Escape Speed and Motion of Satellites — NEET Physics PYQ MCQs with Solutions

Solution: $v_e=\sqrt{\dfrac{2GM}{R}}$ with $M=\rho\cdot\dfrac{4}{3}\pi R^3$, giving $v_e=\sqrt{2G\rho\dfrac{4}{3}\pi R^2}\propto R$ for constant density. With $R_p=4R_e$, $v_p=4v_e=4v$.

Solution: By energy conservation, $\dfrac{1}{2}m(kv_e)^2-\dfrac{GMm}{R}=-\dfrac{GMm}{R+h}$, and using $v_e^2=\dfrac{2GM}{R}$ gives $\dfrac{k^2}{R}-\dfrac{1}{R}=-\dfrac{1}{R+h}$. Solving, $h=R\left(\dfrac{k^2}{1-k^2}\right)=\dfrac{Rk^2}{1-k^2}$.

Solution: $v_e=\sqrt{2gR}=R\sqrt{\dfrac{8}{3}\pi G\rho}$. For the planet, radius and density are twice, so $v_p=2R\sqrt{\dfrac{8}{3}\pi G(2\rho)}$. Dividing, $\dfrac{v_e}{v_p}=\dfrac{R\sqrt{\rho}}{2R\sqrt{2\rho}}=\dfrac{1}{2\sqrt{2}}$.

Solution: Total energy of a satellite at height $h$ is $E=\text{KE}+\text{PE}=\dfrac{GMm}{2(R+h)}-\dfrac{GMm}{(R+h)}=\dfrac{-GMm}{2(R+h)}=\dfrac{-GMmR^2}{2R^2(R+h)}=\dfrac{-mg_0R^2}{2(R+h)}$, using $g_0=\dfrac{GM}{R^2}$.

Solution: Orbital speed $v_0=\sqrt{\dfrac{GM_e}{R_e+h}}=\sqrt{\dfrac{gR_e^2}{R_e+h}}=\sqrt{\dfrac{gR_e}{1+h/R_e}}$. Substituting $g=9.8$, $R_e=6.38\times10^6$ m, $h=0.25\times10^6$ m gives $v_0=\sqrt{60\times10^6}=7.76\times10^3$ m/s $=7.76$ km/s.

Solution: The only force on the satellite is the gravitational force, which is always directed towards the centre of the earth. Hence the acceleration of $S$ is always directed towards the centre of the earth.

Solution: For a black hole the escape speed must be at least equal to $c$. $v_e=\sqrt{\dfrac{2GM}{R'}}$; setting $v_e\approx c$ gives $c^2=\dfrac{2GM}{R'}$, so $R'=\dfrac{2GM}{c^2}=\dfrac{2\times6.67\times10^{-11}\times6\times10^{24}}{9\times10^{16}}\approx0.889\times10^{-2}\approx10^{-2}$ m.

Solution: Orbital velocity $v=\sqrt{\dfrac{GM}{r}}\propto\dfrac{1}{\sqrt{r}}$, so $\dfrac{v_A}{v_B}=\sqrt{\dfrac{r_B}{r_A}}=\sqrt{\dfrac{R}{4R}}=\dfrac{1}{2}$. With $v_A=3v$, $\dfrac{3v}{v_B}=\dfrac{1}{2}$ gives $v_B=6v$.

Solution: At a platform of height $h=R$, escape energy equals binding energy: $\dfrac{1}{2}m(fv_e)^2=\dfrac{GMm}{R+h}$, so $fv_e=\sqrt{\dfrac{2GM}{2R}}$ ...(i). At the surface, $v_e=\sqrt{\dfrac{2GM}{R}}$ ...(ii). Dividing (ii) by (i): $\dfrac{fv_e}{v_e}=\dfrac{\sqrt{GM/R}}{\sqrt{2GM/R}}\Rightarrow f=\dfrac{1}{\sqrt{2}}$.

