Gravitation Practice Questions

$h=2 R$ $\therefore g^{'}=\;\frac{G M}{(R+h)^2}$ $=\;\frac{G M}{9 R^2}$ $=\;\frac{g}{9}$

Total gravitational potential energy$=- \{\;{4 G M m}/{d / {\sqrt{2}}}+\;\frac{4 G m^2}{d}+\;{2 G m^2}/{{\sqrt{2}} d}\}$ $=-\;\frac{G m}{d}\{M 4 {\sqrt{2}}+(4+{\sqrt{2}}) m\}$ $=-\;\frac{G m}{d}\{4 {\sqrt{2}} M+(4+{\sqrt{2}}) m\}$

$g=\;{ {\text{GM}}}/{ {\text{R}}^2}$ ${\text{M}}=\;\text{ constant }\; {\text{g}}< \;{1}/{ {\text{R}}^2}$ $100 \;{\Delta {\text{g}}}/{ {\text{g}}}=-2 \;{\Delta {\text{R}}}/{ {\text{R}}} 100$ $\%$ change $=-2(-2)$ $\%$ change in ${\text{g}}=4 \%$increase by $4 \%$

$v_{e s c}-\sqrt{\;\frac{2 G M}{R}}=\sqrt{\;\frac{2 G}{R} \times \rho \times \;\frac{4}{3} \pi R^3}$ $\Rightarrow v_{e s c} \propto R {\sqrt{\rho }}$ $\Rightarrow \;\frac{ (v_{e s c}) _B}{ (v_{e s c}) _A}=1$ $\Rightarrow (v_{e s c}) _B=12 {\text{km}} / {\ s}$

Given, distance between centre of ring and sphere, $d=\sqrt{8} R$ Since, gravitational field at the axis of ring, $E=\;\frac{G m d}{(d^{2}+R^{2})^{3 / 2}}$ Here, $G$ is the gravitational constant. $\Rightarrow \;\; E=\;\frac{G m R \sqrt{8}}{(8 R^{2}+R^{2})^{3 / 2}}=\;\frac{G m R \sqrt{8}}{(3 R)^{3}}$ $\Rightarrow \;\; E=\;\frac{G m R \sqrt{8}}{27 R^{3}}=\;\frac{G m \sqrt{8}}{27 R^{2}}$ Force between ring and sphere, $F=M E$ . . . (i) Substituting the value of $E$ in Eq. (i), we get $F=\;\frac{\sqrt{8}}{27} \;\frac{G m M}{R^{2}}$

Let the mass of planet be $m_{p}$ and that of earth be $m$ .Given, $m_{p}=2 m$Density of planet $(\rho _{p})=$ Density ofearth $(\rho )$As we know that,Weight of object on earth is W $w=m g \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$where, $g$ is acceleration due to gravity.and $\;\; g=\;\frac{G m}{R^{2}}$where, $\;\; G=$ universal gravitationalconstant, $R \;\; =\;\text{ radius of earth }\;$ $\;\text{ and }\; \;\; m_{e} \;\; =\;\text{ mass of earth. }\;$ $\Rightarrow g \;\; =\;\frac{G \rho \frac{4}{3} \pi R^{3}}{R^{2}}$ $\Rightarrow g \;\; =G \rho \;\frac{4}{3} \pi R$Putting this value in Eq. (i), we get $w=m G \rho \;\frac{4}{3} \pi R \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$As densities of planets earth is same.i.e., $\;\; \rho _{p}=\rho$ $\Rightarrow \;\frac{m_{p}}{\frac{4}{3} \pi R_{p}^{3}} \;\; =\;\frac{m}{\frac{4}{3} \pi R^{3}}$ $\Rightarrow \;\; R_{p} \;\; =(\;\frac{m_{p}}{m} R^{3})^{1 / 3}$ $\;\; =(\;\frac{2 m}{m} R^{3})^{1 / 3}$ $\Rightarrow \;\; R_{p} \;\; =(2)^{1 / 3} R$Weight on planet, $w_{p}=\;\frac{4}{3}(2)^{1 / 3} \pi m G \rho R \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (iii)$Dividing Eq. (iii) by Eq. (ii), we get $\;\frac{w_{p}}{w} \;\; =\;\frac{\;\frac{4}{3}(2)^{1 / 3} \pi m G \rho R}{m \rho G \;\frac{4}{3} \pi R}$ $\Rightarrow \;\; w_{p} \;\; =2^{1 / 3} w$

