Physics Gravitation

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

Time period of Satellite = T = $\frac{2\pi R}{v}={\left(\frac{4{\pi }^2{R}^3}{GM}\right)}^{\frac{1}{2}}$  

$\Rightarrow T\alpha R^{\frac{3}{2}}\Rightarrow \frac{T_2}{T_1}=\frac{{\left(1.02R\right)}^{\frac{3}{2}}}{R^{\frac{3}{2}}}={\left(1.02\right)}^{\frac{3}{2}}={\left(1+0.02\right)}^{\frac{3}{2}}$     

The percentage difference in the time periods of the two satellites = $\frac{\Delta T}{T_1}$ × 100

= (0.03) × 100 = 3%

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

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

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

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

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

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

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

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

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

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

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

Acceleration due to gravity at r distance above the surface = $\frac{GM}{{\left(R+r\right)}^2}$  

Acceleration due to gravity at r distance below the surface = $\frac{GM}{R^3}\left(R-r\right)$

So, ratio = $\frac{\left(R-r\right){\left(R+r\right)}^2}{R^3}=\frac{\left(R-r\right)\left(R^2+r^2+2Rr\right)}{R^3}=1+\frac{r}{R}-\frac{r^2}{R^2}-\frac{r^3}{R^3}$

at point P,

$=K{E}_P+P{E}_P$

$=\frac{1}{2}m{v}_P^2+\left\{\frac{-G{M}_{1}m}{\left(r/2\right)}-\frac{G{M}_{2}m}{r/2}\right\}$

at infinity (ie for escaping from both masses)

$\Rightarrow v_P=\sqrt[]{\frac{4G\left(M_1+M_2\right)}{r}}$

$U=-\frac{Gm\left(M-m\right)}{d}\times 4+2x-\left(\frac{G{m}^2}{\sqrt[]{2}d}\right)$

For Umax

$\frac{dU}{dm}=0\Rightarrow m=\frac{M}{2}\Rightarrow \frac{M}{m}=2$

$V_A=-\frac{G{M}_1}{r}-\frac{G{M}_2}{R}=-\frac{50G}{25}-\frac{100G}{50}=-4G$

Inside a uniform spherical shell, electric field is zero every where & electric potential is constant but not zero, every where