Gravitation Practice Questions

Given, $M_{A}=200 {kg}, M_{B}=400 {kg}, H_{A}=600 {km}, H_{B}=1600 {km}$ and $\;\; R_{A}=R_{E}+H_{A}=6400+600=7000 {km}$ $R_{B}=R_{E}+H_{B}=6400+1600=8000 {km}$ Let $T_{A}, T_{B}, \omega _{A}, \omega _{B}, R_{A}$ and $R_{B}$ be the time period, angular frequencies and radii of satellites $A$ and $B$, respectively. Force on satellite $A, F_{A}=m_{A} \omega _{A}^{2} R_{A}=\;\frac{G M m_{A}}{R_{A}^{2}}$ $\Rightarrow \;\; \omega _{A}^{2}=\;\frac{G M}{R_{A}^{3}}$ but, $\;\; \omega _{A}=\;\frac{2 \pi }{T_{A}}$ $\therefore \;\; (\;\frac{2 \pi }{T_{A}})^{2}=\;\frac{G M}{R_{A}^{3}} \Rightarrow T_{A}=\sqrt{\;\frac{4 \pi ^{2} R_{A}^{3}}{G M}}$ Similarly, $\;\; T_{B}=\sqrt{\;\frac{4 \pi ^{2} R_{B}^{3}}{G M}}$ $\therefore \;\; T_{B}-T_{A} \;\; =\sqrt{\;\frac{4 \pi ^{2}}{G M}}(\sqrt{R_{B}^{3}}-\sqrt{R_{A}^{3}})$ $\;\; =\;\frac{2 \pi }{\sqrt{G M}}[\sqrt{(8 \times 10^{3})^{3}}-\sqrt{(7 \times 10^{3})^{3}}]$ $=\;\frac{2 \pi \times 10^{9}}{\sqrt{G M}}(8 \sqrt{8}-7 \sqrt{7})$ $=\;{2 \pi \times 10^{9}}/{\sqrt{6.67 \times 10^{-11} \times 6 \times 10^{24}}}(4.107)$ $=\;{2 \pi }/{\sqrt{4 \times 10^{14}}} \times 10^{9}(4.107)$ $=\pi \times 10^{2} \times 4.107=12.9 \times 10^{2} {s}$ $=1.33 \times 10^{3} {s}$

Given that, radius of Earth $=\text{R}_{\text{E}}$ Height above Earth's surface, h = r Depth below Earth surface, d = r We know that, acceleration due to gravity at height (r) and depth (r) is (for $r < \text{R}_{\text{E}}$) $g_{h}=g({\text{R}_{\text{E}}}/{\text{R}_{\text{E}}+h})^{2}=g({\text{R}_{\text{E}}}/{\text{R}_{\text{E}}+r})^{2}$ ...(i) and$g_{d}=g({\text{R}_{\text{E}}-d}/{\text{R}_{\text{E}}})=g({\text{R}_{\text{E}}-r}/{\text{R}_{\text{E}}})$ ...(ii) $\therefore \frac{g_{d}}{g_{h}}={g({\text{R}_{\text{E}}-r}/{\text{R}_{\text{E}}})}/{g({\text{R}_{\text{E}}}/{\text{R}_{\text{E}}+r})^{2}}$ [from Eqs. (i) and (ii)] $\Rightarrow \frac{g_{d}}{g_{h}}={(\text{R}_{\text{E}}-r)(\text{R}_{\text{E}}+r)^{2}}/{\text{R}^{3}}={\text{R}_{\text{E}}^{3}+\text{R}_{\text{E}}^{2} r-\text{R}_{\text{E}} r^{2}-r^{3}}/{\text{R}_{\text{E}}^{3}}$ $1+{r}/{\text{R}_{\text{E}}}-{r^{2}}/{\text{R}_{\text{E}}^{2}}-{r^{3}}/{\text{R}_{\text{E}}^{3}}$