Solution: Potential energy $U=-\dfrac{GM_em}{R_e}$, so $|U|=\dfrac{GM_em}{R_e}$ and kinetic energy $K=\dfrac{1}{2}\dfrac{GM_em}{R_e}$. Thus $\dfrac{K}{|U|}=\dfrac{1}{2}\dfrac{GM_em}{R_e}\times\dfrac{R_e}{GM_em}=\dfrac{1}{2}$.

Solution: $v_{es}=\sqrt{\dfrac{2GM_e}{R_e}}$. For the planet $v'_{es}=\sqrt{\dfrac{2GM_p}{R_p}}$, so $\dfrac{v'_{es}}{v_{es}}=\sqrt{\dfrac{M_p}{M_e}\times\dfrac{R_e}{R_p}}$. Given $M_p=M_e$ and $R_p=\dfrac{R_e}{4}$, so $\dfrac{v'_{es}}{v_{es}}=\sqrt{4}=2$. Hence $v'_{es}=2\times11.2=22.4$ km/s.

Solution: The binding energy of the sphere of mass $m$ on the earth's surface is $\dfrac{GM_em}{R_e}$. To escape, this much energy is supplied as kinetic energy: $\dfrac{1}{2}mv_e^2=\dfrac{GM_em}{R_e}$, giving escape velocity $v_e=\sqrt{\dfrac{2GM_e}{R_e}}$.

Solution: $v_{es}=\sqrt{\dfrac{2GM_e}{R_e}}$, so $\dfrac{v'_{es}}{v_{es}}=\sqrt{\dfrac{M'_e}{M_e}\times\dfrac{R_e}{R'_e}}$. With $M'_e=2M_e$ and $R'_e=\dfrac{R_e}{2}$, $\dfrac{v'_{es}}{v_{es}}=\sqrt{2\times2}=\sqrt{4}=2$. Hence $v'_{es}=2\times11.2=22.4$ km/s.

Solution: When the ball is dropped from the orbiting satellite it already has the satellite's speed due to inertia. Since a satellite's orbit does not depend on its mass, the ball continues to move along the same original orbit as the satellite.

Solution: $v_e=\sqrt{2gR_e}=\sqrt{\dfrac{2GM_e}{R_e}}$ (using $g=\dfrac{GM_e}{R_e^2}$), so $v_e\propto\sqrt{\dfrac{M_e}{R_e}}$. Then $\dfrac{v_e}{v_p}=\sqrt{\dfrac{M_e}{R_e}\times\dfrac{R_p}{M_p}}$. With $R_p=3R_e$ and $M_p=3M_e$, $\dfrac{v_e}{v_p}=\sqrt{\dfrac{3R_e}{3R_e}}=1$, so $v_p=v_e$.

Solution: $v_e=\sqrt{2gR}=\sqrt{\dfrac{2GM}{R}}$. Escape velocity does not depend on the angle of projection, so it remains $11.2$ km/s.

Solution: For a geostationary satellite $T=\dfrac{2\pi r}{v_o}$ with $v_o=\sqrt{\dfrac{GM}{r}}=\sqrt{\dfrac{gR^2}{r}}$ (since $g=\dfrac{GM}{R^2}$). So $T=\dfrac{2\pi r^{3/2}}{\sqrt{gR^2}}$. Also $T=\dfrac{2\pi}{\omega}$. Equating gives $r^3=\dfrac{gR^2}{\omega^2}$, so $r=\left(\dfrac{R^2 g}{\omega^2}\right)^{1/3}$.

Solution: Total energy $=$ PE $+$ KE $=-\dfrac{GMm}{R}+\dfrac{1}{2}mv^2$. With $v=\sqrt{\dfrac{GM}{R}}$, $\dfrac{1}{2}mv^2=\dfrac{GMm}{2R}$, so total energy $=-\dfrac{GMm}{R}+\dfrac{GMm}{2R}=-\dfrac{GMm}{2R}=-\dfrac{1}{2}mv^2$.

Solution: Escape velocity $v_e=\sqrt{\dfrac{2GM}{R}}=\sqrt{2gR}=\sqrt{\dfrac{8\pi\rho GR^2}{3}}$ is the minimum velocity to just escape the gravitational field. It does not depend on the mass of the body or the angle of projection, so it remains $11$ km/s.