Gravitation

Gravitational force applied on the satellite,

= ${{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,$

For satellite, gravitational force = centripetal force

$\therefore$ ${{m{v^2}} \over {\left( {R + x} \right)}} = {{GMm} \over {{{\left( {R + x} \right)}^2}}}\,\,$

where $v$ is the orbital speed of satellite.

also $\,\,g = {{GM} \over {{R^2}}}$ $ \Rightarrow GM = g{R^2}$

$\therefore$ ${v^2} = {{g{R^2}} \over {R + x}} $

$\Rightarrow v = {\left( {{{g{R^2}} \over {R + x}}} \right)^{1/2}}$

For satellite, gravitational force = centripetal force

$\therefore$ ${{m{v^2}} \over {R + x}} = {{GmM} \over {{{\left( {R + x} \right)}^2}}}$

$x=$ height of satellite from earth surface

$m=$ mass of satellite

$ \Rightarrow {v^2} = {{GM} \over {\left( {R + x} \right)}}$ or $v = \sqrt {{{GM} \over {R + x}}} $

We know, $T = {{2\pi } \over \omega }$

$T = {{2\pi \left( {R + x} \right)} \over v} $ [ as $\omega = {v \over r}$ ]

$= {{2\pi \left( {R + x} \right)} \over {\sqrt {{{GM} \over {R + x}}} }}$

which is independent of mass of satellite

For moving a planet around the sun in the circular orbit,

The necessary centripetal force = Gravitational force exerted on it

$\therefore$ ${{m{v^2}} \over r} = {{GMm} \over {{R^n}}}$

$ \Rightarrow $ $v = \sqrt {{{GM} \over {{R^{n - 1}}}}} $

We know, $T = {{2\pi R} \over v}$

$ = 2\pi R \times \sqrt {{{{R^{n - 1}}} \over {GM}}} $

= $2\pi \times \sqrt {{{{R^2} \times {R^{n - 1}}} \over {GM}}} $

= $2\pi \times {{{R^{{{n + 1} \over 2}}}} \over {\sqrt {GM} }}$

$\therefore$ $T \propto {R^{{{ \left( {n + 1} \right)} \over 2}}}$

We know, Escape velocity, ${v_e} = \sqrt {2gR} $

So the escape velocity is independent of the angle at which the body is projected, hence it will remain same as 11 km/s.

According to kepler's law,

${T^2} \propto {R^3}$

$\therefore$ ${{T_1^2} \over {T_2^2}} = {{R_1^3} \over {R_2^3}}$

$ \Rightarrow $ ${T_2} = {T_1}{\left( {{{{R_2}} \over {{R_1}}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} = 5 \times {\left[ {{{4R} \over R}} \right]^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$

$ = 5 \times {2^3} = 40\,\,$ hour

Let $t$ be the time taken for the two masses to collide and ${x_{5M,}}\,{x_M}$ be the distance travelled by the mass $5M$ and $M$ respectively.

The gravitational force acting between two sphere when the distance between them (12R - x) where x is a variable,

$F = {{GM \times 5M} \over {{{\left( {12R - x} \right)}^2}}}$

Acceleration of mass M, ${a_M} = {{G \times 5M} \over {{{\left( {12R - x} \right)}^2}}}$

Acceleration of mass 5M, ${a_{5M}} = {{GM} \over {{{\left( {12R - x} \right)}^2}}}$

For mass $5M$

$u = 0,\,\,S = {x_{5M}},\,\,t = t,\,\,a{ = a_{5M}}$

$S = ut + {1 \over 2}a{t^2}$

$\therefore$ ${x_{5M}} = {1 \over 2}{a_{5M}}{t^2}\,\,\,\,\,\,\,\,\,\,...\left( {i} \right)$

For mass $M$

$u = 0,\,\,s = {x_M},\,\,t = t,\,\,a = {a_M}$

$\therefore$ $s = ut + {1 \over 2}a{t^2} \Rightarrow $

${x_M} = {1 \over 2}{a_M}{t^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

Dividing $(ii)$ by $(iii)$

${{{x_{5M}}} \over {{x_M}}} = {{{1 \over 2}{a_5}_M{t^2}} \over {{1 \over 2}{a_M}{t^2}}}$

$ = {{{a_{5M}}} \over {{a_M}}} = {1 \over 5}$

$\therefore$ $5{x_{5M}} = {x_M}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left....( {iii} \right)$

From the figure,

${x_{5M}} + {x_M} = 12R - 2R - R = 9R\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( iv \right)$

From $(iii)$ and $(iv)$

${{{x_M}} \over 5} + {x_M} = 9R$

$\therefore$ $6{x_M} = 45R$

$\therefore$ ${x_M} = {{45} \over 6}R = 7.5R$

So two sphere collide when the sphere of mass M covered the distance of 7.5R.

Gravitational potential energy E = $ - {{GMm} \over r}$

where M = mass of earth

m = mass of body

r = radius of earth

Energy required to move a body of mass $m$ from an orbit of radius $2R$ to $3R$

$=$ (Potential energy of the Earth-mass system when mass is at distance $3R$ ) $-$ (Potential energy of the Earth-mass system when mass is at distance $2R$)

$ = {{ - GMm} \over {3R}} - \left( {{{ - GMm} \over {2R}}} \right)$

$ = {{ - GMm} \over {3R}} + {{GMm} \over {2R}}$

$ = {{ - 2GMm + 3GMm} \over {6R}} = {{GMm} \over {6R}}$

Escape velocity,

${v_e} = \sqrt {2gR} = \sqrt {{{2GM} \over R}} \Rightarrow {V_e}\, \propto \,{m^0}$

Where $M,R$ are the mass and radius of the planet respectively. In this expression the mass of the body $(m)$ is not present. The escape velocity is independent of the mass m or it depends on m⁰.

The required velocity is called escape velocity (${v_e}$) to leave the earth surface of a body.

${v_e} = $ escape velocity $ = \sqrt {2gR} $

Kinetic Energy $K.E = {1 \over 2}mv_e^2$

$\therefore$ $K.E = {1 \over 2}m \times 2gR = mgR$

When gravitational force of attraction between Earth and a satellite revolving around it becomes zero, then the centripetal force becomes zero. So the satellite will move tangentially to the original orbit with the same velocity as it has at the instant when gravitational force becomes zero.