${F}=\frac{{Gm}^{2}}{(2 {R})^{2}}={m} {R} \omega ^{2}$ $\omega =\frac{1}{2} \\sqrt{\frac{{G}}{{R}^{3}}}$

${T}^{2} \propto{R}^{3}$ ${T}={kR}^{3 / 2}$ $\frac{{dT}}{{T}}=\frac{3}{2} \frac{{dR}}{{R}}$ $=\frac{3}{2} \\times 0.02=0.03$ $\%$ Change $=3 \%$

Given, total mass of body is 2M. Potential energy is maximum at $\text{M} / m=x / 1$ The arrangement of masses to form a square is shown in diagram. The gravitational potential energy of a body is given $\text{U}=-\frac{\text{GMm}}{\text{r}}$ For the given system, the potential energy will be $\text{U}_{\text{T}}=-{{\text{Gm}}(\text{M}-\text{m})}/{d}-{{\text{Gm}}(\text{M}-\text{m})}/{d}-{{\text{Gm}}(\text{M}-\text{m})}/{d} -\frac{\text{Gm}(\text{M}-\text{m})}{d}-\frac{\text{Gm}^{2}}{(\\sqrt{2} d)^{2}}-\frac{\text{G}(\text{M}-\text{m})^{2}}{(\\sqrt{2} d)^{2}}$ $\text{U}_{\text{T}}=-\frac{4 \text{Gm}(\text{M}-\text{m})}{d}-\frac{\text{Gm}^{2}}{\\sqrt{2} d}-\frac{\text{G}(\text{M}-\text{m})^{2}}{\\sqrt{2} d}$ For maximum potential energy, ${\text{dU}_{\text{T}}}/{\text{dm}}=0$ $-\frac{4 \text{G}}{d}[\text{M}-2 \text{m}]-\frac{\text{G}}{\\sqrt{2} d}[2 m]$ $-\frac{\text{G}}{\\sqrt{2} d}[2(\text{M}-\text{m}) \times -1]=0$ $\Rightarrow 4 \text{M}-8 \text{m}+\\sqrt{2} \text{m}=\\sqrt{2}(\text{M}-\text{m})$ $(4-\\sqrt{2}) \text{M}=(8-2 \\sqrt{2}) \text{m}$ $\frac{\text{M}}{\text{m}}=\frac{2(4-\\sqrt{2})}{4-\\sqrt{2}}=2$ Comparing it with the given value, we get x = 2 Thus, potential energy will be maximum when x = 2.

Given, $g_{A}=g_{C}$ Radius of Earth, $R=6400 {km}$ Height, $h=3200 {km}=\;\frac{R}{2}$ Since, $g=\;\frac{G M}{R^{2}}$ where, $G$ is gravitational constant. $\therefore \;\; g_{A}=\;\frac{G M(O A)}{R^{3}}$ . . . (i) and $g_{C} \;\; =\;\frac{G M}{(O C)^{2}}=\;\frac{G M}{R+h}$ $\;\; =\;\frac{G M}{(R+R / 2)^{2}}=\;\frac{G M}{(3 R / 2)^{2}}$ . . . (ii) According to given information from Eqs. (i) and (ii), we get $\Rightarrow \;\; O A \;\; =\;\frac{4}{9} R=\;\frac{4}{9} \times 6400$ $A B \;\; =O B-O A$ $\;\text{ and }\;\;A B=6400-\;\frac{4}{9} \times 6400$ $\Rightarrow \;\; =6400(1-4 / 9)=6400 \times 5 / 9$ $\;\text{ Now, }\; O A: A B=\;\frac{6400 \times 4 / 9}{5 / 9 \times 6400}=\;\frac{4}{5}$ $\;\text{ Hence, }\; x=4$

