A mass of 50 kg is placed at the centre of a uniform spherical shell of mass 100kg and radius 50m. If the gravitational potential at a point, 25m from the centre is V kg/m. The value of V is:

For A satellite T₁ = 1hour

$So{\omega }_1=2\pi red/hour$                

for B satellite T₂ = 8 hour

given R₁ = 2 × 10³ Km

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

$\Rightarrow \frac{\pi }{3}$ rad/hour

 x = 3

For circular motion F_{net   $=\frac{M{V}^2}{R}$}

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

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

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

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

M₁ ->Mass of satellite (1)

M₂ -> Mass of satellite (2)

M_P -> Mass of planet    

Now NLM (2) on S₁

_{$\frac{G{M}_1{M}_P}{R_1^2}=\frac{m_1{v}_1^2}{R_1}$}

Similarly v₂ =   $\sqrt[]{\frac{G{M}_P}{R_2}}$

$\frac{v_1}{v_2}=\sqrt[]{\frac{R_2}{R_1}}=\sqrt[]{\frac{800}{3200}}=\frac{1}{2}=\frac{1}{x}$