Acceleration Due to Gravity and Gravitational PE — NEET Physics PYQ MCQs with Solutions

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.

Solution: Potential at $a/2$ = potential due to the shell (constant inside, $-\dfrac{GM}{a}$) + potential due to the central mass $\left(-\dfrac{GM}{a/2}\right)$ $=-\dfrac{GM}{a}-\dfrac{2GM}{a}=-\dfrac{3GM}{a}.$

Solution: For weightlessness the centripetal force equals gravity: $\dfrac{mv^2}{r}=mg\Rightarrow v=\sqrt{rg}=\sqrt{20\times10}=14.14$ m/s, i.e. between $14$ and $15$ m/s.

Solution: $g=\dfrac{GM}{R^2}$ with $M=\dfrac{4}{3}\pi R^3\rho$, so $g=\dfrac{4}{3}\pi G R\rho\propto R$ for the same density. Hence $\dfrac{g'}{g}=\dfrac{R'}{R}=3\Rightarrow g'=3g.$

Solution: $g=\dfrac{GM}{R^2}=\dfrac{4}{3}\pi G R\rho$. With $g_p=g_e$: $R_p\rho_p=R_e\rho_e$. Since $\rho_p=2\rho_e$, $R_p=\dfrac{R_e}{2}=\dfrac{R}{2}.$

Solution: Given $g_A=9g_B$. From $v^2=2gh$, for the same take-off speed $h\propto\dfrac{1}{g}$, so $\dfrac{h_A}{h_B}=\dfrac{g_B}{g_A}=\dfrac{1}{9}$. Hence $h_B=9h_A=9\times2=18$ m.

Solution: $U=-\dfrac{GMm}{r}$. At the surface $U_e=-\dfrac{GMm}{R}$; at $h=3R$, $r=4R$ so $U_h=-\dfrac{GMm}{4R}$. $\Delta U=U_h-U_e=-\dfrac{GMm}{4R}+\dfrac{GMm}{R}=\dfrac{3}{4}\dfrac{GMm}{R}=\dfrac{3}{4}mgR$ (using $GM=gR^2$).

Solution: Energy conservation with $h=R$: $\dfrac{1}{2}mu^2-\dfrac{GMm}{R}=-\dfrac{GMm}{R+R}$. So $u^2=\dfrac{2GM}{R}-\dfrac{2GM}{2R}=\dfrac{GM}{R}\Rightarrow u=\sqrt{\dfrac{GM}{R}}.$

Solution: For a body of mass $m$ on a sphere of mass $M$, radius $R$: $mg=\dfrac{GMm}{R^2}\Rightarrow g=\dfrac{GM}{R^2}\Rightarrow M=\dfrac{gR^2}{G}.$

Solution: When the rocket accelerates upward with acceleration $a$, the effective gravity becomes $(g+a)$. Since $T=2\pi\sqrt{\dfrac{l}{g+a}}$, the period decreases when the rocket moves up with uniform acceleration.

Solution: For a central attractive force the torque is zero, so angular momentum $L$ is conserved in both magnitude and direction. The potential energy $U=-\dfrac{GMm}{R}$ is negative, and for a bound orbit the total energy $E$ is always negative.