Gravitation — NEET Physics PYQ MCQs with Solutions

Solution: From Kepler's third law, the time period of revolution of satellite around earth is $T^2 \propto r^3$ or $T \propto r^{3/2}$, where $r$ is the radius of the satellite's orbit. Here, $r_1 = 6R_E + R_E$, $T_1 = 24$ h and $r_2 = 2.5R_E + R_E$. So, $\dfrac{T_1}{T_2} = \left(\dfrac{r_1}{r_2}\right)^{3/2}$, $\dfrac{24}{T_2} = \left(\dfrac{6R_E + R_E}{2.5R_E + R_E}\right)^{3/2} = \left(\dfrac{7}{3.5}\right)^{3/2}$. $\Rightarrow T_2 = \dfrac{24}{(2)^{3/2}} = \dfrac{24}{2\sqrt{2}} = \dfrac{12}{\sqrt{2}} = 6\sqrt{2}$ h.

Solution: The closed point $A$ is the perihelion and the farthest point $C$ is the aphelion. As per Kepler's second law of areas, the planet moves slowly ($v_{min}$) when farthest from the Sun and more rapidly ($v_{max}$) when nearest to the Sun. Thus $v_A = v_{max}$, $v_C = v_{min}$, so $v_A > v_B > v_C$. Kinetic energy $K = \dfrac{1}{2}mv^2$, hence $K_A > K_B > K_C$.

Solution: The gravitational force of attraction between the planet and sun provides the centripetal force, i.e. $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ $\Rightarrow v = \sqrt{\dfrac{GM}{r}}$. The time period of the planet is $T = \dfrac{2\pi r}{v}$ $\Rightarrow T^2 = \dfrac{4\pi^2 r^2}{GM/r} = \dfrac{4\pi^2 r^3}{GM}$ ...(i). Also from Kepler's third law $T^2 = Kr^3$ ...(ii). From Eqs. (i) and (ii), $\dfrac{4\pi^2 r^3}{GM} = Kr^3 \Rightarrow GMK = 4\pi^2$.

Solution: Suppose the smaller body (mass $M$) covers a distance $x$ before collision. The gap between the surfaces is $12R - R - 2R = 9R$. Since the centre of mass stays fixed, $Mx = 5M(9R - x)$, or $x = 45R - 5x$, or $x = \dfrac{45R}{6} = 7.5R$.

Solution: From Kepler's third law $T^2 \propto r^3$, where $r$ is the radius of the orbit (semi-major axis). Hence $T_1^2 \propto r_1^3$ and $T_2^2 \propto r_2^3$. So $\dfrac{T_2^2}{T_1^2} = \dfrac{r_2^3}{r_1^3} = \dfrac{(3R)^3}{(6R)^3} = \dfrac{1}{8}$. $\therefore T_2^2 = \dfrac{1}{8}T_1^2 \Rightarrow T_2 = \dfrac{24}{2\sqrt{2}} = 6\sqrt{2}$ h (taking $T_1 = 24$ h).

Solution: Apply conservation of angular momentum. From the law of conservation of angular momentum, $L_1 = L_2$, so $mr_1v_1 = mr_2v_2$ $\Rightarrow r_1v_1 = r_2v_2 \Rightarrow \dfrac{v_1}{v_2} = \dfrac{r_2}{r_1}$.

Solution: When two satellites of the earth are moving in the same orbit, the time period of both is equal. From Kepler's third law $T^2 \propto r^3$, the time period is independent of mass, hence their time periods are equal. The potential energy and kinetic energy are mass dependent, hence they are not equal for the two satellites. But since they orbit in the same orbit, they have equal orbital speed.

Solution: Apply Kepler's second law: the line joining the sun to the planet sweeps out equal areas in equal time intervals, i.e. areal velocity is constant. $\dfrac{dA}{dt} = $ constant, or $\dfrac{A_1}{t_1} = \dfrac{A_2}{t_2}$, where $A_1$ = area under $SCD$ and $A_2$ = area under $ABS$. $\Rightarrow t_1 = \dfrac{A_1}{A_2}t_2$. Given $A_1 = 2A_2$, $\therefore t_1 = 2t_2$.