Let us consider the gravitational force acting on each mass M by adjacent particles be F. and the gravitational force acting on each mass M diagonally be $\text{F}_1$ The net force, $\text{F}_{{\text{net} }}=\frac{\text{Mv}^{2}}{\text{R}}$ Along the centre of circle, $\\sqrt{2} \text{F}+\text{F}_{1}=\frac{\text{Mv}^{2}}{\text{R}}$ $\\sqrt{2}(\frac{\text{GMM}}{(\\sqrt{2} \text{R})^{2}})+(\frac{\text{GMM}}{(2 \text{R})^{2}})$ $=\frac{\text{Mv}^{2}}{\text{R}}$ \Rightarrow $\frac{\text{GM}}{\text{R}^{2}}(\frac{1}{\\sqrt{2}}+\frac{1}{4})=\frac{v^{2}}{\text{R}}$ \Rightarrow $v=\frac{1}{2} \\sqrt{\frac{\text{GM}(2 \\sqrt{2}+1)}{\text{R}}}$ Hence, the speed of each particle is $\frac{1}{2} \\sqrt{\frac{\text{GM}(2 \\sqrt{2}+1)}{\text{R}}}$.

Option D is correct ${T}^{2}=\frac{4 \pi ^{2}}{{GM}} \.{r}^{3}$ ${M}=\frac{4 \pi ^{2}}{{G}} \.\frac{{r}^{3}}{{ T}^{2}}$ by putting values ${M}=6 \\times 10^{23}$

The uniform spherical shell is shown in the figure below Inside the shell, there is no mass, so by applying Gauss's law forgravitation $\∮{g} .{d} {a}=-4 \pi \text{GM}_{{\text{enclosed} }}$ where, g = gravitational field intensity. da = area enclosed G = gravitational constant and $\text{M}_{{\text{enclosed} }}=$ mass enclosed . As, $\text{M}_{{\text{enclosed} }}=0$ So, gravitational field intensity is zero inside the shell, i.e. g = 0 ...(i) We also know that, $g=-\frac{\text{dV}}{\text{dr}}$ where, dV is change in potential due to gravity ⇒ V = constant [using Eq. (i)] So, gravitational field is zero everywhere inside the shell and gravitational potential (V) is constant everywhere.

Applying energy conservation from (1) to (2) $\frac{1}{2} m \.(\frac{2 G M}{R_{e}})-\frac{G M m}{R_{e}}$ $=\frac{1}{2} m v^{2}-\frac{G M m}{R+r}$ $\Rightarrow \frac{1}{2} m v^{2}=\frac{G M m}{R+r}$ $v=\\sqrt{\frac{2 G M}{R+r}}=\frac{d r}{d t}$ $\\sqrt{2 G M} \int _{0} ^{t} d t=\int _{R_{e}} ^{R_{e}+h}(\\sqrt{R+r}) d r$ \\sqrt{2 G M} \.t=\frac{2}{3}\[(R+r)^{3 / 2}]{R{e}}^{R_{e}+h}$$t=\frac{2}{3} \sqrt{\frac{R_{e}^{3}}{2 G M}}[(1+\frac{h}{R_{e}})^{3 / 2}-1]$$\frac{G M}{R_{e}^{2}}=g$$t=1/3\sqrt{{{2R}{{e}}}/{ {g}}}[(1+{{h}}/{{R}{{e}}})^{3 / 2}-1]$

Assume that at a distance $x$ from the planet of mass $M$ , the net gravitational field becomes zero. $\therefore \;\;\frac{G M}{x^{2}}=\;\frac{G \times 9 M}{(8 R-x)^{2}}$ $\Rightarrow \;\frac{1}{x^{2}}=\;\frac{9}{(8 R-x)^{2}}$ $\Rightarrow (\;\frac{1}{x})=(\;\frac{3}{8 R-x})^{2} \Rightarrow \;\frac{1}{x}=\;\frac{3}{8 R-x}$ $\Rightarrow 3 x=8 R-x$ $\Rightarrow 4 x=8 R$ $\Rightarrow x=2 R \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$Now, a satellite should be projected in such a way that its covers a minimum distance of $2 R$ . $\therefore \;\frac{1}{2} m v^{2} \;\; - \;\frac{G M m}{R}-\;\frac{G(9 M) m}{7 R}$ $\;\; =\;\frac{-G M m}{2 R}-\;\frac{G(9 M) m}{6 R}$where, $m$ is the mass of satellite. $\Rightarrow \;\; \;\frac{1}{2} v^{2}=\;\frac{2 G M}{7 R} \Rightarrow v=\sqrt{\;\frac{4}{7} \;\frac{G M}{R}} \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$According to question, the minimum speed $v$ required for the satellite to reach the surface of the second planet is $\sqrt{\;\frac{a G M}{7 R}}$ . So, on comparing it with Eq. (ii), we can write $a=4$

