T₁ = 1 hour

$$\Rightarrow$$ $$\omega$$₁ = 2$$\pi$$ rad/hour

T₂ = 8 hours

$$\Rightarrow$$ $$\omega$$₂ = $${\pi \over 4}$$ rad/hour

R₁ = 2 $$\times$$ 10³ km

As T² $$\propto$$ R³

$$ \Rightarrow {\left( {{{{R_2}} \over {{R_1}}}} \right)^3} = {\left( {{{{T_2}} \over {{T_1}}}} \right)^2}$$

$$ \Rightarrow {{{R_2}} \over {{R_1}}} = {\left( {{8 \over 1}} \right)^{2/3}} = 4 \Rightarrow {R_2} = 8 \times {10^3}$$ km


V₁ = $$\omega$$₁R₁ = 4$$\pi$$ $$\times$$ 10³ km/h

V₂ = $$\omega$$₂R₂ = 2$$\pi$$ $$\times$$ 10³ km/h

Relative $$\omega$$ = $${{{V_1} - {V_2}} \over {{R_2} - {R_1}}} = {{2\pi \times {{10}^3}} \over {6 \times {{10}^3}}}$$

$$ = {\pi \over 3}$$ rad/hour

x = 3

Energy is maximum when mass is split equally so $${M \over m} = 2$$

Acceleration due to gravity will be zero at P therefore,

$${1 \over 2}m{v^2} - {{GMm} \over R} - {{G9Mm} \over {7R}} = 0 - {{GMm} \over {2R}} - {{G9Mm} \over {6R}}$$

$$\therefore$$ $$V = \sqrt {{4 \over 7}{{GM} \over R}} $$

$${V_{es}} = \sqrt {{{2GM} \over R}} $$

$${V_{es}}\sqrt R $$ = const

$$ \therefore $$ $${V_{es}}.\sqrt R = 10{V_{es}}\sqrt {R'} $$

$$ \Rightarrow $$ $$R' = {R \over {100}}$$ = 64 km