Q: Two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

Given,Masses of the two particles are \$m_1=m_2=m\$And radius of the two particles are \$r_1=r_2=R\$So ,The Gravitational force of attraction between the particles \$F=\dfrac{G\times{m}\times{m}}{R^2}--------(1)\$Ans, centripital force \$F=\dfrac{m\times{v^2}}{R}------(2)\$From given condition (1) and (2)\$\dfrac{Gmm}{(2R)^2}=\dfrac{mv^2}{R}\$, \$v=\dfrac{Gm}{4R}=\dfrac{1}{2}\sqrt{\dfrac{Gm}{R}}\$So the speed of the each particle is \$v=\dfrac{1}{2}\sqrt{\dfrac{Gm}{R}}\$

Q: A ball thrown up vertically returns to the thrower after \$6\$s. Find\$(a)\$ The velocity with which it was thrown up,\$(b)\$ The maximum height it reaches, and\$(c)\$ Its position after \$4\$s.

The ball returns to the ground after \$6\$ seconds.Thus the time taken by the ball to reach to the maximum height \$(h)\$ is 3 seconds i.e \$t = 3\$ sLet the velocity with which it is thrown up be \$u\$\$(a)\$. For upward motion, \$v = u+at\$\$\therefore\$ \$0 = u + (-10) \times 3\$ \$\implies u = 30\$ \$m/s\$ \$(b)\$. The maximum height reached by the ball\$h = ut+ \dfrac{1}{2}at^2\$\$h = 30 \times 3+ \dfrac{1}{2} (-10) \times 3^2\$ \$h = 45\ m\$ \$(c)\$. After \$3\$ second, it starts to fall down. Let the distance by which it fall in \$1\$ s be \$d\$\$d = 0 + \dfrac{1}{2}a t'^2\$ where \$t' = 1\$ s\$d = \dfrac{1}{2} \times 10 \times (1)^2 \$ \$ = 5\ m\$ \$\therefore\$ Its height above the ground, \$h' = 45 - 5 = 40\ m\$ Hence after \$4\ s\$, the ball is at a height of \$40\ m\$ above the ground.

Q: A stone is released from the top of a tower of height \$19.6\$m. Calculate its final velocity just before touching the ground.

Initial Velocity \$ u = 0\$Fianl velocity \$ v = ?\$Height, \$s = 19.6 m\$By third equation of motion\$v^{2} = u^{2} + 2 gs\$\$v^{2} = 0 + 2 \times 9.8\times19.6\$\$v^{2} = 384.16\$\$\Rightarrow v = 19.6 m/s\$

Q: The time period of oscillation of a simple pendulum is \$T = 2\pi\sqrt{(l/g)}\$ .measured value of length is 20cm known to have 1mm accuracy and time for 100 oscillation of the pendulum is found to be 90s using wrist watch of 1s resolution.what is the accuracy in determining the value of acceleration due to gravity?

\$T=\$ Time period of one oscillation\$t=\$ Time period of 100 oscillationFor 100 oscillation, \$t=100\,T\$Take ln and differentiate.\$\dfrac{\Delta t}{t}=\dfrac{\Delta T}{T}=\dfrac{2}{90}\,\,\,.....\,\,(1)\,\$Time period of one oscillation\$T=2\pi \sqrt{\dfrac{L}{g}}\$\$g=4{{\pi }^{2}}\dfrac{L}{{{T}^{2}}}\$Take ln on both side and differentiate\$ \dfrac{\Delta g}{g}=\pm \left( \dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T} \right) \$\$ \dfrac{\Delta g}{g}=\pm \left( \dfrac{0.1}{20}+\dfrac{2}{90} \right)==\pm 2.7% \$Hence, error in gravity is \$\pm 2.7\,\$%

Q: Derive an expression for acceleration due to gravity at a depth 'd' below the earth's earth surface.