Given Height of satellite from planet's surface, $h=11 R$ So, the total distance from the centre of planet $=11 R+R=12 R$ Time period of planet $=24 {h}$ Similarly, the total distance of second satellite from the centre of planet $=2 R+R=3 R$ Using the Kepler's law $T^{2} \propto R^{3}$ $\Rightarrow \;\; (\;\frac{T_{1}}{T_{2}})=(\;\frac{R_{1}}{R_{2}})^{3 / 2}$ Substituting the values in the above equation, we get $(\;\frac{24}{T})=(\;\frac{12 R}{3 R})^{3 / 2}$ $\therefore$ The time period of another satellite is $3 {h}$.

Angular momentum conservation equation ${v}_{0} {x}_{2}={v}_{1} {x}_{1}$ ${v}_{1}=\frac{{v}_{0} {x}_{2}}{{x}_{1}}$

Given, $m=1 {kg}, R=1 {m}$ We know that, $F \;\; =\;\frac{G m_{1} m_{2}}{r^{2}}$ $\because \;\; F_{1} \;\; =\;\frac{G m m}{(2 R)^{2}}=\;\frac{G m^{2}}{4 R^{2}}$ $\;\text{ and }\; \;\; F_{2} \;\; =\;\frac{G m m}{(\sqrt{2} R)^{2}}=\;\frac{G m^{2}}{2 R^{2}}$ Net force on one particle, $F_{\;\text{net }\;} \;\; =F_{1}+F_{2} \cos 45^{\circ}+F_{2} \cos 45^{\circ}$ $\;\; =F_{1}+2 F_{2} \cos 45^{\circ}$ $\;\; =\;\frac{G m^{2}}{4 R^{2}}+2(\;\frac{G m^{2}}{2 R^{2}}) \cdot \;\frac{1}{\sqrt{2}}$ $\;\; =\;\frac{G m^{2}}{4 R^{2}}+\;\frac{G m^{2}}{\sqrt{2} R^{2}}$ $\;\; =\;\frac{G m^{2}}{R^{2}}[\;\frac{1}{4}+\;\frac{1}{\sqrt{2}}]$ As the gravitational force provides the necessary centripetal force, so $F_{\;\text{net }\;}=F_{C}=\;\frac{m v^{2}}{R}$ Here, $F_{C}=$ centripetal force. $\Rightarrow \;\frac{G m^{2}}{R^{2}}[\;\frac{1}{4}+\;\frac{1}{\sqrt{2}}] \;\; =\;\frac{m v^{2}}{R}$ $\Rightarrow \;\; v \;\; =\;\frac{1}{2} \sqrt{\;\frac{G m}{R}(1+2 \sqrt{2})}$ $\Rightarrow \;\; v \;\; =\;\frac{1}{2} \sqrt{G(1+2 \sqrt{2})}$

Given, radius of sphere $=R$ Distance of particle from centre of Earth $=3 R$ Force between sphere and particle $=F_{1}$ Radius of cavity $=R / 2$ Let mass of sphere $=M$ Mass of sphere with cavity $=M^{'}$ Mass of particle $=m$ Now, $\;\; M^{'}=\rho \cdot \;\frac{4}{3} \pi (\;\frac{R}{2})^{3}=\;\frac{M}{8}$ By using concept of gravitational force, $F_{1} \;\; =\;\frac{G M m}{(3 R)^{2}}=\;\frac{G M m}{9 R^{2}}$ . . . (i) and $F_{\;\text{cavity }\;} \;\; =\;\frac{G M^{'} m}{(A B)^{2}}=\;\frac{(2}{25 \times 8 R^{2}}$ $\;\; =\;\frac{4 G M m}{25}$ . . . (ii) $\therefore \;\; F_{2} \;\; =F_{1}-F_{\;\text{cavity }\;}=\;\frac{G M m}{9 R^{2}}-\;\frac{4}{25 \times 8} \;\frac{G M m}{R^{2}}$ $\;\; =\;\frac{G M m}{R^{2}}(\;\frac{1}{9}-\;\frac{1}{50})=\;\frac{41}{50 \times 9} \;\frac{G M m}{R^{2}}$ $\because \;\; \;\frac{F_{1}}{F_{2}} \;\; =\;\frac{50}{41}$

