will come towards the equator and moment of inertia of earth will increase because mass particle at equator are at more distant from rotational axis as compare to pole.
Time period (T) = \(\frac { 2\pi }{ \omega } \) (∵ Iω= constant)
So if I increases then ω decreases ⇒ T increases. Due to increment of time period the duration of day and night will increase.

I₁ω₁=I₂ω₂
⇒ \(\frac{2}{5}\)MR² \(\times\) \(\frac { 2\pi }{ T_{ 1 } } \) =\(\frac{2}{5}\)M \(\left( \frac { R }{ 2 } \right) ^{ 2 }\)\(\times\) \(\frac { 2\pi }{ T_{ 2 } } \)
⇒ T₂ = \(\frac { T_{ 1 } }{ 4 } \) ∴T₁ = 24 hr
∴ T₂ = 6hr

Given: M = 20 kg, ω= 100 rad s^-I, R = 0.25m
M.I. of the cylinder about its own axis, I = \(\frac{1}{2}\)MR² = \(\frac{1}{2}\) \(\times\) 20 \(\times\) (25)² = 625 kgm²
Rotational K.E  I = \(\frac{1}{2}\)Iω² = \(\frac{1}{2}\) \(\times\) 0.625 \(\times\) (100)² =3125J
Angular momentum, L= Iω​​​​​​​ = 0.625 \(\times\)​​​​​​​100 = 62.5 kgm² s^-1

Rotational K.E., K = \(\frac { 1 }{ 2 } \) Iω²   ∴ I=\(\frac { 2K }{ \omega ^{ 2 } } \)
Angular momentum, L = Iω =  \(\frac { 2K }{ \omega ^{ 2 } } \).ω = \(\frac { 2K }{ \omega } \)
When angular frequency is doubled and kinetic energy is halved, the angular momentum becomes,
L' = \(\frac { 2\left( \frac { K }{ 2 } \right) }{ 2\omega } \) = \(\frac { 1 }{ 4} \)\(\frac { 2K }{ \omega } \)= \(\frac { L }{4} \)

Given: L = Iω= Constant
Formula: Rotational K.E. is given by, 
K = \(\frac { 1 }{ 2 } \) Iω²=\(\frac { 1 }{ 2 } \frac { { I }^{ 2 }\omega ^{ 2 } }{ I } \)= \(\frac { 1 }{ 2 } .\frac { L^{ 2 } }{ I } \)
For constant L, \(K\propto\frac{1}{1}\) 
So when the moment of inertia decreases, the rotational K.E. increases. Hence rotational K.E. is not conserved.

Formula: By conservation of angular momentum
l₁ω ₁ = I₂ω₂
or mr². \(\frac { 2\pi }{ T_{ 1 } } \) = m \(\left( \frac { r }{ 2 } \right) ^{ 2 }.\frac { 2\pi }{ T_{ 2 } } \)
\({ T }_{ 2 }=\frac { 1 }{ 4 } { T }_{ 1 }\)
= \(\frac { 1 }{ 4 } \times 4\)=1s

The bottom of a ship is made heavy so that its centre of gravity remains low. This ensures the stability of its equilibrium.

No. Angular velocity is not conserved but angular momentum is conserved.

The ladder can rotate about the point of contact of the ladder with the ground. When the labourer is at the top of the ladder, the lener arm of the force is large. Hence the turning effect on the ladder will be large.

For a given force, torque can be increased if the perpendicular distance of the point of application of the force from the axis of rotation is increased. Hence the handle put near the circumference produces maximum torque.

No. The centre of mass of a system is a hypothetical point which acts as a single mass particle of the system for an external force.

Not necessarily. For example, the eM of a ring lies in its hollow portion.