Let \$M\$ be mass of the earth\$,\$ \$R\$ be the radius of the earth\$g_d\$ be gravitational acceleration at depth \$'d'\$ from the earth surface\$g\$ be gravitational acceleration on the earth surfaces\$.\$\$p\$ be the density of the earth\$.\$\$'p'\$ be the point inside the earth at depth \$'d'\$ from earth surfaces\$.\$\$\therefore CS - CP = d,\ \, \, \,\therefore CP = R - d\$-----------\$(1)\$ \$(\$since CS=R\$)\$\$g = \dfrac{{GM}}{{{R^2}}}\$\$\therefore g = \dfrac{{G\frac{4}{3}\pi {R^3}p}}{{{R^2}}}\$\$\therefore g = \dfrac{{4G\pi Rp}}{3}\$--------------\$(2)\$\$gd=\$ acceleration due to gravity at depth \$'d'\$\$g_d=\dfrac{{G \times Mass\,of\,the\,sphere\,with\,radius\,CP}}{{C{p^2}}}\$\$\therefore g_d = \dfrac{{G\frac{4}{3}\pi C{P^3}\rho }}{{C{P^2}}}\$\$\therefore g_d = \dfrac{{4G\pi CP\rho }}{3}\$-----------\$(3)\$Dividing eq\$.\$ \$(3)\$ by eq\$(2)\$\$\dfrac{{g_d}}{g} = \dfrac{{CP}}{R} = \dfrac{{R - d}}{R}\$\$\therefore g_d = g\left( {1 - \dfrac{d}{R}} \right)\$

Q: A body weighs \$63 \$N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to half the radius of the earth ?

Weight of the body,\$W=63\ N\$Acceleration due to gravity at height \$h\$ from the Earths surface is given by the relation:\$g'=\dfrac{g}{(1+\dfrac{h}{R_e})^2}\$Substituting \$h=R_e/2\$,\$g'=4g/9\$Weight of body of mass m at height h is given as:\$W'=mg'\$ \$=4/9\ mg\$ \$=4/9\times 63=28\ N \$

Q: A planet is having a mass twice to that of earth's mass and its radius as 4 times that of the earth's radius. Determine four times the acceleration due to gravity at the surface of this planet. Acceleration due to gravity at the earth's surface is \$10m{s}^{-2}\$.

\$q'=\dfrac{GM}{R^2}\$ \$=\dfrac{G\times 2me}{\left( 4ve\right)^2}\$ \$=\dfrac{1}{8}\dfrac{Gme}{ve^2}\$ \$=\dfrac{g}{8}\$To find \$4\$ times the acceleration due to gravity \$=4\times \dfrac{1}{8}g=\dfrac{g}{2}=\dfrac{10}{2}\$\$=5m/s^2\$Hence, the answer is \$5m/s^2.\$

Q: If \$v_e\$ and \$v_0\$ represent the escape velocity and orbital velocity of a satellite corresponding to circular orbit of radius \$R\$, then

escape velocity \$v_e=\sqrt{GM/R}\$Orbital velocity \$v_o=\sqrt{GM/2R}\$so, \$v_e=\sqrt 2v_o\$

Question 1

Two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

Accepted Answer

Given,Masses of the two particles are $m_1=m_2=m$And radius of the two particles are $r_1=r_2=R$So ,The Gravitational force of attraction between the particles $F=\dfrac{G\times{m}\times{m}}{R^2}--------(1)$Ans, centripital force $F=\dfrac{m\times{v^2}}{R}------(2)$From given condition (1) and (2)$\dfrac{Gmm}{(2R)^2}=\dfrac{mv^2}{R}$, $v=\dfrac{Gm}{4R}=\dfrac{1}{2}\sqrt{\dfrac{Gm}{R}}$So the speed of the each particle is $v=\dfrac{1}{2}\sqrt{\dfrac{Gm}{R}}$

Question 2

A ball thrown up vertically returns to the thrower after $6$s. Find$(a)$ The velocity with which it was thrown up,$(b)$ The maximum height it reaches, and$(c)$ Its position after $4$s.

