Gravitation

According to modified Ampere's law

$?B.dI={\mu }_0\left(I_C+I_D\right)$

For Loop  $L_1{I}_C\ne 0$ and $I_D=0$

For Loop  $L_2{I}_C=0$ and $I_D\ne 0$

Due to  $KCL{I}_C=I_D$

For A satellite T₁ = 1hour

$So{\omega }_1=2\pi red/hour$                

for B satellite T₂ = 8 hour

given R₁ = 2 × 10³ Km

Relative  $\omega =\frac{V_1-V_2}{R_2-R_1}=\frac{2\pi \times 10^3}{6\times 10^3}$  

$\Rightarrow \frac{\pi }{3}$ rad/hour

 x = 3

$g^{\prime }=\frac{G{M}^{\prime }}{R^{\text{'}2}}=\frac{GM}{10{\left(\frac{R}{2}\right)}^2}$

$=\frac{4}{10}\frac{GM}{R^2}=0.4\times 9.8$

$=3.92 m s^{-2}$

Using conservation of energy :

$?\frac{?GMm}{R}+\frac{1}{2}m{\left(\sqrt[]{\frac{2GM}{R}}\right)}^2=\frac{?GMm}{r}+\frac{1}{2}m{v}^2$

$?\frac{1}{2}m{v}^2=\frac{GMm}{r}$

$?{\displaystyle \underset{R}{\overset{R+h}{?}}{r}^{1/2}dr=\sqrt[]{2GM}{\displaystyle \underset{0}{\overset{t}{?}}dt}}$

$?\frac{2}{3}{R}^{3/2}\left[{\left(1+\frac{h}{R}\right)}^{3/2}?1\right]=\sqrt[]{2GM}t$

$t=\frac{1}{3}\sqrt[]{\frac{2R}{g}}\left[{\left(1+\frac{h}{R}\right)}^{3/2}?1\right]$

Since binary mass system performs circular motion about is common centre of mass, so

$m_A{\omega }_A^2{r}_A=\frac{G{m}_B{m}_A}{{\left(r_A+r_B\right)}^2}=\frac{G{m}_B{m}_A}{r^2}$

$m_A{\omega }_A^2\times \frac{m_B}{\left(m_A+m_B\right)}r=\frac{G{m}_B{m}_A}{r^2}$

${\omega }_A=\sqrt[]{\frac{G\left(m_A+m_B\right)}{r^3}}$

Similarly we can show that

${\omega }_B=\sqrt[]{\frac{G\left(m_A+m_B\right)}{r^3}}$

Hence their angular velocity will be same, time period will be same, i.e. TA = TB

For circular motion F_{net   $=\frac{M{V}^2}{R}$}

_{$\sqrt[]{2F}+F\text{'}=\frac{M{V}^2}{R}$}

$\sqrt[]{2}\frac{GMM}{{\left(\sqrt[]{2}R\right)}^2}+\frac{GMM}{{\left(2R\right)}^2}=\frac{M{V}^2}{R}$                

$\frac{GM}{R}\left[\frac{1}{\sqrt[]{2}}+\frac{1}{4}\right]=V^2$              

$V=\frac{1}{2}\sqrt[]{\frac{GM}{R}\left(2\sqrt[]{2}+1\right)}$

$\frac{E_A}{E_B}=\frac{-\frac{GM{m}_A}{2\times 3r}}{\frac{-GM{m}_B}{2\times 4r}}=\frac{m_A}{m_B}\times \frac{4}{3}=\frac{4}{3}\times \frac{4}{3}=\frac{16}{9}$

$\mathrm{}{W}_{\text{equator}}=W_{\text{pole}}-m{\omega }^{2}R$

$\begin{array}{rr}& =196-19.6\times {\left(\frac{2\pi }{24\times 60\times 60}\right)}^2\times 6400\times {10}^3\\ & =195.33\mathrm{N}\end{array}$

$g=\frac{9}{{\left(1+\frac{h}{R}\right)}^2}$

$\Rightarrow \frac{g}{3}=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$

$\Rightarrow {\left(1+\frac{h}{R}\right)}^2=3$

1 + $\frac{h}{R}=\sqrt[]{3}$ $\frac{}{}\text{exception 25:}$

$\Rightarrow \frac{h}{R}=\sqrt[]{3}-1$

$\frac{h}{R}=1.732-1$

$\frac{h}{R}=0.732$

h = 0.732 × 6400

= 4684.8 km » 4685 km

F = $\frac{G{m}_2}{d^2}$ - (i)

Now,  $F\text{'}=\frac{G{m}_1{m}_2}{d^2}=\frac{G2m}{3}\times \frac{4m}{3d^2}$

$k\text{'}=\frac{8}{9}G\frac{m^2}{d^2}$

$\therefore F\text{'}=\frac{8}{G}F$