\(m_1=m_2=m\)
v = 2v and v₂ = -v
\(\therefore\)   \({V}_{CM}={m_1v_1+m_2v_2\over m+m}={m\times(2v)+m(-v) \over m+m}={v \over 2}\)

\({x}_{CM}={m(2.5)+m(3.5)+2m(3)\over m+m+2m}=3\)
\({y}_{CM}={m(1.5)+m(1.5)+2m(3)\over m+m+2m}=2.25\)

\({x}_{CM}={m_1+m_2+m_2x_2\over m_1+m_2}={12\times0+16\times1.13 \over 12+16}=0.6457\ \mathring{A}\)

For symmetrically shaped bodies of uniform composition (Eg: spheres, cylinders, rectangular solids), the centre of mass is located at the geometrical centre of the body.

The angular momentum of a particle of mass m moving along a circular path of radius r with linear velocity v is given by
\(\bar{L}=\bar{r}\times m \bar{v}\)
If the speed (v) of the particle decreases along the circular path, the magnitude of angular momentum (= mvr) decreases. However, the direction of angular momentum remains unchanged.

The centripetal force cannot do work on the object since it is always perpendicular to the velocity.

No power is needed to maintain uniform circular motion.
Power required = \(\bar{F}.\bar{v}=F v\cos\theta\)
 = \(Fv\cos90°=0\)

The moment of inertia of the system (circular table + two boys) decreases. To conserve angular momentum (L = 100 = constant), the speed of rotation of the circular table increases.

When a cat falls to ground from a height, it stretches its body along with the tail so that its moment of inertia becomes high. Since angular momentum (L = Iω) remain constant, the value of angular speed ω decreases and therefore the cat is able to land on the ground gently.

The moment of inertia of a body is minimum about an axis passing through its centre of mass.

The heavy boxes should be loaded first so that the CG of the loaded cart remains in the lowest position. This ensures stability of equilibrium.

Here \(\tau \) = 2.0 \(\times\) 10^-4Nm,
α = 4 rad s^-2, I = ?
Formula: \(\tau\) = Iα
I = \(\frac { \tau }{ \alpha } =\frac { 2.0\times 10^{ -4 } }{ 4 } \) = 0.5 \(\times\) 10^-4 kgm²

Here R = 0.25m
M= 2kg
V= 2ms^-1
Rotational K.E = \(\frac { 1 }{ 2 } I\omega ^{ 2 }\) = \(\frac { 1 }{ 2 } MR ^{ 2 }\) \(\times \left( \frac { V }{ R } \right) ^{ 2 }\)
\(\frac { 1 }{ 2 } MV ^{ 2 }\) = \(\frac { 1 }{ 2 }\) ｘ2ｘ4 =4J

Given: m = 50g = 0.05 kg, r = 50cm = 0.50m
Formula: I = mr²
= 0.05 \(\times\) (0.50)²
= 0.0125 kgm²

M = 1025kg
R = 6400 km = 6.4\(\times\)10⁶ m
M.I of the earth about its diameter,
I =\(\frac { 2 }{ 5 } \)MR² = \(\frac { 2 }{ 5 } \)\(\times\)10²⁵ \(\times\)(6.4\(\times\)10⁶)²     
=1.64 \(\times\)10³⁸ kgm²

M = 8 kg, k = 25cm = 0.25m
∴ I = Mk²
= 8 \(\times\) (0.25)²
=0.5 kgm²

Given: ​​​​\(\widehat { F } =7\widehat { i } -3\widehat { j } -5\widehat { k } \) and \(\widehat { r } =\widehat { i } +\widehat { j } -\widehat { k } \)
Formula: ∴ \(\tau =\widehat { r } \times \widehat { F } =\)\(\left| \begin{matrix} \widehat { i } & \widehat { j } & \widehat { k } \\ 1 & 1 & -1 \\ 7 & -3 & -5 \end{matrix} \right| \)
or  \(\tau =-8\widehat { i } -2\widehat { j } -10\widehat { k } \)

