Class 11 Physics | Gravitation MCQs with Answers – Part 5

Explanation: Surface gravity for a uniform-density spherical planet is \[ g_s=\frac{4}{3}\pi G\rho R \] So, \[ g_s\propto \rho R \] For the new planet: \[ \rho'=2\rho_E \] \[ R'=\frac{R_E}{2} \] Therefore, \[ \frac{g'}{g}=\frac{\rho'R'}{\rho_ER_E} \] \[ \frac{g'}{g}=\frac{(2\rho_E)(R_E/2)}{\rho_ER_E} \] \[ \frac{g'}{g}=1 \] Weight is \(W=mg\), and the body’s mass remains unchanged. \( \textbf{Final answer:} \) \(W\), because the density increase and radius decrease exactly balance.

Explanation: Time period just above the surface is \[ T=2\pi\sqrt{\frac{R^3}{GM}} \] For a planet of density \(\rho\), \[ M=\frac{4}{3}\pi R^3\rho \] Substitute: \[ T=2\pi\sqrt{\frac{R^3}{G\left(\frac{4}{3}\pi R^3\rho\right)}} \] Cancel \(R^3\): \[ T=2\pi\sqrt{\frac{1}{\frac{4}{3}\pi G\rho}} \] \[ T=2\pi\sqrt{\frac{3}{4\pi G\rho}} \] Square the outside factor into the root: \[ T=\sqrt{4\pi^2\cdot\frac{3}{4\pi G\rho}} \] \[ T=\sqrt{\frac{3\pi}{G\rho}} \] \( \textbf{Final answer:} \) \(T=\sqrt{\frac{3\pi}{G\rho}}\), so for equal-density planets, the near-surface orbital period is the same.

Explanation: For a satellite just above the surface: \[ T=2\pi\sqrt{\frac{R^3}{GM}} \] For equal density, \[ M=\frac{4}{3}\pi R^3\rho \] Substitution gives \[ T=\sqrt{\frac{3\pi}{G\rho}} \] This expression contains density \(\rho\) but not radius \(R\). Since both planets have the same mean density, their near-surface orbital periods are equal. The larger planet has larger orbital speed and larger orbit circumference, and these effects balance in the period. \( \textbf{Final answer:} \) \(1:1\), because same density gives the same near-surface orbital period.

Explanation: Escape speed is \[ v_e=\sqrt{\frac{2GM}{R}} \] For a uniform sphere, \[ M=\frac{4}{3}\pi R^3\rho \] Substitute: \[ v_e=\sqrt{\frac{2G\left(\frac{4}{3}\pi R^3\rho\right)}{R}} \] \[ v_e=\sqrt{\frac{8\pi G}{3}\rho R^2} \] Take the square root: \[ v_e=R\sqrt{\frac{8\pi G\rho}{3}} \] So, \[ v_e\propto R\sqrt{\rho} \] \( \textbf{Final answer:} \) \(R\sqrt{\rho}\), combining radius and density scaling.

Explanation: For a uniform-density planet, \[ v_e\propto R\sqrt{\rho} \] For planet \(P\): \[ v_{eP}\propto R\sqrt{\rho} \] For planet \(Q\): \[ R_Q=2R \] \[ \rho_Q=4\rho \] So, \[ v_{eQ}\propto (2R)\sqrt{4\rho} \] \[ v_{eQ}\propto (2R)(2\sqrt{\rho}) \] \[ v_{eQ}\propto4R\sqrt{\rho} \] Therefore, \[ \frac{v_{eQ}}{v_{eP}}=4 \] \( \textbf{Final answer:} \) \(4\), because both larger radius and larger density increase escape speed.

Explanation: Inside a uniform sphere, only the enclosed mass within radius \(r\) contributes to the field. The enclosed mass is \[ M_r=\frac{4}{3}\pi r^3\rho \] The gravitational field at radius \(r\) is \[ g(r)=\frac{GM_r}{r^2} \] Substitute \(M_r\): \[ g(r)=\frac{G\left(\frac{4}{3}\pi r^3\rho\right)}{r^2} \] \[ g(r)=\frac{4}{3}\pi G\rho r \] This is linear in \(r\), so it becomes zero at the centre. \( \textbf{Final answer:} \) \(\frac{4}{3}\pi G\rho r\), valid inside a uniform solid sphere.

