Gravitation Practice Questions

$\;\omega =\;{2 \pi }/{ {\text{T}}} \Rightarrow \omega \propto \;{1}/{ {\text{T}}}$ $\; {\text{T}}_{\;\text{moon }\;}=27 \;\text{ days }\;$ $\; {\text{T}}_{\;\text{earth }\;}=365 \;\text{ days }\; 4 \;\text{ hour }\;$ $\;\Rightarrow \omega _{\;\text{moon }\;}>\omega _{\;\text{earth }\;}$

We have ${\text{R}}=\;{ {\text{L}}}/{\pi }$ $\; {\text{g}}_0=\;{2 {\text{G}} { {\text{M}}}/{ {\text{L}}}}/{ {\text{R}}}=\;{2 {\text{GM}} \pi }/{ {\text{L}}^2}$ $\; \therefore {\text{F}}_{ {\text{m}}}= {\text{mg}}_0=\;{2 {\text{GM}} \pi {\text{m}}}/{ {\text{L}}^2}$Hence option (4)

$\; {\text{KE}}=\;\frac{1}{2} {\ mv}^2=\;{ {\text{GMm}}}/{2 {\text{a}}}$ $\; {\text{PE}}=-2 {\text{KE}}$ $\; {\text{TE}}=- {\text{KE}}$

$\; {\text{g}}^{'}=\;{ {\text{GMe}}}/{(3 {\text{R}})^2}=\;\frac{1}{9} {\text{g}}$ $\; {\text{T}}=2 \pi \sqrt{\;{ℓ}/{ {\text{g}}^{'}}}$Since the time period of second pendulum is $2 {\ sec}$.$\; {\text{T}}=2 {\ sec}$ $\;2=2 \pi \sqrt{\;{ℓ}/{ {\text{g}}} 9}$ $\;ℓ=\;\frac{1}{9} {\text{m}}$

${\text{T}}=\;{2 \pi {\text{r}}^{3 / 2}}/{\sqrt{ {\text{GM}}}}$ $\;{ {\text{T}}_1}/{ {\text{T}}_2}= (\;{ {\text{r}}_1}/{ {\text{r}}_2}) ^{3 / 2} (\;{ {\text{M}}_2}/{ {\text{M}}_1}) ^{1 / 2}$ $\;\frac{6}{24}=\;{ ( {\text{r}}_1) ^{3 / 2}}/{ (4.2 \times 10^4) ^{3 / 2}} (\;{ {\text{M}}}/{ {\text{M}} / 4}) ^{1 / 2}$ ${\text{r}}_1=1.05 \times 10^4 {\text{km}}$

Acceleration due to gravity $g^{'}=\;\frac{g}{4}$ $\; {\text{T}}=2 \pi \sqrt{\;{4 ℓ}/{ {\text{g}}}}$ $\; {\text{T}}=2 \pi \sqrt{\;{4 \times 4}/{ {\text{g}}}}$ $\; {\text{T}}=2 \pi \;\frac{4}{\pi }=8 {\ s}$

$\; {\text{V}}_{\;\text{escape }\;}=\sqrt{\;{2 {\text{GM}}}/{ {\text{R}}}}$ $\; {\text{V}}_{\;\text{planet }\;}=\sqrt{\;{2 {\text{GM}}}/{ {\text{R}}}}= {\text{V}}$ $\; {\text{V}}_{\;\text{Moon }\;}=\sqrt{\;{2 {\text{GM}} \times 16}/{144 {\text{R}}}}$ $=\;\frac{1}{3} \sqrt{\;{2 {\text{GM}}}/{ {\text{R}}}}$ $\; {\text{V}}_{\;\text{Moon }\;}=\;{ {\text{V}}_{\;\text{Planet }\;}}/{3}=\;{ {\text{V}}}/{3}$

