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 total mechanical energy of a satellite of mass m in a circular orbit of radius r around a planet of mass M is given by
\(E=-\frac{G M m}{2 r}\)
So, \(E \propto \frac{m}{r}\)
Given,
Ratio of the masses of the satellites, \(\frac{m_A}{m_B}=\frac{3}{1}\)
Ratio of the orbits of the satellites,
\(\frac{r_A}{r_B}=\frac{r}{4 r}=\frac{1}{4}\)
Now, the ratio of the total meachanical energy,
\(\frac{E_A}{E_B}=\frac{m_A}{m_B} \times \frac{r_B}{r_A}\)
\( \Rightarrow \frac{E_A}{E_B}=\frac{3}{1} \times \frac{4}{1} \)
\( \Rightarrow \frac{E_A}{E_B}=12: 1 \)

\(\because\) Total energy, \(E_0=-\frac{G M m}{2 r}\)
Potential energy, \(U=-\frac{G M m}{r}=2 E_0\)
Kinetic energy, \(K=+\frac{G M m}{2 r}=-E_0\).

Step 1: Given
1. The equal mass of the particles is m
2. The radius of the circle is R.
Step 2: Formula used
1. Gravitational force \(=\frac{G M m}{R^2}\)
2. Centrifugal force \(=m \omega^2 R\)
Step 3: Solution
Let the speed of rotation of each particle be v.
By equating gravitational force with centrifugal force
\(\frac{G m^2}{(2 R)^2}=m \omega^2 R \)
\(G=\text { Gravitational constant }\)
2R =  The distance between the center of the two particles.
\(\omega=\) Angular velocity of the particles
Now,
\(\Rightarrow  \frac{G m^2}{4 R^2}=m \omega^2 R \)
\(\Rightarrow  \frac{G m^2}{4 R^2 \cdot m R}=\omega^2 \)
\(\Rightarrow \frac{G m}{4 R^3}=\omega^2 \)
\(\Rightarrow  \omega =\sqrt{\frac{G m}{4 R^3}}\)
From the relation between angular velocity and linear velocity
\( v=\omega R \)
\( \Rightarrow v=\sqrt{\frac{G m}{4 R^3} \times R} \)
\( \Rightarrow v=\sqrt{\frac{G m R^2}{4 R^3}} \)
\( \Rightarrow v=\sqrt{\frac{G m}{4 R}}\)
[Putting value of \(\omega\) from (1)]
Hence the required answer is \(v=\sqrt{\frac{G m}{4 R}}\).

Yes, because kepler's third law is the relation between distance and time.

Contact forces have electrical region.

The law of conservation of angular momentum.

\(T^{ 2 }\alpha R^{ 3 }\)
Hence, \(\frac { { T }_{ 1 }^{ 2 } }{ { T }_{ 2 }^{ 2 } } =\frac { { R }_{ 1 }^{ 3 } }{ { R }_{ 2 }^{ 3 } } \Rightarrow { T }_{ 2 }^{ 2 }=\left[ \frac { { R }_{ 2 } }{ { R }_{ 1 } } \right] ^{ 3 }T_{ 1 }^{ 2 }\)
\(\Rightarrow T_{ 2 }=T_{ 1 }\left[ \frac { R_{ 2 } }{ { R }_{ 1 } } \right] ^{ 3/2 }=365\left( \frac { R/2 }{ R } \right) ^{ 3/2 } \)
\(=365\times \frac { 1 }{ \sqrt [ 2 ]{ 2 } } =129\ days\)

Earth retains most of the atmosphere due to force of gravity.

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
\(\omega=\frac{v_r}{r}=\frac{\pi\times10^4}{(4-1)10^4}\ rad/hour\)
\(\omega=\frac{\pi}{3}\ rad/hour\)

Gravitational attraction between two point objects are given as:
\(F = \dfrac{{G{M_1}{M_2}}}{{{r^2}}}\)…….(1)
Where,
F is the attractive force,
G is gravitational constant,
M₁ and M₂ are the masses of two objects,
r is the distance between the center of masses of the two objects.
The volume of a sphere of radius R is given by:
\(V = \dfrac{4}{3}\pi {R^3}\)……. (2)
Where,
V is the volume of the sphere,
R is the radius of the sphere.
Mass of an object with given density and volume:
\(M = \rho .V\)……. (3)
Where,
M is the mass of the object,
ρ is the density of the object,
Complete step by step solution:
Given:
The radius of the smaller sphere is R.
The radius of a larger sphere is 2R.
The density of both spheres is ρ.
The spheres are kept with their surface touching each other.
To find: Contact force between the spheres.
Step 1:
Use eq.(2) in eq.(3) to get the mass of the first sphere of R as:
\(M_1=\rho \times\left(\frac{4}{3} \pi R^3\right) \)
\(\therefore M_1=\frac{4}{3} \pi \rho R^3\)
Step 2:
Similarly, Use eq.(2) in eq.(3) to get the mass of the first sphere of $2 R$ as:
\(M_2=\rho \times\left(\frac{4}{3} \pi(2 R)^3\right) \)
\(\therefore M_2=\frac{32}{3} \pi \rho R^3\)
Step 3:
For a uniform sphere, its center of mass always stays at its center. As they are kept just in touch so the distance between their center is r=R+2R=3R. Now, substitute r and the values of  M₁ and M₂ obtained from eq.(4) and eq.(5) in eq.(1) to get the attractive force value as:
\(F=\frac{G \times\left(\frac{4}{3} \pi \rho R^3\right) \times\left(\frac{32}{3} \pi \rho R^3\right)}{(3 R)^2} \)
\( \therefore F=\frac{128}{81} G \pi^2 R^4 \rho^2 \)

