Overview

$U_i=-\frac{GMm}{R}$

${\left(T.E.\right)}_f=-\frac{GMm}{8R}$

$\Rightarrow Energy={\left(T.E\right)}_f-\left(U_i\right)$

$=\frac{7GMm}{8R}$

Given $F\propto \frac{1}{r^n}$

$m{\omega }^{2}R=\frac{K}{R^n}$

$?\frac{m.4{\pi }^{2}R}{T^2}=\frac{K}{R^n}$

$?T^2\propto R^{n+1}$

$T\propto R^{\frac{n+1}{2}}$

$F=mr{\omega }^2=\frac{k}{r}$

$\begin{array}{l}\Rightarrow \omega \propto \frac{1}{r}\\ \Rightarrow T\propto r\end{array}$

Inside a spherical shell gravitational field is zero, so net force acting on a particle placed inside it is also zero.

Total energy of a bound system is always negative.

Since the water removed from the pond will be equal to the water displacement by the (man + boat) system, the level of water in the pond remains the same.

Gravitational force does not depend on the surrounding medium.

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

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

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

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

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

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

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