$\;\Rightarrow \;{ {\text{GMm}}}/{( {\text{R}}+ {\text{h}})^2}=\;{ {\ mv}^2}/{( {\text{R}}+ {\text{h}})}$ $\;\Rightarrow \;{ {\text{GM}}}/{( {\text{R}}+ {\text{h}})}= {\text{v}}^2 . . .(1)$ $\;\Rightarrow {\text{v}}=( {\text{R}}+ {\text{h}}) \omega$ $\;\Rightarrow {\text{v}}=( {\text{R}}+ {\text{h}}) \;{2 \pi }/{ {\text{T}}} . . . . . . .$ (2)$\;\Rightarrow \;{ {\text{GM}}}/{ {\text{R}}^2}= {\text{g}}$ $\;\Rightarrow {\text{GM}}= {\text{gR}}^2 . . . . . . . .(3)$Put value from (2) & (3) in eq. (1)$\;\Rightarrow \;{ {\text{gR}}^2}/{( {\text{R}}+ {\text{h}})}=( {\text{R}}+ {\text{h}})^2 (\;{2 \pi }/{ {\text{T}}}) ^2$ $\;\Rightarrow \;{ {\text{T}}^2 {\text{R}}^2 {\text{g}}}/{(2 \pi )^2}=( {\text{R}}+ {\text{h}})^3$ \;\Rightarrow \[;{ {\text{T}}^2 {\text{R}}^2 {\text{g}}}/{(2 π)^2}]$ ^{1 / 3}- {\text{R}}= {\text{h}}$

$\; {\text{h}}=318.5 \approx (\;{ {\text{R}}_{ {\text{e}}}}/{20})$ $\; {\text{T}} \cdot {\text{E}}_{ {\text{i}}}=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{ {\text{R}}_{ {\text{e}}}}$ $\; {\text{T}} \cdot {\text{E}}_{ {\text{f}}}=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{2 ( {\text{R}}_{ {\text{e}}}+ {\text{h}}) }=\;{- {\text{GM}}_{ {\text{e}}} {\text{m}}}/{2 ( {\text{R}}_{ {\text{e}}}+\;{ {\text{R}}_{ {\text{e}}}}/{20}) }$ $\;\Rightarrow {\text{T}} \cdot {\text{E}}_{ {\text{f}}}=\;{-10 {\text{GM}}_{ {\text{e}}} {\text{m}}}/{21 {\text{R}}_{ {\text{e}}}}$Change in total mechanical energy$\;= {\text{TE}}_{ {\text{f}}}- {\text{TE}}_{ {\text{i}}}$ $\;=\;{ {\text{GM}}_{ {\text{e}}} {\text{m}}}/{{\text{Re}}} [1-\;\frac{10}{21}] =\;{11 {\text{GM}}_{ {\text{e}}} {\text{m}}}/{21 {\text{Re}}}$

$\;\;{\pi {\text{r}}_1^2}/{ {\text{T}}_{ {\text{A}}}}=\;{ {\text{L}}}/{2 {\text{m}}_1} . . . . . . . .(1)$ $\;\;{\pi {\text{r}}_2^2}/{ {\text{T}}_{ {\text{B}}}}=\;{3 {\text{L}}}/{2 {\text{m}}_2} . . . . . . .(2)$ $\;\Rightarrow \;{ {\text{T}}_{ {\text{A}}}}/{ {\text{T}}_{ {\text{B}}}}=3 \cdot \;{ {\text{m}}_1}/{ {\text{m}}_2} \cdot (\;{ {\text{r}}_1}/{ {\text{r}}_2}) ^2$ $\; (\;{ {\text{T}}_{ {\text{A}}}}/{ {\text{T}}_{ {\text{B}}}}) ^2= (\;{ {\text{r}}_1}/{ {\text{r}}_2}) ^3 \Rightarrow (\;{ {\text{r}}_1}/{ {\text{r}}_2}) ^2= (\;{ {\text{T}}_{ {\text{A}}}}/{ {\text{T}}_{ {\text{B}}}}) ^{\;\frac{4}{3}}$ $\;\Rightarrow \;\frac{1}{27} \cdot (\;{ {\text{m}}_2}/{ {\text{m}}_1}) ^3= (\;{ {\text{T}}_{ {\text{A}}}}/{ {\text{T}}_{ {\text{B}}}})$

Total energy $=\;{- {\text{GMm}}}/{2(2 R)}$if energy $=\;{10^4 {\text{R}}}/{6}$ is added then$\;{- {\text{GMm}}}/{4 {\text{R}}}+\;{10^4 {\text{R}}}/{6}=\;{- {\text{GMm}}}/{2 {\text{r}}}$where $r$ is new radius of revolving and $g=\;\frac{G M}{R^2}$ $\;-\;{ {\text{mgR}}}/{4}+\;{10^4 {\text{R}}}/{6}=-\;{ {\text{mgR}}^2}/{2 {\text{r}}} ( {\text{m}}=10^3 {\text{kg}})$ $\;-\;{10^3 \times 10 \times {\text{R}}}/{4}+\;{10^4 {\text{R}}}/{6}=-\;{10^3 \times 10 \times {\text{R}}^2}/{2 {\text{r}}}$ $\;-\;\frac{1}{4}+\;\frac{1}{6}=-\;{ {\text{R}}}/{2 {\text{r}}}$ $\; {\text{r}}=6 {\text{R}}$