Explanation: Inside a uniform sphere: \[ g(r)=\frac{4}{3}\pi G\rho r \] So the slope of \(g\) versus \(r\) is \[ \frac{4}{3}\pi G\rho \] Use \(\frac{4}{3}\pi\approx4.19\). Substitute: \[ \text{slope}=4.19(6.67\times10^{-11})(5.5\times10^3) \] First multiply \(6.67\times10^{-11}\) and \(5.5\times10^3\): \[ 6.67\times5.5=36.685 \] \[ 36.685\times10^{-8}=3.6685\times10^{-7} \] Now multiply by \(4.19\): \[ 4.19(3.6685\times10^{-7})\approx1.54\times10^{-6} \] The unit is \[ \frac{\text{m s}^{-2}}{\text{m}}=\text{s}^{-2} \] \( \textbf{Final answer:} \) \(1.54\times10^{-6}\,\text{s}^{-2}\), the linear coefficient inside a uniform planet.

Explanation: \( \textbf{Surface gravity scaling:} \) For a uniform planet, \[ g_s\propto \rho R \] If \(\rho'=2\rho\) and \(R'=2R\), then \[ \frac{g'_s}{g_s}=2\times2=4 \] So, \[ g'_s=4g_s \] Near-surface orbital speed is \[ v_o=\sqrt{g_sR} \] Therefore, \[ \frac{v'_o}{v_o}=\sqrt{\frac{g'_sR'}{g_sR}} \] \[ \frac{v'_o}{v_o}=\sqrt{(4)(2)}=\sqrt{8}=2\sqrt{2} \] Escape speed is \[ v_e=\sqrt{2g_sR} \] It has the same \(g_sR\) scaling as \(v_o\), so \[ \frac{v'_e}{v_e}=2\sqrt{2} \] \( \textbf{Final answer:} \) \(4g_s\), \(2\sqrt{2}v_o\), and \(2\sqrt{2}v_e\).

Explanation: Circular orbital speed is \[ v_o=\sqrt{\frac{GM}{r}} \] With \(M'=4M\) and the same \(r\): \[ v'_o=\sqrt{\frac{G(4M)}{r}} \] \[ v'_o=2\sqrt{\frac{GM}{r}}=2v_o \] The period is \[ T=2\pi\sqrt{\frac{r^3}{GM}} \] With \(M'=4M\): \[ T'=2\pi\sqrt{\frac{r^3}{G(4M)}} \] \[ T'=\frac{1}{2}2\pi\sqrt{\frac{r^3}{GM}} \] \[ T'=\frac{T}{2} \] \( \textbf{Final answer:} \) speed doubles and period becomes half, because the stronger central mass increases orbital speed at the same radius.

Explanation: Escape speed from the surface satisfies \[ v_e^2=\frac{2GM}{R} \] Given launch speed: \[ v=\frac{3}{4}v_e \] So, \[ v^2=\frac{9}{16}v_e^2 \] \[ v^2=\frac{9}{16}\cdot\frac{2GM}{R} \] \[ v^2=\frac{9GM}{8R} \] Initial total energy: \[ E_i=\frac{1}{2}mv^2-\frac{GMm}{R} \] Substitute \(v^2\): \[ E_i=\frac{1}{2}m\left(\frac{9GM}{8R}\right)-\frac{GMm}{R} \] \[ E_i=\frac{9GMm}{16R}-\frac{GMm}{R} \] \[ E_i=-\frac{7GMm}{16R} \] At maximum centre-distance \(r_{\max}\), speed is zero: \[ E_f=-\frac{GMm}{r_{\max}} \] Energy conservation gives \[ -\frac{GMm}{r_{\max}}=-\frac{7GMm}{16R} \] Cancel negative \(GMm\): \[ \frac{1}{r_{\max}}=\frac{7}{16R} \] \[ r_{\max}=\frac{16R}{7} \] \( \textbf{Final answer:} \) \(\frac{16R}{7}\), so the maximum height above the surface is \(\frac{9R}{7}\).

Explanation: Let the zero-field point be at distance \(x\) from mass \(M\). Then its distance from \(9M\) is \[ d-x \] At the zero-field point: \[ \frac{GM}{x^2}=\frac{G(9M)}{(d-x)^2} \] Cancel \(GM\): \[ \frac{1}{x^2}=\frac{9}{(d-x)^2} \] Taking positive square root: \[ \frac{1}{x}=\frac{3}{d-x} \] \[ d-x=3x \] \[ x=\frac{d}{4} \] So the distance from \(9M\) is \[ d-x=\frac{3d}{4} \] Potential is scalar, so potentials add algebraically: \[ V=-\frac{GM}{d/4}-\frac{G(9M)}{3d/4} \] \[ V=-\frac{4GM}{d}-\frac{12GM}{d} \] \[ V=-\frac{16GM}{d} \] \( \textbf{Final answer:} \) \(-\frac{16GM}{d}\), and the potential is not zero even though the net field is zero.

