Gravitation

Orbital velocity of a satellite in a circular orbit of radius a is given by

$v = \sqrt {{{GM} \over a}} $

$ \Rightarrow v \propto \sqrt {{1 \over a}} $

$ \Rightarrow {{{v_2}} \over {{v_1}}} = \sqrt {{{{a_1}} \over {{a_2}}}} $

$ \therefore $ ${v_2} = {v_1}\sqrt {{{4R} \over R}} $ = 2v₁ = 6V

Here, Mass of a particle = M
Mass of a spherical shell = M
Radius of a spherical shell = a
Let O be centre of a spherical shell.
Gravitational potential at point P due to particle at O is

${V_1} = - {{GM} \over {a/2}}$

Gravitational potential at point P due to spherical shell is

${V_2} = - {{GM} \over a}$

Hence, total gravitational potential at point P is V = V₁ + V₂

$ = - {{GM} \over {a/2}} + \left( { - {{GM} \over a}} \right) = - {{2GM} \over a} - {{GM} \over a} = - {{3GM} \over a}$

$\left| V \right| = {{3GM} \over a}$

Since the man is in gravity free space, force on man + stone system is zero.

Therefore centre of mass of the system remains at rest. Let the man goes x m above when the stone reaches the floor, then

M_man × x = M_stone × 10

$x = {{0.5} \over {50}} \times 10$

x = 0.1 m

Therefore final height of man above floor = 10 + x = 10 + 0.1 = 10.1 m

According to Kepler’s law, the areal velocity of a planet around the sun always remains constant.

SCD : A₁– t₁ (areal velocity constant)
SAB : A₂ – t₂

${{{A_1}} \over {{t_1}}} = {{{A_2}} \over {{t_2}}}$

${t_1} = {t_2}.{{{A_1}} \over {{A_2}}}$   (given A₁ = 2A₂)

$ = {t_2}.{{2{A_2}} \over {{A_2}}}$

$ \therefore $ ${t_1} = 2{t_2}$

Since orbital velocity of satellite is $v = \sqrt {{{GM} \over r}} $

it does not depend upon the mass of the satellite.

Therefore, both satellites will move with same speed.

Escape velocity of the body from the surface of earth is v = $\sqrt {2gR} $

For escape velocity of the body from the platform potential energy + kinetic energy = 0

$ - {{GMm} \over {2R}} + {1 \over 2}m{v^2} = 0$

$ \Rightarrow f{v_{escape}} = \sqrt {{{GM} \over {{R^2}}}.R} = \sqrt {gR} = fv$

From the surface of the earth, ${v_{escape}} = \sqrt {2gR} $

$ \therefore f{v_{escape}} = {{{v_{escape}}} \over {\sqrt 2 }}$, $ \therefore $ $f = {1 \over {\sqrt 2 }}$

K.E. of satellite moving in an orbit around the earth is

$K = {1 \over 2}m{v^2} = {1 \over 2}m{\left( {\sqrt {{{GM} \over r}} } \right)^2} = {{GMm} \over {2r}}$

P.E. of satellite and earth system is

$U = {{GMm} \over r}$

$ \Rightarrow {K \over U} = {{{{GMm} \over {2r}}} \over {{{GMm} \over r}}} = {1 \over 2}$

We know that

$g = {{GM} \over {{R^2}}} = {{G\left( {{4 \over 3}\pi {R^3}} \right)} \over {{R^2}}} = {4 \over 3}\pi GR\rho $

${{g'} \over g} = {{R'} \over R} = {{3R} \over R} = 3$

$ \therefore $ g' = 3g

From equation of acceleration due to gravity.

${g_e} = {{G{M_e}} \over {R_e^2}} = {{G\left( {4/3} \right)\pi R_e^3} \over {R_e^2}}{\rho _e}$

${g_e} \propto {R_e}{\rho _e}$

Acceleration due to gravity of planet

${g_p} \propto {R_p}{\rho _p}$

$ \therefore $ ${R_e}{\rho _e} = {R_p}{\rho _p} \Rightarrow {R_e}{\rho _e} = {R_p}2{\rho _e}$

$ \Rightarrow {R_e} = {1 \over 2}R$   ($ \because $ R_e = R)

Gravitational potential energy on earth’s surface = $ - {{GMm} \over R}$, where M and R are the mass and radius of the earth respectively, m is the mass of the body and G is the universal gravitational constant.

Gravitational potential energy at a height h = 3R

$ = - {{GMm} \over {R + h}} = - {{GMm} \over {R + 3R}} = - {{GMm} \over {4R}}$

$ \therefore $ Change in potential energy

$ = - {{GMm} \over {4R}} - \left( { - {{GMm} \over R}} \right)$

$ = - {{GMm} \over {4R}} + {{GMm} \over R}$

$ = {3 \over 4}{{GMm} \over R}$ $ = {3 \over 4}mgR$

The velocity of the mass while reaching the surface of both the planets will be same.

i.e., $\sqrt {2g'h'} = \sqrt {2gh} $

$\sqrt {2 \times g \times h'} = \sqrt {2 \times 9g \times 2} $

$ \Rightarrow 2h' = 36$

$ \Rightarrow h' = 18\,m$

Gravitational force is independent of medium, Hence, this will remain same.

Using ${v^2} = {{2gh} \over {1 + {h \over R}}}$ and given h = R

$ \therefore $ $v = \sqrt {gR} = \sqrt {{{GM} \over R}} $

Gravitational force is required for convection

${F_{surface}} = G{{Mm} \over {R_e^2}}$

${F_{{R_e}/2}} = G{{Mm} \over {{{\left( {{R_e} + {R_e}/2} \right)}^2}}} = {4 \over 9} \times {F_{surface}}$

$ = {4 \over 9} \times 72 = 32\,N$

${v_{earth}} = \sqrt {{{2G{M_e}} \over {{R_e}}}} $

${v_{planet}} = \sqrt {{{2G{M_p}} \over {{R_p}}}} = \sqrt {{{2G{M_e}} \over {{R_e}/4}}} = \sqrt {{{8G{M_e}} \over {{R_e}}}} $

${{{v_{planet}}} \over {{v_{earth}}}} = \sqrt {{{8G{M_e}} \over {{R_e}}}} \times \sqrt {{{{R_e}} \over {2G{M_e}}}} = 2$

$ \therefore $ ${v_{planet}} = 2 \times {v_{earth}}$

= 2 $ \times $ 11.2 = 22.4 km/s