$\;{v}=\sqrt{\;\frac{G M}{R}}$ $\;\;\frac{v_A}{v_B}=\sqrt{\;\frac{R_B}{R_A}}=\sqrt{\;\frac{R}{4 R}}=\;\frac{1}{2}$ $\;v_B=2 v_A=6 v$

$\; {\text{F}}=\;{ {\text{GMm}}}/{ {\text{R}}^{3 / 2}}= {\text{m}} \omega ^2 {\text{R}}$ $\;\omega ^2 \propto \;{1}/{ {\text{R}}^{5 / 2}} \;\;$ $\because {\text{T}}=\;\frac{2 \pi }{\omega } \;\; \;\text{ so }\;$ $\; {\text{T}}^2 \propto {\text{R}}^{5 / 2}$

$\; {\text{g}}=\;{ {\text{GM}}}/{ {\text{R}}^2} \Rightarrow {\text{g}} \propto \;{1}/{ {\text{R}}^2}$ $\;\;{ {\text{g}}_2}/{ {\text{g}}_1}=\;{ {\text{R}}_1^2}/{ {\text{R}}_2^2}$ $\; {\text{g}}_2=4 {\text{g}}_1 ( {\text{R}}_2=\;{ {\text{R}}_1}/{2})$

$\;\;\frac{1}{2} {\ mv}_{ {\text{e}}}^2=\;{ {\text{GMm}}}/{ {\text{R}}_{ {\text{E}}}}$ $\; {\text{g}}=\;{ {\text{GM}}}/{ {\text{R}}_{ {\text{E}}}^2}$ $\; {\text{K}}= {\text{mgR}}_{ {\text{E}}}$

By conservation of angular momentum$\; {\text{I}}_1 \omega _1= {\text{I}}_2 \omega _2$ $\; (\;\frac{2}{5} {\text{MR}}^2) \;{2 \pi }/{ {\text{T}}_1}=\;\frac{2}{5} {\text{M}} (\;\frac{3}{4} {\text{R}}) ^2 \;{2 \pi }/{ {\text{T}}_2}$ $\;\;{1}/{ {\text{T}}_1}=\;{9}/{16 {\text{T}}_2}$ $\;\;{1}/{ {\text{T}}_2}=\;\frac{9}{16} \times {\text{T}}_1=\;\frac{9}{16} \times 24 {\text{hr}}=\;\frac{27}{2} {\text{hr}}=13 {\text{hr}} 30 {\text{mins}} .$

At surface: ${\text{mg}}=300 {\text{N}}$ $\; {\text{m}}=\;{300}/{ {\text{g}}_{ {\ s}}}$ $\;\;\text{ At Depth }\; \;{ {\text{R}}}/{4}: {\text{g}}_{ {\text{d}}}= {\text{g}}_{ {\ s}} [1-\;{ {\text{d}}}/{ {\text{R}}}]$ $\; {\text{g}}_{ {\text{d}}}= {\text{g}}_{ {\ s}} [1-\;{ {\text{R}}}/{4 {\text{R}}}]$ $\; {\text{g}}_{ {\text{d}}}=\;{3 {\text{g}}_5}/{4}$ $\;\;\text{ weight at depth }\; \;\; = {\text{m}} \times {\text{g}}_{ {\text{d}}}$ $\;= {\text{m}} \times \;{3 {\text{g}}_{ {\ s}}}/{4}$ $\;=\;\frac{3}{4} \times 300$ $\;=225 {\text{N}}$ $\;$

$-\;\frac{G M_E}{R_E+h}=-5.12 \times 10^{-7}$ ..........(i) $\;\frac{G M_E}{ (R_E+h) ^2}=6.4 . . . . ....$ (ii) By (i) and (ii) $\Rightarrow h=16 \times 10^5 {\text{m}}=1600 {\text{km}}$