Overview

  $\frac{GM}{{\left(3R/2\right)}^2}=\frac{GM}{R^3}\times r$

$\Rightarrow OA=\frac{4R}{9}=r$

$AB=R-\frac{4R}{9}=\frac{5R}{9}\Rightarrow OA:AB=4:5=x:y\Rightarrow x=4$

$T=2\pi \sqrt[]{\frac{r^3}{G{M}_e}}$

$T_B-T_A=\frac{2\pi }{\sqrt[]{G{M}_e}}\left(r_B^{3/2}-r_A^{3/2}\right)=\frac{2\times 3.14}{\sqrt[]{6.67\times 10^{-11}\times 6\times 10^{24}}}\left[{\left(8\times 10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(7\times 10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$F_{1}cos45°+F_{2}cos45°+F_3=\frac{m{v}^2}{r}$

$\Rightarrow \frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(2r\right)}^2}=\frac{m{v}^2}{r}$

$\Rightarrow v=\sqrt[]{\frac{Gm}{4r}\left(2\sqrt[]{2}+1\right)}$

Putting m = 1 kg and r = 1 m,

$v=\frac{1}{2}\sqrt[]{G\left(1+2\sqrt[]{2}\right)}$

According to question, we can write

  $d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2\times 125=500m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

$g\propto r0\le r\le R$

$g\propto \frac{1}{r^2}r>R$

Kindly go through the solution 

$b=a\sqrt[]{1-e^2}$

$\frac{dA}{dt}=\frac{L}{2m}$

$\frac{A}{T}=\frac{L}{2m}$

$T=\frac{2mA}{L}=\frac{2m\pi ab}{L}=\frac{2m\pi a^2\sqrt[]{1-e^2}}{L}$

$V_{in}=\frac{-GM}{2R}\left[3-{\left(\frac{r}{R}\right)}^2\right],$

$V_{surface}=\frac{-GM}{R},V_{out}=\frac{-GM}{r}$

$T=2\pi \sqrt[]{\frac{r^3}{G{M}_e}}$

$T_B-T_A=\frac{2\pi }{\sqrt[]{G{M}_e}}\left(r_B^{3/2}-r_A^{3/2}\right)=\frac{2\times 3.14}{\sqrt[]{6.67\times 10^{-11}\times 6\times 10^{24}}}\left[{\left(8\times 10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(7\times 10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

  $\frac{1}{2}m{V}_a^2-\frac{GMm}{2x}$  

           =   $\frac{1}{2}m\frac{V_a^2}{4}-\frac{GMm}{4x},\boxed{V_a=\sqrt[]{\frac{2}{3}\frac{GM}{x}}}$

ROC at point – B

$ROC=\frac{V_b^2}{a\perp }=\frac{\frac{2GM}{3R}\times \frac{1}{4}}{\frac{GM}{{\left(4R\right)}^2}}=\frac{8R}{3}$