Given, masses and radii of Earth and Moon are $(m_{1}, \text{R}_{1}),(m_{2}, \text{R}_{2})$ and separation between their centre of mass = r If v be the minimum escape velocity of particle of mass m, then by using law of conservation of energy, Initial energy = Final energy = 0 $\frac{1}{2} mv^{2}-\frac{\text{G} m_{1} m}{r / 2}-\frac{\text{G} m_{2} m}{r / 2}=0$ $\Rightarrow \frac{1}{2} m v^{2}=\frac{2 \text{G} m}{r}(m_{1}+m_{2})$ $\Rightarrow v=\\sqrt{\frac{4 \text{G}}{r}(m_{1}+m_{2})}$

According to the Kepler's second law, $\frac{d A}{d t}=$ constantArea covered from $P$ to $P^{'}$ is $d A$, then $d A=\;\frac{1}{2} r^{2} d \theta \Rightarrow \;\frac{d A}{d t}=\;\frac{1}{2} r^{2} \;\frac{d \theta }{d t}$ $\Rightarrow \;\frac{d A}{d t}=\;\frac{1}{2} r^{2} \omega$...(i) Angular momentum, $L=M v r=M(\omega r) r=M \omega r^{2}$ $\Rightarrow \omega =\;\frac{L}{m r^{2}}$ $\therefore$ From Eqs. (i) and (ii), we get $\;\frac{d A}{d t}=\;\frac{1}{2} r^{2}(\;\frac{L}{M r^{2}})$ $\Rightarrow \;\frac{d A}{d t}=\;\frac{L}{2 M}$

The gravitational potential at a point due to multiple masses is the scalar sum of the potentials due to individual masses.
The point of interest is $r = 25 \text{ m}$ from the center.

1. The potential due to the central point mass $m_p = 50 \text{ kg}$ at distance $r = 25 \text{ m}$ is:
   $V_p = -\frac{G m_p}{r} = -\frac{G (50 \text{ kg})}{25 \text{ m}} = -2G$

2. The potential due to the spherical shell of mass $M_s = 100 \text{ kg}$ and radius $R_s = 50 \text{ m}$ at a point inside the shell ($r < R_s$) is constant and equal to the potential on its surface:
   $V_s = -\frac{G M_s}{R_s} = -\frac{G (100 \text{ kg})}{50 \text{ m}} = -2G$

3. The total gravitational potential $V_{\text{total}}$ at $r = 25 \text{ m}$ is the sum of these potentials:
   $V_{\text{total}} = V_p + V_s = -2G - 2G = -4G$
   Thus, $V = -4G \text{ kg/m}$.

The given situation is shown below The gravitational force between these two stars provide the required centripetal force for rotation in a circle about their common centre. Assuming 2 mat origin, the centre of mass of the system lies at $x=\;\frac{2 m \times 0+m \times d}{2 m+m}=\;\frac{d}{3}$ Hence, $F_{G}=F_{C}$ where, $F_{C}$ is gravitational force between them and $F_{C}$ is centripetal force. $\Rightarrow \;\; \;\frac{G m_{1} m_{2}}{r^{2}} \;\; =2 m \omega ^{2} x$ $\Rightarrow \;\; \;\frac{G(2 m)(m)}{d^{2}} \;\; =2 m \omega ^{2} \times \;\frac{d}{3} \Rightarrow \omega ^{2}=\;\frac{3 G m}{d^{3}}$ $\omega \;\; \omega \;\; =\sqrt{\;\frac{3 G m}{d^{3}}}$ We know that, $\;\omega =\;\frac{2 \pi }{T}$ $\;T=\;\frac{2 \pi }{\omega }$ $\Rightarrow$ $T \;\; =\;{2 \pi }/{\sqrt{\;\frac{3 G m}{d^{3}}}}$ $\;\; =2 \pi \sqrt{\;\frac{d^{3}}{3 G m}}$ [using Eq. (i)]

Condition of weightlessness, $g_{\;\text{equator }\;}=0$ $g_{\;\text{equazar }\;}=g-\omega ^{2} R$ The radius of the Earth, $R=6400 {km}=6.4 \times 10^{6} {m}$ $\Rightarrow$ $\omega =\sqrt{\;\frac{g}{R}}$ $\omega =\sqrt{\;\frac{10}{6.4 \times 10^{6} {m}}} \Rightarrow \omega =\;\frac{1}{800} \frac{rad}{s}$ The duration of the day, $T=\;\frac{2 \pi }{\omega } \Rightarrow T=\;\frac{2 \pi }{\frac{1}{800}}$ $\Rightarrow T=5024 {s} \Rightarrow T=\;\frac{5024}{60} {min}$ $\Rightarrow \;\; T=84 {min}$ Hence, the duration of the day would be 84 min.