\(\alpha =\frac { \tau }{ I } =\frac { (5+10){ r }_{ 2 }-15{ r }_{ 1 } }{ I } \)
=\(\frac { 15(20-10)\times { 10 }^{ 2 } }{ 1500 } \)
= 10^-3rad s^-2

P = 200 hp = 200 hp \(\times\) \(\frac { 76\omega }{ 1hp } \) = 1.49 \(\times\) 10⁵ \(\omega \)
\(\omega \) = 6000 rev/min
 = 6000 \(\times\) \(\frac { 2\pi }{ 60 } \) = 628 rad/sec
\(\tau =\frac { \rho }{ \omega } =\frac { 1.49\times 10^{ 5 } }{ 628 } \) = 237.5 Nm

\(\frac { I_{ 1 } }{ { I }_{ 2 } } =\frac { M_{ 1 }R_{ 1 }^{ 2 } }{ M_{ 2 }R^{ 2 }_{ 2 } } \)
\(\Rightarrow \frac { M_{ 1 } }{ M_{ 2 } } =\frac { I_{ 1 } }{ { I }_{ 2 } } \times \left[ \frac { R_{ 2 } }{ R_{ 1 } } \right] ^{ 2 }\)
\(\frac { 4 }{ 1 } \times \left( \frac { 1 }{ 2 } \right) ^{ 2 }=\ \frac { 1 }{ 4 } \)

I_system = 2 \(\times\) I_upper + I_inner
= 2 \(\times\)\(\frac { 5 }{ 4 } \) MR² + \(\frac { MR^{ 2 } }{ 4 } \) = \(\frac { 11 }{ 4 } \)MR²

I_x = 3 \(\times\) (3)² + 4 \(\times\) (2)² = 43 unit
I_y = 2\(\times\) (2)² + 4 \(\times\) (2)² = 24 unit
I_z = 2 \(\times\) (2)² + 3 \(\times\) (3)² + 4 \(\times\) (\(\sqrt { 2 } \))²
= 8 + 27 + 32 = 67 unit

(a) I_xx' = 4 \(\times\) (0.3)² + 1 \(\times\) (0.8)² = 1 kgm² 
(b) I_xx'= 4 \(\times\) (3)² + 2 \(\times\) (2)² + 3 \(\times\) (4)² = 92 kgm²

When the round object moves, it always tends to roll on any surface which has a coefficient of friction any value greater than zero (μ > 0). The friction that enabling the rolling motion is called rolling friction.

Power delivered is the work done per unit time. If we differentiate the expression for work done with respect to time, we get the instantaneous power (P).
p =\(\frac { dw }{ dt } =\tau \frac { d\theta }{ dt } \)   \(\because (dw=\tau d\theta )\)
p =ፒω

(i) A rigid body which has non-zero. external torque (ፒ) about the axis of rotation would have an angular acceleration (α) about that axis.
(ii) The scalar relation between the torque and angular acceleration is,
ፒ=Iα
where, I is the moment of inertia of the rigid body. The torque in rotational motion is equivalent to the force in linear motion.

In the absence of external torque, the angular momentum of the rigid body or system of particles is conserved.
If \(\tau\) = 0 then, \(\frac { dL }{ dt } =0\) ; L= constant

(i) The torque of a force about an axis is independent of the choice of the origin as long as it is chosen on that axis itself.
(ii) Let O be the origin on the axis AB, which is the rotational axis of a rigid body. F is the force acting at the point P. Now, choose another point O' anywhere on the axis.
(iii) The torque of F about O' is,
\(\overset { \rightarrow }{ O'P } \times \overset { \rightarrow }{ F } =(\overset { \rightarrow }{ O'O } +\overset { \rightarrow }{ OP } )\times \overset { \rightarrow }{ F } \)
=\((\overset { \rightarrow }{ O'O } \times \overset { \rightarrow }{ F } )+(\overset { \rightarrow }{ OP } \times \overset { \rightarrow }{ F } )\)
(iv) As \(\overset { \rightarrow }{ O'O } \times \overset { \rightarrow }{ F } \)  is perpendicular to \(\overset { \rightarrow }{ O'O } \)  this term will not have a component along AB. Thus, the component of \(\overset { \rightarrow }{ O'P } \times \overset { \rightarrow }{ F } \)  is equal to that of \(\overset { \rightarrow }{ OP } \times \overset { \rightarrow }{ F } \).

