Escape speed v_e = \(\sqrt 2g R\)
if g' = 4g 
then \(v'_e\) = \(\sqrt (4g) R\)
= \(\sqrt 2g R \) \(\times\)2
= 2v_e

\(\text { Escape speed } v_{e}=\sqrt{2 g R}\)
\(\text { If } g^{\prime}=4 \mathrm{~g}\)
\(\text { Then } v_{e}^{\prime}=\sqrt{2(4 g) R}\)
\(=\sqrt{2 g R} \times 2 \)
\(=2 v_{e} \)

Moment of inertia of a disc passing through
\(\text { centre } I=\frac{1}{2} M R^{2}\)
\(\text { Energy of the first disc }=\frac{1}{4} I \omega_{1}^{2}\)
\(\text { Energy of the second disc }=\frac{1}{4} I \omega_{2}^{2}\)
\(\text { Loss of energy }=\frac{1}{4} I\left(\omega_{1}-\omega_{2}\right)^{2}\)

\(v_{0}=r \omega ; \quad \therefore v_{0} \alpha r\)
For a wheel (uniform ring) the distance of a point on the rim in level with the center 
\(\text { [i.e., radius] is } \sqrt{2} r\)
\(\therefore \text { The speed of the center is } \sqrt{2} v_{0}\)

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 \(V=\frac{1}{2} \sqrt{\frac{G M}{R}(1+2 \sqrt{2})}\)

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.

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

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

The conditions are
(i) When position vector \(\vec{r}\) and force \(\vec{F}\) are parallel or antiparallel.
(ii) If the force acts at the reference point.

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

Stable equilibrium	Unstable equilibrium
The body tries to come back to equilibrium if slightly disturbed and released.	The body cannot come back to equilibrium if slightly disturbed and released.
The center of mass of the body shifts slightly higher if disturbed from equilibrium.	The center of mass of the body shifts slightly lower if disturbed from equilibrium.
Potential energy of the body is minimum and it increases if disturbed.	Potential energy of the body is not minimum and it decreases if disturbed.

The expression for magnitude of angular momentum of a rigid body, L = I\(\omega\) . The expression for magnitude of torque on a rigid body is, \(\tau\) = l\(\alpha\)
(ii) We can further write the expression for torque as,
\(r=1{{d\omega}\over{dt}}\because\left(\alpha={{d\omega}\over{dt}} \right)\)
where, \(\omega\) is angular velocity and a is angular acceleration.
(iii) We can also write equation (1) as,
\(\tau={{d(I\omega)}\over{dt}}\)
\(\tau={{dL}\over{dt}}\)
An external torque on a rigid body fixed to an axis produces rate of change of angular momentum in the body about that axis.

Examples of torque in day-to-day life are
1. The opening and closing of a door about the hinges and
2. Turning of a nut using a wrench.