Explanation: The distance from the centroid of an equilateral triangle to each vertex is \[ r=\frac{a}{\sqrt{3}} \] The field at the centroid due to one mass \(M\) has magnitude \[ g_0=\frac{GM}{r^2} \] \[ g_0=\frac{GM}{a^2/3} \] \[ g_0=\frac{3GM}{a^2} \] If all three masses were \(M\), the net field at the centroid would be zero by symmetry. Replacing one of the \(M\) masses by \(2M\) is equivalent to adding one extra mass \(M\) at that vertex. So the remaining net field equals the field due to that extra \(M\) alone. Therefore, \[ g_{\text{net}}=g_0=\frac{3GM}{a^2} \] The direction is toward the vertex that has the extra mass. \( \textbf{Final answer:} \) \(\frac{3GM}{a^2}\), directed toward the \(2M\) vertex.

Explanation: Distance from the centre of a square to any corner is \[ r=\frac{a}{\sqrt{2}} \] Gravitational potential is scalar, so the potentials due to all four masses are added directly. The total mass factor is \[ M+2M+3M+4M=10M \] So the total potential at the centre is \[ V=-\frac{G(10M)}{r} \] Substitute \(r=\frac{a}{\sqrt{2}}\): \[ V=-\frac{10GM}{a/\sqrt{2}} \] \[ V=-\frac{10\sqrt{2}GM}{a} \] No vector cancellation is used for potential. \( \textbf{Final answer:} \) \(-\frac{10\sqrt{2}GM}{a}\), because potential depends only on scalar distance and source mass.

Explanation: Escape speed from the surface satisfies \[ v_e^2=\frac{2GM}{R} \] Given launch speed: \[ v=\frac{2}{3}v_e \] So, \[ v^2=\frac{4}{9}v_e^2 \] \[ v^2=\frac{4}{9}\cdot\frac{2GM}{R} \] \[ v^2=\frac{8GM}{9R} \] Initial total energy at the surface is \[ E_i=\frac{1}{2}mv^2-\frac{GMm}{R} \] \[ E_i=\frac{1}{2}m\left(\frac{8GM}{9R}\right)-\frac{GMm}{R} \] \[ E_i=\frac{4GMm}{9R}-\frac{GMm}{R} \] \[ E_i=-\frac{5GMm}{9R} \] At maximum distance \(r_{\max}\), the speed becomes zero: \[ E_f=-\frac{GMm}{r_{\max}} \] By conservation of energy: \[ -\frac{GMm}{r_{\max}}=-\frac{5GMm}{9R} \] Cancel negative \(GMm\): \[ \frac{1}{r_{\max}}=\frac{5}{9R} \] \[ r_{\max}=\frac{9R}{5} \] Maximum height above the surface is \[ h=r_{\max}-R \] \[ h=\frac{9R}{5}-R=\frac{4R}{5} \] \( \textbf{Final answer:} \) \(\frac{4R}{5}\), measured above the surface, not from the centre.

Explanation: Use conservation of mechanical energy. Initial radius: \[ r_i=4R \] Final radius: \[ r_f=R \] The body starts from rest, so \[ K_i=0 \] Initial energy: \[ E_i=-\frac{GMm}{4R} \] Final energy: \[ E_f=\frac{1}{2}mv^2-\frac{GMm}{R} \] Set \(E_i=E_f\): \[ -\frac{GMm}{4R}=\frac{1}{2}mv^2-\frac{GMm}{R} \] Move the potential term: \[ \frac{1}{2}mv^2=\frac{GMm}{R}-\frac{GMm}{4R} \] \[ \frac{1}{2}mv^2=\frac{3GMm}{4R} \] Cancel \(m\): \[ \frac{1}{2}v^2=\frac{3GM}{4R} \] \[ v^2=\frac{3GM}{2R} \] \[ v=\sqrt{\frac{3GM}{2R}} \] \( \textbf{Final answer:} \) \(\sqrt{\frac{3GM}{2R}}\), obtained from the loss of gravitational potential energy.

