Gravitation Practice Questions

$\;\;{G {\text{m}}^2}/{4 {\text{r}}^2}=\;{ {\ mv}^2}/{ {\text{r}}}$ $\;v=\sqrt{\;{ {\text{Cm}}}/{4 {\text{r}}}}$

${\text{V}}_{ {\text{e}}}=\sqrt{\;{2 {\text{GM}}}/{ {\text{R}}}}$As, $\;\frac{M}{R}$ increases $\Rightarrow V_e$ increases${\text{V}}_{ {\text{e}}} \propto \;{1}/{\sqrt{ {\text{R}}}}$As, $V_e$ depends on $R$

Sol. ${\text{T}}^2 \propto {\text{R}}^3$ $\;\;{ {\text{T}}_1^2}/{ {\text{T}}_2^2}=\;{ {\text{R}}_1^3}/{ {\text{R}}_2^3} \Rightarrow (\;{ {\text{T}}_1}/{ {\text{T}}_2}) ^2= (\;{ {\text{R}}}/{{ {\text{R}}}/{4}}) ^3$ $\; \therefore \;\; \;{ {\text{T}}_1^2}/{ {\text{T}}_2^2}=64$ $\; \therefore \;\; {\text{T}}_2^2=\;{ {\text{T}}_1^2}/{64}$ $\; \therefore \;\; T_2=\;\frac{24}{8}=3$ $\;$

$\; {\text{V}}_{ {\text{e}}}=\sqrt{\;{2 {\text{GM}}}/{ {\text{R}}}}=\sqrt{\;{2 {\text{G} \rho } \frac{4}{3} \pi {\text{R}}^3}/{ {\text{R}}}}= {\text{C}} {\sqrt{\rho }} \cdot {\text{R}}$ $\;\;{ {\text{V}}_{ {\text{e}}_1}}/{ {\text{V}}_{ {\text{e}}_2}}=\;{ {\text{R}}_1}/{ {\text{R}}_2} \sqrt{\;\frac{\rho _1}{\rho _2}}=\;\frac{1}{2}$ $\;\;{ {\text{R}}_1^2}/{ {\text{R}}_2^2} \times \;\frac{\rho _1}{\rho _2}=\;\frac{1}{4}$ $\;\;{ {\text{R}}_1}/{ {\text{R}}_2}=\;\frac{1}{3}$ $\; {\text{g}}=\;{ {\text{GM}}}/{ {\text{R}}^2}=\;{ {\text{G}} \frac{4}{3} \pi {\text{R}}^3 \times \rho }/{ {\text{R}}^2} {\text{C}} \cdot \rho {\text{R}}$ $\;\;{ {\text{g}}_1}/{ {\text{g}}_2}=\;{\rho _1 {\text{R}}_1}/{\rho _2 {\text{R}}_2}=\;\frac{1}{4} \;{ {\text{R}}_2^2}/{ {\text{R}}_1^2} \times \;{ {\text{R}}_1}/{ {\text{R}}_2}$ $\;=\;\frac{1}{4} \times \;{ {\text{R}}_2}/{ {\text{R}}_1}=\;\frac{3}{4}$

$\; {\text{mg}}=200 {\text{N}}$ $\; {\text{g}}^{'}= {\text{g}} (1-\;{ {\text{d}}}/{ {\text{R}}}) = {\text{g}} (1-\;{ {\text{R}}}/{2 \times {\text{R}}}) =\;{ {\text{g}}}/{2}$ $\;\;\text{ weight }\;= {\text{mg}}^{'}=\;{ {\text{mg}}}/{2}=\;\frac{200}{2}=100 {\text{N}}$

$\;\Delta {\text{V}}=\sqrt{\;{ {\text{GM}}}/{ {\text{R}}}}({\sqrt{2}}-1)$ $\;=\sqrt{\;{ {\text{GM}}}/{ {\text{R}}^2} \times {\text{R}}}({\sqrt{2}}-1)$ $\;=\sqrt{ {\text{gR}}}({\sqrt{2}}-1)=8000({\sqrt{2}}-1) {\text{ms}}^{-1}$ $\;=8({\sqrt{2}}-1) {\text{kms}}^{-1}$ $\;$

$\; {\text{g}}=\;{ {\text{GM}}}/{ {\text{R}}^2}=\;{ {\text{G}} \times \frac{4}{3} \pi {\text{R}}^3 \times \rho }/{ {\text{R}}^2}=\;\frac{4}{3} {\text{G}} \pi {\text{PR}}$ $\;\;{ {\text{g}}_2}/{ {\text{g}}_1}=\;{ {\text{R}}_2}/{ {\text{R}}_1} \times \;\frac{\rho _2}{\rho _1}=1.5 \times \;\frac{1}{2}=\;\frac{3}{4}$

$g_h\;=\;{g}/{ (1+\;{ {\text{h}}}/{ {\text{R}}}) ^2}$ ${\text{h}}\;=9 {\text{R}}$ ${\text{g}}_{ {\text{h}}}\;=\;{ {\text{g}}}/{(1+9)^2}=\;{ {\text{g}}}/{100}$ ${\text{w}}_{ {\text{h}}}\;=\;{ {\text{mg}}}/{100}=\;{ {\text{w}}}/{100}$