\({ T }_{ j }=11.6yr,{ r }_{ j }=?,{ T }_{ e }=1yr,{ r }_{ e }=1.5\times { 10 }^{ 11 }m\)
\(\\ \frac { { T }_{ j }^{ 2 } }{ { T }_{ e }^{ 2 } } =\frac { { r }_{ j }^{ 3 } }{ { r }_{ e }^{ 3 } } \)
\(\\ \Rightarrow \quad { r }_{ j }={ r }_{ e }\left( \frac { { { T }_{ j } } }{ { T }_{ e } } \right) ^{ 2/3 }=1.5\times 10^{ 11 }\times \left( \frac { 11.6 }{ 1 } \right) ^{ 2/3 }\)
\(\\ { r }_{ j }=7.68\times 10^{ 11 }m\) 

Let us consider that the rockey sphere has mass M and radius R.The escape speed for such a sphere is given by
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}{ V }_{ e }=\sqrt { \frac { 2GM }{ R } } =\sqrt { \frac { 2G\left( \frac { 4\pi }{ 3 } \right) { R }^{ 2 }P }{ R } } =\sqrt { \frac { 8\pi Gp }{ 3 } } R\quad or\quad R={ v }_{ e }=\sqrt { \frac { 3 }{ 8\pi Gp } } \)
Here,  \(p=3.0g/cm^{ 2 }=3.0\times { 10 }^{ 3 }kg/{ m }^{ 3 },{ v }_{ e }=40\quad m/s\)
Thus \(R=40\sqrt { \frac { 3 }{ 8\times 3.14\times 6.67\times { 10 }^{ -11 }\times 3\times { 10 }^{ 3 } } } m\)
or       \(R=40\times 776.6m=30904m=30.904km\)

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

Here \(R=6400 \mathrm{~km}=6.4 \times 10^6 \mathrm{~m}\) 
g = 9.8 m s^-2
Orbital velocity near the earth's surface is \(v_0=\sqrt{g R}=\sqrt{9.8 \times 6.4 \times 10^6}=7290 \mathrm{~ms}^{-1}\) Time period,
\( T=\frac{2 \pi R}{v_0}=\frac{2 \times 22 \times 6.4 \times 10^6}{7 \times 7290}=5079 \mathrm{~s} \)
\( =1.411 \mathrm{~h}\)

Here, \(M_e=9 M_m\), and \(R_e=2 R_m\)
v_e (escape speed on surface of Earth) \(=11.2 \mathrm{~km} / \mathrm{s}\)
Let V_m be the speed required to pull out of the gravitational force of mars.
We know that
\(v_e=\sqrt{\frac{2 G M_e}{R_e}}\)  and \(v_m=\sqrt{\frac{2 G M_m}{R_m}}\)
Dividing, we get \(\frac{v_m}{v_e}=\sqrt{\frac{2 G M_m}{R_m} \times \frac{R_e}{2 G M_e}}\)
\( =\sqrt{\frac{M_m}{M_e} \times \frac{R_e}{R_m}}=\sqrt{\frac{1}{9} \times 2}=\frac{\sqrt{2}}{3} \)
\( \Rightarrow v_m=\frac{\sqrt{2}}{3}(11.2 \mathrm{~km} / \mathrm{s})=5.3 \mathrm{~km} / \mathrm{s} \)

7.417 km/s, 1 h 43 min 3 s

Given mass of satelite m=400 kg
initial energy is given by \(E_i=\frac{-G M m}{4 R}\)
final energy is givne by \(E_f=\frac{-G M m}{8 R}\)
\( \triangle E=\frac{G M m}{8 R}=\frac{G M}{R^2} \frac{m R}{8} \)
\(=\frac{g m R}{8} \)
\(=\frac{9.81 \times 400 x 6.37 \times 10^6}{8} \)
\( \triangle E=3.13 \times 10^9 j\)
\(-3.13\times 10^{ 9 }J,\ -6.26\times 10^{ 9 }J\)