Gravitation

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

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

$=3.92 m s^{-2}$

Using conservation of energy :

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

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

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

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

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

W₁ = 1000 N ->Weight of person at surface of earth W₂ = 400 N Þ Weight of person at surface of Mars. There will be a neutral point -N where weight will be zero because at this point net gravitational field due to earth and mars is zero.

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

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

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

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

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   $=\frac{M{V}^2}{R}$}

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

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

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

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

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

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

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

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

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

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

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

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

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