$\;\;\text{ Speed }\; V=\sqrt{\;\frac{G M_e}{r}}$ $\; {\text{m}}_e=\;\text{ mass of earth }\;$ $\;r=\;\text{ radius of orbit }\;$Independent on mass of satellite so speed is same Other three quantities are mass dependent$\;\to \;\; {\text{KE}}=\;{ {\text{Gm}}_{ {\text{e}}} {\text{m}}}/{2 {\text{r}}}$ $\;\to \;\; P E=\;{- {\text{Gm}}_{ {\text{e}}} {\text{m}}}/{ {\text{r}}}$ $\;\to \;\; {\text{TE}}=\;{- {\text{Gm}}_{ {\text{e}}} {\text{m}}}/{2 {\text{r}}}$ $\;$

$g_{e f f}=g-\omega ^{2} R_{e} sin \;^{2} \theta , \theta \to$ co-latitude angle $g_{\;\text{eff }\;}=g (1-\;\frac{d}{R_{e}}) , d$ here depth

Due to rotation of earth, $g_{\;\text{eff }\;}=g-\omega ^2 R \cos ^2 \theta$ Where ' $\theta$ ' is angle made with equator Also, At poles, $\theta =90^{\circ}$ $\;\Delta {\text{g}}=\omega ^2 {\text{R}} \cos ^2 \theta$ $\; \;\; =\omega ^2 {\text{R}} \cos ^2 90=0$[no effect on poles]${\text{g}}_{\;\text{eff }\;}= {\text{g}}-\omega ^2 {\text{R}} \cos ^2 \theta$for equator $\theta =0^{\circ}$So, $g_{\;\text{eff }\;}=g-w^2 R$ $\ \Delta {\text{g}}=\omega ^2 {\text{R}}$ (Which is maximum change)

$\;\;\text{ Loss \in PE = Gain \in KE }\;$ $\; (-\;{ {\text{GMm}}}/{2 {\text{R}}}) - (-\;{ {\text{GMm}}}/{ {\text{R}}}) =\;\frac{1}{2} {\ mv}^2$ $\;\Rightarrow {\text{v}}^2=\;{ {\text{GM}}}/{ {\text{R}}}= {\text{gR}}$ $\;\Rightarrow {\text{v}}=\sqrt{ {\text{gR}}}$

$\;\;{ {\text{GM}}}/{ {\text{R}}^2} [1-\;{ {\text{d}}}/{ {\text{R}}}] =\;{4 \times {\text{GM}}}/{(4 {\text{R}})^2}$ $\;1-\;{ {\text{d}}}/{ {\text{R}}}=\;\frac{1}{4} \Rightarrow \;{ {\text{d}}}/{ {\text{R}}}=\;\frac{3}{4} \Rightarrow {\text{d}} \;\frac{3}{4} {\text{R}}$ $\; {\text{d}}=4800 {\text{km}}$

$\;g=\;\frac{4 \pi }{3} {\text{GR}} \delta$ $\;g \propto \delta R$ $\;\;\frac{g_A}{g_B}=\;\frac{\delta _A R_A}{\delta _B \cdot R_B}=\;\frac{\delta \cdot R}{\frac{\delta }{3} \cdot 4 R}=\;\frac{3}{4}$

$\;F_C=m \omega ^2 a$ $\;\;\frac{G m^2}{(2 a)^2}=m \omega ^2 a$ $\;\omega =\sqrt{\;\frac{G m}{4 a^3}}$

Since planets revolve around the sun in an elliptical orbit its linear speed is not constant, hence option 1 not correct (and right choice).Other statement are correct as per theory.

Average Density of planet $=$ average density of earth$\;\;\frac{M_e}{\frac{4}{3} \pi R_e^3}=\;\frac{M_p}{\frac{4}{3} \pi R_p^3}$ $\;\Rightarrow \;\frac{M_e}{R_e^3}=\;\frac{2 M_e}{R_p^3}$ $\;\Rightarrow R_p=2^{\;\frac{1}{3}} R_e-------(i)$ $\;\;\text{ Now, }\; g=\;\frac{G M}{R^2}$ $\;\;\frac{g_e}{g_p}=\;\frac{M_e}{R_e^2} \times \;\frac{R_p^2}{2 M_e}=2^{\;\frac{2}{3}-1}=2^{-\;\frac{1}{3}}$ $\;\Rightarrow g_p=2^{\;\frac{1}{3}} g_e$ $\;\Rightarrow W_p=2^{\;\frac{1}{3}} W_e$

$\; {\text{g}}^{'}=\;{ {\text{g}}}/{ (1+\;{ {\text{h}}}/{ {\text{R}}}) ^2}$ $\; {\text{g}}^{'}= {\text{g}} \{1-\;{ {\text{d}}}/{ {\text{R}}}\}$Statement I is correct & Statement II is incorrect