When r₁ > R, the point lies outside the sphere. Then sphere can be considered to be a point mass body whose whole mass can be supposed to be concentrated at its centre. Then gravitational intensity at a point distance r₁ from the centre of sphere will be,
\(I_1=\frac{GM}{r_1^2}\)........(1)
When r₂ < R, the point P lies inside the sphere. The unit mass body placed at P, will experience gravitational pull due to sphere of radius r₂, whose mass is
\(M'=\frac{M\times\frac{4}{3}\pi r_2^3}{\frac{4}{3}\pi R^3}=\frac{Mr_2^3}{R^3}\)
Therefore the gravitational intensity at P will be \(l_2=\frac{GMr_2^3}{R^3}\times\frac{1}{r_2^2}=\frac{GMr_2}{R^3}\) ........(2)
\(\frac{I_1}{I_2}=\frac{GM}{r_1^2}\times\frac{R^3}{GMr_2}=\frac{R^3}{r_1^2r_2}\)

For a satellite revolving just near the Earth's surface, the orbital velocity has a magnitude given by
\(v_{orb}=\sqrt {gR}\)
\(\therefore v_{orb}=\sqrt {9.8\times6400\times1000}\) = 7.92 x 10³ms^-1 or 7.92 km s^-1.

Consider a satellite of mass m revolving around the earth at a height h from its surface so that radius of its orbit r = R + h. If v_o be the orbital velocity of satellite then centripetal force needed by it for its uniform circular motion is
\(F=\frac{mv_0^2}{r}\)
This value of centripetal force is provided by the gravitational pull of the earth acting on the satellite i.e.,
\(F=\frac{GMm}{r^2}\)
For equilibrium, \(\frac{mv_0^2}{r}=\frac{GMm}{r^2}\)
\(\Rightarrow v_0=\sqrt {\frac{GM}{r}}=\sqrt {\frac{GM}{(R+h)}}\)
But \(g=\frac{GM}{R^2},\ hence\ GM=gR^2\)
\(\therefore v_0=\sqrt {\frac{gR^2}{(R+h)}}=R\sqrt {\frac{g}{(R+h)}}.\)

Escape speed from the earth's surface is
\(v_e=[\frac{2GM}{R}]^{\frac{1}{2}}\)
or \([\frac{2GM}{R}]^{\frac{1}{2}}=11.2\) ...........(1)
Escape speed from Saturn's surface will be,
\(v=[\frac{2GM'}{R'}]^{\frac{1}{2}}\)
Now,M' = 95M, R' = 9.5R
\(v=[\frac{2\times95GM}{9.5R}]^{\frac{1}{2}}=3.16\times\sqrt {2\frac{GM}{R}}\)
\([\frac{2GM}{R}]^{\frac{1}{2}}=11.2\ kms^{-1} \)
\(\therefore\) v = 3.16 x 11.2 = 35.4 km s^-1.

Let x be the distance of a point from the earth where resultant gravitational field intensity is zero. So
\(\frac{GM_e}{x^2}=\frac{GM_e}{(r-x)^2}\ or\ \frac{81Mm}{x^2}=\frac{Mm}{(r-x)^2}\ or\frac{9}{x}=\frac{1}{(r-x)}\)
or 9r = 10x or x = 9r/10 = 0.9 r.

Let v₀, v_e be the orbital and escape speeds of the satellite, then \(v_e=\sqrt 2v_0.\)
Energy in the given orbit \(E_1=\frac{1}{2}mv_0^2=E\) .............(1)
Energy for the escape speed, \(E_2=\frac{1}{2}mv_e^2=\frac{1}{2}m(\sqrt 2v_0)^2=2E\)
\(\therefore\) Energy required to be supplied = E₂ - E₁ = E.

Suppose that a satellite of mass m described a circular orbit around a planet of Mass M. The force of attraction between the planet and its satellite is
\(F=-g\frac{Mm}{r^2}\)
This force must be mass times the centripetal acceleration, i.e.,
\(\frac{v^2}{r}=\omega^2r\)
Thus \(m\omega^2r=\frac{4\pi^2mr}{T^2}=G\frac{mM}{r^2}\)
\(M=\frac{4\pi^2r^3}{GT^2}\)

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

The magnitude of the angular momentum at P is L_p = m_pT_pv_p, since inspection tells us that r_p and v_p are mutually perpendicular. Similarly, L_A = m_pr_Av_A From angular momentum conservation
m_pr_pv_p =  m_pr_Av_A
or \(\frac { { v }_{ p } }{ { v }_{ A } } =\frac { { r }_{ A } }{ { r }_{ p } } \)
Since rA > rp, vp > vA . The area SBAC bounded by the ellipse and the radius vectors SB and SC is larger than SBPC in Fig.. From Kepler’s second law, equal areas are swept in equal times. Hence the planet will take a longer time to traverse BAC than CPB.

Imagine a satellite of mass m moving with a velocity v in an orbit of radius r around a planet mass M.
PE of the satellite  ,  \(U=-\frac { GMn }{ r } \)
KE of the satellite, \(K=\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { GMn }{ 2r } \quad \quad [as\ v=\sqrt { Gm/r] } \)
Total energy of the satellite,i.e
\(E=K+U=\frac { GMn }{ 2r } -\frac { GMn }{ r } =-\frac { GMn }{ 2r } \)
For the sake of clarity,take  \(\frac { GMn }{ 2r } =x\)
Clearly,   \(U=-2x,K=x,E=-x\)
The orbitting satellite losses energy due to viscous force acting on it due to Let the new orbital radiuatmosphere and as such it loses height.
Let the new orbital radius be \(\frac { r }{ 2 } \) (say)
Clearly,    \({ U }^{ ' }=-4x\) 
\({ K }^{ ' }=2x\)
\({ E }^{ ' }=-2x\)
Clearly \(\ { E }^{ ' } \) and \({ K }^{ ' }>K\) ,Since ,kinetic energy has incresed,the velocity of the satellite increases.

Mass of the earth,
\(M=6.0\times 10^{ 24 }kg,\quad m=67kg\)
\(G=6.67\times { 10 }^{ -11 }N{ m }^{ 2 }{ kg }^{ -2 }\)
Gravitational potential, 
\(V=-\frac { GM }{ R }\)
\( =-\frac { 6.67\times { 10 }^{ -11 }\times 6\times { 10 }^{ 24 } }{ 6.6\times { 10 }^{ 10 } } \)
\( V=-6.1\times { 10 }^{ 3 }J{ kg }^{ -1 }\) 

\(W=0-\left[ \frac { -GMm }{ R } \right] =\frac { GMm }{ R } \)
\(=g{ R }^{ 2 }\times \frac { m }{ R } =mgR\)
\(\\ =1000\times 10\times 6400\times { 10 }^{ 3 }=64\times { 10 }^{ 9 }J=6.4\times 10^{ 10 }J\) 

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

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

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,
\( T=\frac{2 \pi R}{v_0}=\frac{2 \times 22 \times 6.4 \times 10^6}{7 \times 7290}=5079 \mathrm{~s} \)
\( =1.411 \mathrm{~h}\)