\(\mathrm{g}_{\text {latitude'}} \ g^{\prime}=g-\omega^{2} R \cos ^{2} \lambda\)
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
\(\mathrm{T} =2 \pi \sqrt{\frac{R^{3}}{G M}}  \)
\(\mathrm{T}^{2} =\frac{4 \pi^{2} \times R^{3}}{G M}  \)
\(\therefore \mathrm{M} =4 \pi^{2} \times G \times \frac{R^{3}}{T^{2}}  \)
\(\mathrm{~T}^{2} \propto \mathrm{R}^{3}  \)
\(\frac{R^{3}}{T^{2}} =\left[\frac{421700 \times 10^{3}}{1.769}\right]^{3 / 2}=\frac{\left(421.7 \times 10^{6}\right)^{3}}{(1.769)^{2}}  \)
\(=\frac{74991.3 \times 10^{3} \times 10^{18}}{3.1293}=\frac{7499.14 \times 10^{22}}{3.1293}  \)
\(=2396.4 \times 10^{22}\)
\(\mathrm{M} =4 \times 3.14 \times 6.67 \times 10^{-11} \times 2396.4 \times 10^{22}  \)
\(=1.898 \times 10^{27} \mathrm{~kg}\)

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.

If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so during new Moon we can observe solar eclipse. But Moon's orbit is tilted 5^o with respect to Earth's orbit. Due to this 5^o tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment.

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.

(i) Implies that the satellite is bound to the Earth and it cannot escape from the Earth
(ii) As h approaches ∝ the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth's gravity and is not bound to Earth at large distances.

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

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

Due to Earth's spinning motion, the stars in sky appear to move in circular motion about the pole star.

(i) Geostationary satellites are appeared to be stationary when seen from Earth. Its time period is same as that of the Earth.
(ii) A polar satellite orbits Earth from North pole to south pole. Its time period is nearly 100 minutes and it completes many revolution in a day.