\(\frac{v_1}{v_2}=\frac{T_1}{T_2}=1\) because both satellites are revolving in same orbit and for a given orbit the orbital velocity, as well as time period, is independent of the mass of satellite.

The gravitation force between satellite and the earth provides the centripetal force required by the satellite to move in a circular orbit.The satellite orbits around earth at such a higher height where air friction is neglible.

Escape velocity, 
\({ v }_{ e }=\sqrt { 2 } { v }_{ 0 }\)
\({ v }_{ e }=\sqrt { 2 } \times 8=11.31kms^{ -1 }\)

From the law of conservation of angular momentum
\(m{ r }_{ 1 }{ v }_{ 1 }=m{ r }_{ 2 }{ v }_{ 2 }\quad \Rightarrow \quad { r }_{ 1 }{ v }_{ 1 }={ r }_{ 2 }{ v }_{ 2 }\ \ or\quad \frac { { v }_{ 1 } }{ { v }_{ 2 } } =\frac { { r }_{ 2 } }{ { r }_{ 1 } } \)

The centripetal force required by a satellite of mass m revolving in a circular orbit of radius r with a speed v is supplied by the gravitational force extended by the planet of mass M on the satellite. Thus
\(\frac{Mv^2}{r}=G\frac{Mm}{r^2}\)
\(v=\sqrt {\frac{Gm}{r}}\)
The period of revolution of the satellite is
\(T=\frac{2\pi r}{v}=2\pi\sqrt {\frac{r^3}{GM}}\)
For satellite S₁ let T=T₁,r=r₁,v=v₁
Then \(T_1^2=\frac{4\pi^2r_1^3}{GM}\)
For satellite S₂ , \(T_2^2=\frac{4\pi^2r_2^2}{GM}\therefore \frac{T_1^2}{T_2^2}=\frac{r_1^3}{r_2^3}\)
\(r_2=r_1(\frac{T_2}{T_1})^{\frac{2}{3}}\)
\(=10^4(\frac{8}{1})^{\frac{2}{3}}\)
\(=4\times10^4\ km\)
\(v_1=\frac{2\pi r_1}{T_1}\)
\(=\frac{2\pi\times10^4}{1}=2\pi\times10^4\ km/hr\)
\(v_2=\frac{2\pi r_2}{T_2}\)
\(=\frac{2\pi\times4\times10^4}{8}\)
\(=\pi\times10^4\ km/hour\)
Velocity of S₂ relative to \(S_1=v_2-c_1=\theta_r(say)\)
\(v_r=(\pi\times10^4-2\pi\times10^4)\ km/hr\)
\(=-\pi\times10^4\ km/hour\)
Let r₂-r₁ = r
The angular velocity of S₂ relative to S₁ is given by
\(\omega=\frac{v_r}{r}=\frac{\pi\times10^4}{(4-1)10^4}\ rad/hour\)
\(\omega=\frac{\pi}{3}\ rad/hour\)

Acceleration due to gravity at depth d from the earth's surface is given by
\({ g }^{ ' }=g\left( 1-\frac { d }{ { R }_{ e } } \right) \)
Therefore, acceleration due to gravity decreases with increasing depth.

Height of the earth's satellite\(h=\left( \frac { { T }^{ 2 }R^{ 2 }g }{ 4\pi ^{ 2 } } \right) -R\)
Given,T= 90  min = \(90\times 60=5400s,R=6380km\)
\(g=9.8m/{ s }^{ 2 }=9.8\times { 10 }^{ -3 }km/{ s }^{ 2 }\)
Thus
 \(h=\left[ \frac { (5400)^{ 2 }\times (6380)^{ 2 }\times (9.8\times { 10 }^{ -3 } }{ 4\times 9.87 } \right] ^{ 1/3 }km-6380\quad km\)
or h = (6655 - 6380)km = 275 km

At depth \(=\frac{R}{2}\) value of acceleration due to gravity,
\(g'=g(1-\frac{R}{2R})=\frac{g}{2}\)
At height x,
\(g'=g(1-\frac{2x}{R})\)
\(\therefore g(1-\frac{2x}{R})=\frac{g}{2}\)
\(\frac{1}{2}=\frac{2x}{R}\Rightarrow x=\frac{R}{4}.\)

No, escape velocity is independent of the direction of projection

Yes,escape velocity depends (through slightly) on the location from where the body is projected because with location g changes and so should \({ v }_{ e }(=\sqrt { 2gR) } \) change.

One-eighth the present year
Since \(T^2 \propto r^3 \therefore\left(\frac{T}{T}\right)^2=\left(\frac{1}{4}\right)^3 \Rightarrow T^{\prime}=\frac{1}{8} T\)

\( T \propto r^{3 / 2} \)
\( \therefore \frac{T_1}{T_2}=\left(\frac{10^{13}}{10^{12}}\right)^{3 / 2}=10 \sqrt{10} . \)