Explanation: Initial circular orbit: \[ K_i=\frac{GMm}{2r} \] \[ U_i=-\frac{GMm}{r} \] \[ E_i=-\frac{GMm}{2r} \] Final circular orbit radius: \[ r_f=4r \] So, \[ K_f=\frac{GMm}{2(4r)}=\frac{GMm}{8r} \] \[ U_f=-\frac{GMm}{4r} \] \[ E_f=-\frac{GMm}{2(4r)}=-\frac{GMm}{8r} \] Change in kinetic energy: \[ \Delta K=K_f-K_i \] \[ \Delta K=\frac{GMm}{8r}-\frac{GMm}{2r} \] \[ \Delta K=-\frac{3GMm}{8r} \] Change in potential energy: \[ \Delta U=U_f-U_i \] \[ \Delta U=-\frac{GMm}{4r}+\frac{GMm}{r} \] \[ \Delta U=\frac{3GMm}{4r} \] Change in total energy: \[ \Delta E=E_f-E_i \] \[ \Delta E=-\frac{GMm}{8r}+\frac{GMm}{2r} \] \[ \Delta E=\frac{3GMm}{8r} \] \( \textbf{Final answer:} \) \(-\frac{3GMm}{8r},+\frac{3GMm}{4r},+\frac{3GMm}{8r}\), so the satellite loses kinetic energy but gains total energy.

Explanation: Circular speed satisfies \[ v_o^2=\frac{GM}{r} \] New speed is \[ v=1.5v_o=\frac{3}{2}v_o \] Initial total energy after speed increase is \[ E=\frac{1}{2}m\left(\frac{3}{2}v_o\right)^2-\frac{GMm}{r} \] \[ E=\frac{1}{2}m\left(\frac{9}{4}v_o^2\right)-mv_o^2 \] because \[ \frac{GMm}{r}=mv_o^2 \] So, \[ E=\frac{9}{8}mv_o^2-mv_o^2 \] \[ E=\frac{1}{8}mv_o^2 \] At infinity, potential energy is zero, so \[ E=\frac{1}{2}mv_\infty^2 \] Thus, \[ \frac{1}{2}mv_\infty^2=\frac{1}{8}mv_o^2 \] Cancel \(m\): \[ v_\infty^2=\frac{1}{4}v_o^2 \] \[ v_\infty=\frac{v_o}{2} \] \( \textbf{Final answer:} \) \(\frac{v_o}{2}\), because \(1.5v_o\) is greater than the escape threshold \(\sqrt{2}v_o\).

Explanation: Surface gravity is \[ g_s=\frac{GM}{R^2} \] For the new planet: \[ M'=8M_E,\qquad R'=2R_E \] So, \[ g'_s=\frac{G(8M_E)}{(2R_E)^2} \] \[ g'_s=\frac{8GM_E}{4R_E^2}=2g \] Near-surface orbital speed is \[ v_o=\sqrt{\frac{GM}{R}} \] Thus, \[ \frac{v'_o}{v_o}=\sqrt{\frac{M'/R'}{M_E/R_E}} \] \[ \frac{v'_o}{v_o}=\sqrt{\frac{8M_E/(2R_E)}{M_E/R_E}} \] \[ \frac{v'_o}{v_o}=\sqrt{4}=2 \] Escape speed is \[ v_e=\sqrt{\frac{2GM}{R}} \] It has the same \(\sqrt{\frac{M}{R}}\) scaling as \(v_o\). So, \[ \frac{v'_e}{v_e}=2 \] \( \textbf{Final answer:} \) \(2g\), \(2v_o\), and \(2v_e\).

Explanation: Surface gravity is \[ g=\frac{GM}{R^2} \] The new planet has the same surface gravity: \[ \frac{GM'}{R'^2}=\frac{GM_E}{R_E^2} \] Given: \[ R'=9R_E \] So, \[ \frac{M'}{(9R_E)^2}=\frac{M_E}{R_E^2} \] \[ \frac{M'}{81R_E^2}=\frac{M_E}{R_E^2} \] \[ M'=81M_E \] Escape speed can be written as \[ v_e=\sqrt{2gR} \] Since \(g\) is the same and \(R'=9R_E\): \[ \frac{v'_e}{v_e}=\sqrt{\frac{2g(9R_E)}{2gR_E}} \] \[ \frac{v'_e}{v_e}=\sqrt{9}=3 \] \( \textbf{Final answer:} \) \(81M_E\) and \(3v_e\), because keeping \(g\) fixed while increasing \(R\) greatly increases the required mass.

Explanation: \( \textbf{Mass of the uniform planet:} \) For density \(\rho\) and radius \(R\), \[ M=\frac{4}{3}\pi R^3\rho \] The surface gravity is \[ g_s=\frac{GM}{R^2} \] Substitute the expression for \(M\): \[ g_s=\frac{G\left(\frac{4}{3}\pi R^3\rho\right)}{R^2} \] Cancel \(R^2\) from \(R^3\): \[ g_s=\frac{4}{3}\pi G\rho R \] This result shows that for planets of the same density, surface gravity is directly proportional to radius. \( \textbf{Final answer:} \) \(\frac{4}{3}\pi G\rho R\), because mass increases as \(R^3\) while the surface formula divides by \(R^2\).