(i) The -torque about the axis will rotate the object about it and the torque perpendicular to the axis will turn the axis of rotation.
(ii) When both exist simultaneously on a rigid body, the body will have a precession.

The direction of torque helps us to find the type of rotation caused by the torque. For example, if the direction of torque is out of the paper, then the rotation produced by the torque is anticlockwise. On the other hand, if the direction of the torque is into the paper, then the rotation is clockwise.

The direction of torque is found using right hand rule. This rule says that if fingers of right hand are kept along the position vector with palm facing the direction of the force and when the fingers are curled the thumb points to the direction of the torque.

A point mass is a hypothetical point particle which has nonzero mass and no size or shape.

The center of mass of a body is defined as a point where the entire mass of the body appears to be concentrated.

A rigid body is the one which maintains its definite and fixed shape even when an external force acts on it.

Internal forces are the forces acting among the particles within a system that constitute the body. External forces are the forces acting on the particles of a system from outside.

Height = Altitude h = 200 km
Depth d = 200 km
The value of g at that altitude is
\(\mathrm{g}_{\mathrm{h}}=g\left[1-\frac{2 h}{R_{e}}\right] \)
\(R_{e}=6400 \mathrm{~km} \)
\(\therefore \quad \mathrm{g}_{\mathrm{h}}=g\left[1-\frac{2 \times 200}{6400}\right] \)
\(\therefore g_{u p}=g\left[1-\frac{2 \times 200}{6400}\right] \)
\(g_{\text {up }} =g\left[\frac{6000 \times 10^{3}}{6400 \times 10^{3}}\right] \)
\(=g \times \frac{15}{16}=g \times 0.9375 \)
\(\mathrm{g}_{\text {up }}=0.94 \mathrm{~g} \)
\(\mathrm{g}_{\text {depth }}=\mathrm{g}_{d}=\mathrm{g}\left[1-\frac{d}{R_{E}}\right] \)
\(g_{\text {down }}=g\left[1-\frac{200 \times 10^{3}}{6400 \times 10^{3}}\right] \)
\(=g\left[1-\frac{200}{6200}\right] \)
\(=g\left[1-\frac{1}{32}\right]=g\left[\frac{32-1}{32}\right] \)
\(=-g \times \frac{31}{32} \)
\(=0.96875 \mathrm{~g} \)
\(\mathrm{g}_{\text {down }} \simeq 0.96 \mathrm{~g}\)

At Earth's surface \(\mathrm{KE}=\frac{1}{2} \mathrm{M} v_{\mathrm{e}}^{2}, \mathrm{P} . \mathrm{E}=-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}\)
At infinity K.E = \(\frac{1}{2} M V_{\infty}^{2}, \mathrm{P} . \mathrm{E}=\mathrm{O}\)
By law of conservation of energy,
\(\frac{1}{2} \mathrm{Mv}_{\mathrm{e}}^{2}-\frac{\mathrm{GMM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}} =\frac{1}{2} \mathrm{M} v_{\infty}^{2}  \)
\(v_{\mathrm{e}}^{2} =v_{\infty}^{2}+\frac{2 \mathrm{GM}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{E}}}  \)
\(\text {But } \mathrm{GM}_{\mathrm{E}} =\mathrm{qR_{ \textrm {E } } ^ { 2 }}  \)
\(\therefore v_{\mathrm{e}}^{2} =v_{\infty}^{2}+2 \mathrm{gR}_{\mathrm{E}}  \)
\(v_{\mathrm{c}} =\sqrt{\mathrm{v}_{\infty}^{2}+2 \mathrm{gR}}\)