Given, the time period of the first satellite, $\text{T}_1$ = 1h The time period of the second satellite, $\text{T}_2$ = 8 h The radius of the orbit of nearer satellite, $\text{R}_1$ = 2000 kmApplying the Kepler's third law, $\text{T}^{2} \propto\text{R}^{3}$ $(\frac{\text{T}_{1}}{\text{T}_{2}})^{2}=(\frac{\text{R}_{1}}{\text{R}_{2}})^{3}$ \Rightarrow $\frac{2000}{\text{R}_{2}}=(\frac{1}{8})^{2 / 3}$ \Rightarrow $\text{R}_2$ = 8000 km As we know the relation between the angular speed and timeperiod $\omega =\frac{\theta }{\text{T}}$ So, the angular speed of the nearer satellite to the orbit, $\omega _{1}=\frac{2 \pi }{1} {\text{rad}} / {\text{h}}$ and the angular speed of the farther satellite to the orbit, $\omega _{2}=\frac{2 \pi }{8}=\frac{\pi }{4} {\text{rad}} / {\text{h}}$ The speed of the nearer satellite to orbit, $v_{1}=\omega _{1} \text{R}_{1}=(2 \pi ) \times 2 \times 10^{3} {\text{km}} / {\text{h}}$ The speed of the farther satellite to the orbit $\text{V}_{2}=\omega _{2} \text{R}_{2}$ $=(\frac{\pi }{4}) \times 8000$ $=\pi \\times 2 \times 10^{3} {\text{km}} / {\text{h}}$ Thus, the relative angular speed of the nearer satellite to the farthersatellite, $\omega =\frac{\text{V}_{1}-\text{V}_{2}}{\text{R}_{1}-\text{R}_{2}}$ $=\frac{2 \pi \times 2 \times 10^{3}-\pi \times 2 \times 10^{3}}{8000-2000}$ $=\frac{2 \pi \times 10^{3}}{6000}=\frac{\pi }{3} {\text{rad}} / {\text{h}}$ Comparing with, $\omega =\frac{\pi }{x}$ The value of x = 3.

Given, maximum distance of comet from the Sun, $r_{1}=1.6 \times 10^{12} {m}$ Minimum distance of comet from the Sun, $r_{2}=8.0 \times 10^{10} {m}$ Speed of the comet at nearest point, $v_{2}=6 \times 10^{4} {ms}^{-1}$ Applying law of conservation of angular momentum, $m v_{1} r_{1}=m v_{2} r_{2}( \because$ mass of comet will remain same $)$ $\Rightarrow \;\; V_{1}=\;\frac{V_{2} r_{2}}{r_{1}}$ $=\;\frac{6 \times 10^{4} \times 8 \times 10^{10}}{1.6 \times 10^{12}}=\;\frac{48 \times 10^{14}}{1.6 \times 10^{12}}$ $=3 \times 10^{3} \frac{m}{s}$

We know that, $v_{e}=\sqrt{\;\frac{2 {Gm}}{R_{1}}}$ ...(i) where, $v_{e}=$ escape velocity, $G=$ gravitational constant, $R=$ radius of earth and $m=$ mass of the body. Now, the escape velocity is increased to 10 times, $\Rightarrow \;\; 10 v_{e}=\sqrt{\;\frac{2 G m}{R_{2}}}$...(ii) Dividing Eq. (ii) by Eq. (i), we get $\Rightarrow$ $10=\sqrt{\;\frac{R_{1}}{R_{2}}} \Rightarrow 100=\;\frac{R_{1}}{R_{2}}$ $\Rightarrow \;\; R_{2}=\;\frac{R_{1}}{100}=\;\frac{6400}{100} \Rightarrow R_{2}=64 {km}$