Solution: According to Newton's law of gravitation, the force between two spheres is $F = \dfrac{GMm}{r^2}$. From this relation, the gravitational force does not depend on the medium between the two spheres, hence it remains the same, i.e. $F$.

Solution: According to Kepler's third law $T^2 \propto r^3$, where $T$ = time period of revolution and $r$ = semi major axis. $\therefore \dfrac{T_A^2}{T_B^2} = \dfrac{r_A^3}{r_B^3}$, so $\dfrac{r_A}{r_B} = \left(\dfrac{T_A}{T_B}\right)^{2/3} = (8)^{2/3} = 2^{3 \times \frac{2}{3}} = 4$, or $r_A = 4r_B$.

Solution: According to Kepler's third law, the square of the time period is directly proportional to the cube of the semi-major axis of the orbit, i.e. $T^2 \propto r^3$. $\therefore \dfrac{T_1^2}{T_2^2} = \dfrac{(r)^3}{(2r)^3} = \dfrac{1}{8} \Rightarrow \dfrac{T_1}{T_2} = \dfrac{1}{2\sqrt{2}}$.

Solution: According to the question, gravitational force between two objects is $F = \dfrac{k}{R}$. In equilibrium, the gravitational force provides the required centripetal force to the particle. $\therefore \dfrac{mv^2}{R} = \dfrac{k}{R}$. Hence, $v \propto R^0$.

Solution: According to Kepler's third law (law of periods), the square of the time period $T$ is proportional to the cube of the semi-major axis $r$, i.e. $T^2 \propto r^3$. Here $r_1 = 10^{13}$ m, $r_2 = 10^{12}$ m. $\therefore \dfrac{T_1^2}{T_2^2} = \dfrac{r_1^3}{r_2^3} = \dfrac{(10^{13})^3}{(10^{12})^3} = \dfrac{10^{39}}{10^{36}} = 10^3$, or $\dfrac{T_1}{T_2} = 10\sqrt{10}$.

Solution: From the property of the ellipse, $\dfrac{2}{R} = \dfrac{1}{r_1} + \dfrac{1}{r_2}$, or $\dfrac{2}{R} = \dfrac{r_1 + r_2}{r_1 r_2} \Rightarrow R = \dfrac{2r_1r_2}{r_1 + r_2}$.

Solution: Let at depth $d$, the gravitational acceleration becomes $\dfrac{g}{n}$, i.e. $g_d = \dfrac{g}{n} \Rightarrow g\left(1 - \dfrac{d}{R}\right) = \dfrac{g}{n} \Rightarrow 1 - \dfrac{d}{R} = \dfrac{1}{n} \Rightarrow 1 - \dfrac{1}{n} = \dfrac{d}{R} \Rightarrow \dfrac{n-1}{n} = \dfrac{d}{R} \Rightarrow d = \left(\dfrac{n-1}{n}\right)R$.

Solution: Given, $w = mg = 72$ N (on the surface of earth). At height equal to half of the radius of the earth, $h = \dfrac{R}{2}$. Acceleration due to gravity, $g' = g\left(\dfrac{R}{R+h}\right)^2 = g\left(\dfrac{R}{R + R/2}\right)^2 = g\left(\dfrac{4R^2}{9R^2}\right) \Rightarrow g' = \dfrac{4}{9}g$. $\Rightarrow mg' = \dfrac{4}{9}mg = \dfrac{4}{9} \times 72 = 32$ N, so $w' = 32$ N.

Solution: The gravitational potential energy at earth's surface is $U_1 = -\dfrac{GMm}{R}$, where $G$ = gravitational constant, $M$ = mass of earth, $m$ = mass of object and $R$ = radius of the earth (the negative sign indicates the work done in bringing the object from infinity to a distance $R$). At height $h$, $U_2 = -\dfrac{GMm}{R+h}$. Change in PE $= U_2 - U_1 = GMm\left(\dfrac{1}{R} - \dfrac{1}{R+h}\right) = \dfrac{GMmh}{R(R+h)}$.