Velocity of the Earth around Sun
y = 30 km/s
Kinetic energy \( \mathrm{K} . \mathrm{E}=\frac{1}{2} \frac{G M_{E} M_{\mathrm{s}}}{\left(\mathrm{R}_{E}+\mathrm{h}\right)}\)
Kinetic Energy \( \mathrm{K} . \mathrm{E} =\frac{1}{2} M_{s} V^{2}=\frac{1}{2} \times 1.9 \times 10^{30} \times\left(30 \times 10^{3}\right)^{2} \)
\(=\frac{1.9 \times 900 \times 10^{30+6}}{2} \)
\(\mathrm{~K} . \mathrm{E} =\mathrm{U}_{\mathrm{K}}=26.5 \times 10^{32} \mathrm{~J} \)
Total energy \(\mathrm{E} =-\frac{G M_{s} M_{E}}{2\left(\mathrm{R}_{E}\right)}  \)
\(\mathrm{E}_{\text {tot }} =-\frac{-6.67 \times 10^{-11} \times 1.9 \times 10^{36} \times 5.9 \times 10^{24}}{2 \times 6.4 \times 10^{6}}\)
\(\mathrm{E}_{\mathrm{tot}}=-23.29 \times 10^{32} \mathrm{~J}\)
Total energy has the negative sign. The negative sign in the total energy implies that the satellite is bound to the Earth and Earth cannot escape from the Sun i,e., -ve sign implies that Earth is bounded with Sun.

The Earth to Sun distance \(\mathrm{R}_{\mathrm{E}}=150 \times 10^{6} \mathrm{~km}=150 \times 10^{9} \mathrm{~m}\)
Mass of the Earth \( \mathrm{M}_{E}=5.9 \times 10^{24} \mathrm{~kg}\)
Mass of the Sun \(\mathrm{M}_{\mathrm{s}}=1.9 \times 10^{30} \mathrm{~kg}\)
Gravitational energy of the Earth and Sun
\(\mathrm{U} =-\frac{G m_{1} m_{2}}{r}  \)
\(\mathrm{U} =-\frac{G m_{E} m_{s}}{R_{E}}  \)
\(=-\left[\frac{6.67 \times 10^{-11} \times 5.9 \times 10^{24} \times 1.9 \times 10^{30}}{150 \times 10^{9}}\right]  \)
\(=-\left[\frac{6.67 \times 5.9 \times 1.9}{150} \times 10^{-11+24+30-9}\right]  \)
\( =-\left[\frac{74.7707}{150} \times 10^{34}\right]  \)
\(=-0.49847 \times 10^{34}  \)
\(\mathrm{U}=-49.847 \times 10^{32} \mathrm{~J}\)

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

Ratio of orbital distances \(\frac{d_{1}}{d_{2}}=2\).
Gravitational field \(\mathrm{E} =-\frac{G M}{r^{2}} \hat{r}  \)
\(\mathrm{E}_{1} =-\frac{G M}{d_{1}^{2}}  \)
\(\mathrm{E}_{2} =-\frac{G M}{d_{2}^{2}}  \)
\(\frac{E_{1}}{E_{2}} =-\frac{G M}{d_{1}^{2}} \times \frac{d_{2}^{2}}{(-G M)}  \)
\(\frac{E_{1}}{E_{2}} =\left(\frac{d_{2}}{d_{1}}\right)^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}  \)
\(\therefore \mathrm{E}_{2} =4 \mathrm{E}_{1}\)

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 \(\quad \mathrm{T}_{5}=6, \quad \mathrm{a}_{5}=72=2 \times 6^{2} \Rightarrow 2 \mathrm{~T}_{5}^{2}\)
\(\therefore \alpha \propto 2 \mathrm{~T}^{2}\)