Overview

$90.5\mathrm{J}$
$$

$\frac{-{\mathrm{G}\mathrm{M}}_{\mathrm{e}}\mathrm{M}}{\mathrm{R}}+\frac{1}{2}{\mathrm{M}\mathrm{u}}^2=\frac{-{\mathrm{G}\mathrm{M}}_{\mathrm{e}}\mathrm{M}}{2\mathrm{R}}+\frac{1}{2}{\mathrm{M}\mathrm{v}}^2$

$\text{exception 25:}$  $v=\sqrt[]{u^2-\frac{G{M}_e}{R}}$ Transverse velocity of rocket

${}_{}$ $V_T\to$  Radial velocity of rocket

$V_R\to$

$V-\sqrt[]{\frac{G{M}_e}{2R}}$

Kinetic energy $\frac{M}{10}{V}_T=\frac{9M}{10}\sqrt[]{\frac{G{M}_e}{2R}}\mathrm{};\frac{M}{10}{V}_r=M\sqrt[]{u^2-\frac{G{M}_e}{R}}$ $\frac{}{}{}_{}\frac{}{}\text{exception 25:}\mathrm{}\frac{}{}{}_{}\text{exception 25:}$

$=\frac{1}{2}\frac{M}{10}\left(V_T^2+V_r^2\right)=\frac{M}{20}\left(81\frac{G{M}_e}{2R}+100u^2-100\frac{G{M}_e}{R}\right)$

$=\frac{\mathrm{M}}{20}\left(100{\mathrm{u}}^2-\frac{119\mathrm{G}\mathrm{M}}{2\mathrm{R}}\right)$

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

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

Kindly consider the following Image

F = Gm²/ (2R)² = mω²R Given m = 1kg
⇒ ω² = Gm/ (4R³)
⇒ ω = (1/2)√ (G/R³)

T² = (4π²r³)/GM
⇒ M = (4π²r³)/ (G T²)
⇒ M = 6 × 10¹¹ × (9 × 10? )³ / (450 × 60)²
= 6.48 x 10²³ kg
(Note: There is a calculation error in the source image. Using the formula and given values: M = (6e11 * (9e6)^3) / (27000)^2 = 6e23 kg)

We have to find the point where the gravitational field must be zero.
EG = 0
GM/x² = G (9M)/ (8R-x)²
1/x² = 9/ (8R-x)²
8R - x = 3x => x = 2R
Potential at A (surface of first planet), VA = -GM/R - G (9M)/7R = -16GM/7R
Potential at point x, Vx = -GM/x - G (9M)/ (8R-x) = -GM/2R - G (9M)/6R = -2GM/R
ΔV = Vx -

g = (4/3)πRρG . (i) (ρ = density)
Now ρ = M/V = M/ (4/3)πR³)
R³ = M/ (4/3)πρ) => R ∝ (M/ρ)¹/³
From equation (i)
g ∝ Rρ ∝ (M/ρ)¹/³ρ = M¹/³ρ²/³
For planet, M' = 2M, ρ' = ρ
g'/g = (M'/M)¹/³ (ρ'/ρ)²/³ = (2)¹/³ (1)²/³ = 2¹/³
W' = mg' = m (2¹/³g) = 2¹/³ (mg) = 2¹/³W

$m{\omega }^2.\frac{2d}{3}=\frac{G.m.2m}{d^2}\Rightarrow \omega =\sqrt[]{\frac{3Gm}{d^3}}$  

$T=\frac{2\pi }{\omega }=2\pi \sqrt[]{\frac{d^3}{3Gm}}$

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

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

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

Putting m = 1 kg and r = 1 m,

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