Let $v_{A}, v_{B}, M_{A}, M_{B}$ and $R_{A}, R_{B}$ be the escape velocities, masses and radii of planet $A$ and $B$, respectively. As we know that, $v=\sqrt{2 g R}=\sqrt{\;\frac{2 G M}{R^{2}} \cdot R}=\sqrt{\;\frac{2 G M}{R}}$ where, $G$ is the gravitational constant. $\Rightarrow \;\; v \propto \sqrt{\;\frac{M}{R}}$ Since, $v_{A}=v_{B}$ $\because \;\; \;\frac{M_{A}}{R_{A}}=\;\frac{M_{B}}{R_{B}} \Rightarrow M_{A} R_{B}=M_{B} R_{A}$ or $M_{1} R_{2}=M_{2} R_{1}$ Hence, option (b) is the correct.

By using law of conservation of energy, Energy on the surface of earth $(E_{\;\text{surface }\;})=$ Energy at height $(h=10 R)$ $\Rightarrow \;\frac{-G M m}{R}+\;\frac{1}{2} m v_{i}^{2}=\;\frac{-G M m}{R+10 R}+0=\;\frac{-G M m}{11 R}$ $\Rightarrow \;\; 1 / 2 m v_{i}^{2}=\;\frac{11}{11} \;\frac{G M m}{R}-\;\frac{G M m}{11 R}$ $\Rightarrow \;\; \;\frac{1}{2} m v_{i}^{2}=\;\frac{10 G M m}{11 R} \Rightarrow v_{i}^{2}=\;\frac{20 G M}{11 R}$ $\;\;\; v_{i}=\sqrt{\;\frac{10}{11}} v_{e} \;\; ( \because v_{e}=\sqrt{\;\frac{2 G M}{R}}=\;\text{ escape velocity) }\;$ $\Rightarrow \;\; x=10$

At neutral point g = 0 so graph (C) is correct Given,Acceleration due to gravity on Earth, $g_{E}=10 {m} / {s}^{2}$and acceleration due to gravity on Mars, $g_{M}=4 {m} / {s}^{2}$We know that,$g_{E}$ (at height $h)=g_{\;\text{Earth }\;}(1-\;\frac{2 h}{R})$ $\therefore$ Weight at Earth,$m g_{E}=100 \times 10=1000 {N}$As the spaceship moves far away from Earth, the value of $g_{E}$ decreases to zero at a point where $g_{E}+g_{M}=0$ and hence weight will also be zero.This point is called neutral point and is shown by graph (c) in the given figure.Then, $g_{E}$ increases $4 {ms}^{-2}$ at Mars surface and weight becomes $400 {N}$ which is also exhibited by graph (c).

According to Kepler's third law, $T^{2} \propto R^{3}$ $\Rightarrow \;\; \;\frac{T_{2}}{T_{1}}=(\;\frac{R_{2}}{R_{1}})^{3 {/} 2}$ $\Rightarrow \;\; \;\frac{T_{2}}{T}=(\;\frac{9 R}{R})^{3 {/}2}$ $[\because R_{1} , =R ; R_{2} , =9 R]$ $\Rightarrow$ $T_{2}=27 T$

We know that binding energy of earth, ${BE}=-\;\frac{3}{5} \;\frac{G M^{2}}{R}$ $\therefore$ Energy required to break the earth into pieces $=- {BE}=\;\frac{3}{5} \;\frac{G M^{2}}{R}$ .......(i) According to question, the amount of energy that needs to be supplied is $\;\frac{x}{5} \;\frac{G M^{2}}{R}$. Comparing it with value in Eq. (i), we get, $x=3$

Given, period of revolution of first satellite, $T_{1}=1 {h}$ Period of revolution of second satellite, $T_{2} \;\; =8 {h}$ $\;\frac{T_{1}}{T_{2}} \;\; =\;\frac{1}{8}$ We know that, $\omega =\;\frac{2 \pi }{T}$ $\Rightarrow$ $\omega \propto \;\frac{1}{T}$ $\because \;\; \;\frac{\omega _{1}}{\omega _{2}}=\;\frac{T_{2}}{T_{1}} \Rightarrow \;\frac{\omega _{1}}{\omega _{2}}=\;\frac{8}{1}$ or $\;\; \omega _{1}: \omega _{2}=8: 1$

According to Kepler's second law of planetary motion, areal velocity of every planet moving around the sun should remain constant in elliptical orbit.