Gravitation Practice Questions

Net force on any one particle$={GM^2}/(2R)^{2+}{GM^2}/(R\sqrt2)^2cos45°$ $={GM}^2/R^2[1/4+1/\sqrt2]$This force will be equal to centripetal force so${Mu^2}/R={GM}^2/R^2[{1+2\sqrt2}/4]$ $u=\sqrt{\frac{GM}{4R}[1+2\sqrt2]}=1/2\sqrt{{GM}/R(2\sqrt2+1)}$

$T.E_f=-\frac{GMm}{6R}$ $T.E_i=-{GMm}/R$ $\Delta W=T.E_f-T.E_i=\frac{5GmM}{6R}$

Potential energy at earth surface $=-\;\frac{G M m}{R}$ and at free space potential energy $=0$Work done for this $=0- (-\;\frac{G M m}{R}) =\;\frac{G M m}{R}$So the required energy for this work is$\;=\;\frac{G M m}{R}$ $\;=\;\frac{g R^2 m}{R} [\;\text{ as }\; g=\;\frac{G M}{R^2}]$ $\;=m g R$ $\;=1000 \times 10 \times 6400 \times 10^3$ $\;=6.4 \times 10^{10} \;\text{ Joules }\;$

Let the gravitational field at $P$, distant $x$ from mass $m$, be zero$\; \therefore \;\frac{G m}{x^2}=\;\frac{4 G m}{(r-x)^2}$ $\;\Rightarrow 4 x^2=(r-x)^2$ $\;\Rightarrow 2 x=r-x$ $\;\Rightarrow x=\;\frac{r}{3}$Gravitational potential at point $P$,$V=-\;\frac{G m}{\frac{7}{3}}-\;\frac{4 G m}{\frac{2 r}{3}}=-\;\frac{9 G m}{r}$

Given that, at height ${\text{h}}$ from ground the acceleration due to gravity becomes $\;\frac{g}{9}$.We know acceleration at earth surface due to gravity ${\text{g}}=\;\frac{G M}{R^2}$and acceleration at height $h$ due to gravity $g^{'}=\;\frac{G M}{(R+h)^2}$ So $\;\frac{g}{9}=\;\frac{G M}{(R+h)^2}$ $\Rightarrow \;\frac{g}{9}=\;\frac{G M}{R^2} \cdot \;\frac{R^2}{(R+h)^2}$ $=g \cdot (\;\frac{R}{R+h}) ^2$ $\Rightarrow \;\frac{1}{9}= (\;\frac{R}{R+h}) ^2$ $\Rightarrow \;\frac{R}{R+h}=\;\frac{1}{3}$ $\Rightarrow 3 R=R+h$ $\therefore h=2 R$

Gravitational field $\vec{g}=-\;\frac{G M}{r^2}$where, $M=$ mass enclosed in the closed surfaceGravitational flux through a closed surface is given by$|\vec{g} \cdot d \vec{S}|=4 \pi r^2 \cdot \;\frac{G M}{r^2}=4 \pi G M$So Statement - 1 is correct.Statement - 2 is also correct because when the shape of the earth is spherical, area of the Gaussian surface is $4 \pi r^2$. This proves inverse square law.

Let $M_e$ is mass of earth then mass of planet $M_p=10 M_e$.And let ${\text{R}}_e$ is radius of earth then radius of planet ${\text{R}}_{ {\text{p}}}=\;\frac{R_c}{10}$Escape velocity of earth, $v_e=\sqrt{\;\frac{2 G M_e}{R_e}}$Escape velocity of planet, $v_p=\sqrt{\;\frac{2 G M_p}{R_p}}$ $\therefore \;\frac{v_p}{v_e}=\;{\sqrt{\frac{2 G M_p}{R_p}}}/{\sqrt{\;\frac{2 G M_e}{R_e}}}$ $=\sqrt{\;\frac{M_p}{M_e} \times \;{ {\text{R}}_e}/{R_p}}$ $=\sqrt{\;\frac{10 M_e}{M_e} \times \;{ {\text{R}}_e}/{ {\text{R}}_e / 10}}=10$ $\therefore v_p=10 \times v_e$ $=10 \times 11=110 {\text{km}} / {\ s}$

electronic charge does does not depend on acceleration due to gravity as it is a universal constant.So, electronic charge on earth= electronic charge on moon$\therefore$ Required ratio $=1$

We know, Work done = Difference in potential energy$\; \therefore W=\Delta U=U_f-U_i=0- [\;\frac{-G M m}{R}]$ $\;\Rightarrow W=\;\frac{6.67 \times 10^{-11} \times 100}{0.1} \times \;\frac{10}{1000}$ $\;=6.67 \times 10^{-10} J$

$\;\;\text{ Mass of earth }\;=\;\text{ Volume }\; \times \;\text{ Density of earth }\;(\rho )$ $\; \therefore {\text{M}}=\;\frac{4}{3} \pi R^3 \times \rho$We know, $g=\;\frac{G M}{R^2}$ $\;\Rightarrow g=\;\frac{G \times \rho \times \frac{4}{3} \pi R^3}{R^2}$ $\;\Rightarrow g=\;\frac{4}{3} \rho \pi G R$ $\; \therefore g \propto \rho \;\text{ or }\; \rho \propto g$