Accepted Answer

The ball returns to the ground after $6$ seconds.Thus the time taken by the ball to reach to the maximum height $(h)$ is 3 seconds i.e $t = 3$ sLet the velocity with which it is thrown up be  $u$$(a)$. For upward motion,               $v = u+at$$\therefore$     $0 = u + (-10) \times 3$                 $\implies u = 30$  $m/s$     $(b)$. The maximum height reached by the ball$h = ut+  \dfrac{1}{2}at^2$$h = 30 \times 3+  \dfrac{1}{2}  (-10)  \times 3^2$              $h = 45\ m$  $(c)$. After $3$ second, it starts to fall down. Let the distance by which it fall in $1$ s   be   $d$$d = 0 +  \dfrac{1}{2}a t'^2$                        where   $t' = 1$ s$d =   \dfrac{1}{2} \times 10 \times (1)^2  $    $ =  5\ m$  $\therefore$ Its height above the ground, $h' = 45 - 5  = 40\ m$ Hence after $4\ s$, the ball is at a height of $40\ m$ above the ground.

Question 3

A stone is released from the top of a tower of height $19.6$m. Calculate its final velocity just before touching the ground.

Accepted Answer

Initial Velocity $ u = 0$Fianl velocity $ v = ?$Height, $s = 19.6 m$By third equation of motion$v^{2} = u^{2} + 2 gs$$v^{2} = 0 + 2 \times 9.8\times19.6$$v^{2}  = 384.16$$\Rightarrow v = 19.6 m/s$

Question 4

Discuss the variation of g with(a) altitude (b) depth

Accepted Answer

( a )  The acceleration due to gravity on the earth is given by&#10;&#10;$g = \dfrac {GM}{R^2} i.e. GM = R^2g$  . . . . . . .( 1 )&#10;&#10;The acceleration due to gravity at height 'h' from the surface of the&#10;earth is given by&#10;&#10;$g_h = \dfrac{GM}{(R+h)^2} i.e. GM = (R+h)^2 g_h$ . . . . . . ( 2 )&#10;&#10;From ( 1 ) and ( 2 ) we have,&#10;&#10;$\dfrac{g_h}{g} = \dfrac {R^2}{(R+h)^2}$&#10;&#10;$\dfrac{g_h}{g} = (1 - \dfrac{2h}{R})$  &#10;&#10;( b )  The acceleration due to gravity on the surface of the earth&#10;is given by.&#10;&#10;$g = \dfrac{GM}{R^2}$ . . . .( 1 )&#10;&#10;let 'Q' be the density of the material of the earth.&#10;&#10; Now, mass = volume $\times$ density&#10;&#10; $M = \dfrac{4}{3} \pi R^3 \times \rho$&#10;&#10; Substituting in equation ( 1 ) we get &#10;&#10; $g = \dfrac{G}{R^2} \times \dfrac{4}{3} \pi R^3 \times \rho = \dfrac{4}{3}&#10;\pi GR \rho$&#10;&#10; $g = \dfrac{4}{3} \pi GR \rho$ . . .. .. ( 2 )&#10;&#10; $g_d = \dfrac{4}{3} \pi G(R-d) \rho$  . . . .. .( 3 )&#10;&#10; Dividing equation( 3 ) by ( 2 ) we get &#10;&#10; $\dfrac{g_d}{g} = \dfrac{R-d}{R} = (1 - \dfrac{d}{R})$&#10;&#10; $g_d = g ( 1 - \dfrac{d}{R})$

Question 5

The time period of oscillation of a simple&#10;pendulum is $T = 2\pi\sqrt{(l/g)}$ .measured value of length is 20cm known to have 1mm&#10;accuracy and time for 100 oscillation of the pendulum is found to be 90s using&#10;wrist watch of 1s resolution.what is the accuracy in determining the value of&#10;acceleration due to gravity?&#10;&#10;

Accepted Answer

$T=$ Time period of one oscillation$t=$ Time period of 100 oscillationFor 100 oscillation, $t=100\,T$Take ln and differentiate.$\dfrac{\Delta t}{t}=\dfrac{\Delta T}{T}=\dfrac{2}{90}\,\,\,.....\,\,(1)\,$Time period of one oscillation$T=2\pi \sqrt{\dfrac{L}{g}}$$g=4{{\pi }^{2}}\dfrac{L}{{{T}^{2}}}$Take ln on both side and differentiate$ \dfrac{\Delta g}{g}=\pm \left( \dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T} \right) $$ \dfrac{\Delta g}{g}=\pm \left( \dfrac{0.1}{20}+\dfrac{2}{90} \right)==\pm 2.7% $Hence, error in gravity is $\pm 2.7\,$%

Question 6

Derive an expression for acceleration due to gravity at a depth 'd' below the earth's earth surface.