Solution: At depth $d$, $g'=g\left(1-\dfrac{d}{R}\right)$, so $mg'=mg\left(1-\dfrac{d}{R}\right)$. Half way down $d=\dfrac{R}{2}$, so $mg'=200\left(1-\dfrac{R/2}{R}\right)=200\times\dfrac{1}{2}=100$ N.

Solution: Work done equals the change in potential energy. $U_1=-\dfrac{GMm}{R}$ and at $h=R$, $U_2=-\dfrac{GMm}{2R}$. So $\Delta U=-\dfrac{GMm}{2R}+\dfrac{GMm}{R}=\dfrac{GMm}{2R}=\dfrac{gR^2m}{2R}=\dfrac{mgR}{2}$ (using $g=\dfrac{GM}{R^2}$).

Solution: New $G'=10G$, so $g'=\dfrac{G'M_E}{R^2}=\dfrac{10GM_E}{R^2}=10g$; thus $g$ increases (it does not stay unchanged). With larger $g$, raindrops fall faster, walking becomes more difficult, and $T=2\pi\sqrt{\dfrac{l}{g}}$ (so $T\propto\dfrac{1}{\sqrt{g}}$) decreases. Hence statement (d) is not correct.

Solution: At height $h$, $g_h=g\left(1-\dfrac{2h}{R}\right)$; at depth $d$, $g_d=g\left(1-\dfrac{d}{R}\right)$. Setting $g_h=g_d$ with $h=1$ km gives $\dfrac{d}{R}=\dfrac{2h}{R}\Rightarrow d=2h=2$ km.

Solution: In space there is no external gravity. Because of the masses of the two astronauts there is a small mutual gravitational attractive force between them, so they slowly move towards each other.

Solution: Potential at height $h$: $V=-\dfrac{GM}{R+h}$ and $g'=\dfrac{GM}{(R+h)^2}$, so $\dfrac{|V|}{g'}=R+h$. Thus $R+h=\dfrac{5.4\times10^7}{6.0}=9\times10^6$ m, giving $h=(9-6.4)\times10^6=2.6\times10^6$ m $=2600$ km.

Solution: Inside the earth $g_{depth}=g_{surface}\left(1-\dfrac{d}{R}\right)$, so $g$ increases linearly from zero at the centre to a maximum at the surface ($r=R$). Outside, $g_{height}=g_{surface}\dfrac{R^2}{(R+h)^2}$, so $g$ decreases with distance. This matches graph (b).

Solution: $V=-2G\left[\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots\right]$. The geometric series sums to $\dfrac{1}{1-\frac{1}{2}}=2$, so $V=-2G\times2=-4G$.

Solution: Height above surface $=2R$, so the final distance from the centre $=R+2R=3R$. $\Delta U=-\dfrac{GMm}{3R}-\left(-\dfrac{GMm}{R}\right)=\dfrac{2GMm}{3R}=\dfrac{2}{3}mgR$ (using $g=\dfrac{GM}{R^2}$).

Solution: $\dfrac{GMm}{(R+h)^2}=\dfrac{1}{16}\dfrac{GMm}{R^2}\Rightarrow \dfrac{R^2}{(R+h)^2}=\dfrac{1}{16}\Rightarrow \dfrac{R}{R+h}=\dfrac{1}{4}\Rightarrow R+h=4R\Rightarrow h=3R$.

Solution: At the geomagnetic North and South poles the needle stays in the North-South direction only.

Solution: Force on the body $F=\dfrac{GM_p m}{(D_p/2)^2}=\dfrac{4GM_p m}{D_p^2}$ (radius $=D_p/2$). Since $F=ma$, $a=\dfrac{F}{m}=\dfrac{4GM_p}{D_p^2}.$

Solution: Power $P=\vec{F}\cdot\vec{V}=FV\cos\theta$. Just before the body hits the ground $\theta=0$ and its speed is maximum, so the power is greatest then.