At height $h$ acceleration due to gravity, $g_h=g [1-\;\frac{2 h}{R}]$;At depth $d$ acceleration due to gravity, $g_d=g [1-\;\frac{d}{R}]$According to the question,$\;g_h=g_d$ $\; \therefore g [1-\;\frac{2 h}{R}] =g [1-\;\frac{d}{R}]$ $\;\Rightarrow \;\frac{2 h g}{R}=\;\frac{d g}{R}$ $\;\Rightarrow d=2 h$

Gravitational force applied on the satellite,$=\;\frac{G M m}{(R+x)^2}$For satellite, gravitational force $=$ centripetal force$\therefore \;\frac{m v^2}{(R+x)}=\;\frac{G M m}{(R+x)^2}$where $v$ is the orbital speed of satellite.also $g=\;\frac{G M}{R^2} \Rightarrow G M=g R^2$ $\; \therefore v^2=\;\frac{g R^2}{R+x}$ $\;\Rightarrow v= (\;\frac{g R^2}{R+x}) ^{1 / 2}$

Gravitational potential energy on the earth surface of a body ${\text{U}}=-\;\frac{G m M}{R}$ And at the height ${\text{h}}$ from the earth surface the potential energy $U_h=-\;\frac{G m M}{R+h}=-\;\frac{G m M}{2 R}$ [as ${\text{h}}= {\text{R}}$] So the gain in the potential energy $\Delta U=U_h-U$ $\; \therefore \Delta U=\;\frac{-G m M}{2 R}+\;\frac{G m M}{R}$ $\;\Rightarrow \Delta U=\;\frac{G m M}{2 R}$Now $\;\frac{G M}{R^2}=g ; \;\; \therefore \;\frac{G M}{R}=g R$ $\therefore \Delta U=\;\frac{1}{2} m g R$

For moving a planet around the sun in the circular orbit, The necessary centripetal force $=$ Gravitational force exerted on it $\therefore \;\frac{m v^2}{r}=\;\frac{G M m}{R^n}$ $\Rightarrow v=\sqrt{\;\frac{G M}{R^{n-1}}}$We know, $T=\;\frac{2 \pi R}{v}$ $\;=2 \pi R \times \sqrt{\;\frac{R^{n-1}}{G M}}$ $\;=2 \pi \times \sqrt{\;\frac{R^2 \times R^{n-1}}{G M}}$ $\;=2 \pi \times \;{R^{\frac{n+1}{2}}}/{{\sqrt{G M}}}$ $\; \therefore T \propto R^{\;\frac{(n+1)}{2}}$

For satellite, gravitational force = centripetal force $\therefore \;\frac{m v^2}{R+x}=\;\frac{G m M}{(R+x)^2}$ $x=$ height of satellite from earth surface $m=$ mass of satellite $\Rightarrow v^2=\;\frac{G M}{(R+x)}$ or $v=\sqrt{\;\frac{G M}{R+x}}$ We know, $T=\;\frac{2 \pi }{\omega }$ $T=\;\frac{2 \pi (R+x)}{v}$ [ as $\omega =\;\frac{v}{r}]$ $=\;{2 \pi (R+x)}/{\sqrt{\;\frac{G M}{R+x}}}$ which is independent of mass of satellite

According to kepler's law,$T^2 \propto R^3$ $\therefore \;\frac{T_1^2}{T_2^2}=\;\frac{R_1^3}{R_2^3}$ $\Rightarrow T_2=T_1 (\;\frac{R_2}{R_1}) ^{3 / 2}=5 \times [\;\frac{4 R}{R}] ^{3 / 2}$ $=5 \times 2^3=40$ hour𝜕

We know, Escape velocity, $v_e={\sqrt{2 g R}}$ So the escape velocity is independent of the angle at which the body is projected, hence it will remain same as $11 \frac{\text{km}}{\ s}$.

Let $t$ be the time taken for the two masses to collide and $x_{5 M}, x_M$ be the distance travelled by the mass $5 M$ and $M$ respectively. The gravitational force acting between two sphere when the distance between them ( $12 {\text{R}}- {\text{x}})$ where $x$ is a variable, $F=\;\frac{G M \times 5 M}{(12 R-x)^2}$ Acceleration of mass ${\text{M}}, a_M=\;\frac{G \times 5 M}{(12 R-x)^2}$ Acceleration of mass $5 {\text{M}}, a_{5 M}=\;\frac{G M}{(12 R-x)^2}$ For mass $5 M$ $u=0, S=x_{5 M}, t=t, a=a_{5 M}$ $S=u t+\;\frac{1}{2} a t^2$ $\therefore x_{5 M}=\;\frac{1}{2} a_{5 M} t^2$.....(i) For mass $M$ $u=0, s=x_M, t=t, a=a_M$ $\therefore s=u t+\;\frac{1}{2} a t^2 \Rightarrow$ $x_M=\;\frac{1}{2} a_M t^2$....(ii) Dividing $(i i)$ by $(i i i)$ $\;\;\frac{x_{5 M}}{x_M}=\;\frac{\;\frac{1}{2} a_{5 M} t^2}{\frac{1}{2} a_M t^2}$ $\;=\;\frac{a_{5 M}}{a_M}=\;\frac{1}{5}$ $\therefore 5 x_{5 M}=x_M$.......(iii) From the figure,$x_{5 M}+x_M=12 R-2 R-R=9 R$......(iv)From $(i i i)$ and $(i v)$ $\;\frac{x_M}{5}+x_M=9 R$ $\therefore 6 x_M=45 R$ $\therefore x_M=\;\frac{45}{6} R=7.5 R$ So two sphere collide when the sphere of mass $M$ covered the distance of $7.5 {\text{R}}$.