Accepted Answer

Let $M$ be mass of the earth$,$ $R$ be the radius of the earth$g_d$ be gravitational acceleration at depth $'d'$ from the earth surface$g$ be gravitational acceleration on the earth surfaces$.$$p$ be the density of the earth$.$$'p'$ be the point inside the earth at depth $'d'$ from earth surfaces$.$$\therefore CS - CP = d,\ \, \, \,\therefore CP = R - d$-----------$(1)$ $($since CS=R$)$$g = \dfrac{{GM}}{{{R^2}}}$$\therefore g = \dfrac{{G\frac{4}{3}\pi {R^3}p}}{{{R^2}}}$$\therefore g = \dfrac{{4G\pi Rp}}{3}$--------------$(2)$$gd=$ acceleration due to gravity at depth $'d'$$g_d=\dfrac{{G \times Mass\,of\,the\,sphere\,with\,radius\,CP}}{{C{p^2}}}$$\therefore g_d = \dfrac{{G\frac{4}{3}\pi C{P^3}\rho }}{{C{P^2}}}$$\therefore g_d = \dfrac{{4G\pi CP\rho }}{3}$-----------$(3)$Dividing eq$.$ $(3)$ by eq$(2)$$\dfrac{{g_d}}{g} = \dfrac{{CP}}{R} = \dfrac{{R - d}}{R}$$\therefore g_d = g\left( {1 - \dfrac{d}{R}} \right)$

Question 7

A body weighs $63 $N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to half the radius of the earth ?

Accepted Answer

Weight of the body,$W=63\ N$Acceleration due to gravity at height $h$ from the Earths surface is given by the relation:$g'=\dfrac{g}{(1+\dfrac{h}{R_e})^2}$Substituting $h=R_e/2$,$g'=4g/9$Weight of body of mass m at height h is given as:$W'=mg'$       $=4/9\ mg$       $=4/9\times 63=28\ N $

Question 8

What is escape velocity? Obtain an expression for it.

Accepted Answer

Escape velocity: -The minimum speed with which a body must be projected in order that it will escape from the earth gravitational filed is called as escape velocity.Expression of it:-consider a body of mass $m$ on the surface of earthwhose mass and radius are respectively represented by $M$ and $R$.and earth is $F = \dfrac{{GMm}}{{{R^2}}}$gravity potential energy $\left( {GPE} \right) = \dfrac{{GMm}}{{{R^2}}}$$w = F.S = \dfrac{{GMm}}{{{R^2}}}$In order to projector body into the space overcoming the earth's gravitational field it must be give a kinetic energy which is equal topotential energy.therefore;kinetic energy is equal to potential energy.

Question 9

A planet is having a mass twice to that of earth's mass and its radius as 4 times that of the earth's radius. Determine four times the acceleration due to gravity at the surface of this planet. Acceleration due to gravity at the earth's surface is $10m{s}^{-2}$.

Accepted Answer

$q'=\dfrac{GM}{R^2}$    $=\dfrac{G\times 2me}{\left( 4ve\right)^2}$    $=\dfrac{1}{8}\dfrac{Gme}{ve^2}$    $=\dfrac{g}{8}$To find $4$ times the acceleration due to gravity $=4\times \dfrac{1}{8}g=\dfrac{g}{2}=\dfrac{10}{2}$$=5m/s^2$Hence, the answer is $5m/s^2.$

Question 10

If $v_e$ and $v_0$ represent the escape velocity and orbital velocity of a satellite corresponding to circular orbit of radius $R$, then

Accepted Answer

escape velocity $v_e=\sqrt{GM/R}$Orbital velocity $v_o=\sqrt{GM/2R}$so, $v_e=\sqrt 2v_o$