Question

A satellite is projected with velocity $ \sqrt{\dfrac{5}{6}} $ times its escape speed from earth’s surface. The initial velocity of the satellite and parallel to the surface of earth. The maximum distance of the satellite from the centre of earth will be:

Answer 1

Hint :By using angular momentum $ \rho $ energy conservations formulas:
(1) Angular momentum conservation $ \text{L= mVr} $
(2) Energy conservation: kinetic energy $ + $ potential energy = constant

Complete Step By Step Answer:
We know that,
Angular momentum of conservation
 $ {{\text{m}}_{1}}{{\text{V}}_{1}}{{\text{r}}_{1}}\,=\,{{\text{m}}_{2}}{{\text{V}}_{2}}{{\text{r}}_{2}}\, $
Here, $ {{\text{m}}_{1}}\text{=}\,{{\text{m}}_{2}}=\text{m} $
Initial velocity $ {{\text{V}}_{1}}=\sqrt{\dfrac{5}{6}}\,{{\text{V}}_{e}} $ where $ {{\text{V}}_{e}} $ is escape velocity
 $ {{\text{V}}_{2}}= $ Final velocity/maximum velocity
 $ \text{m}\sqrt{\dfrac{5}{6}}\,{{\text{V}}_{e}}\,\text{R}\,\text{=}\,\text{m}\,{{\text{V}}_{2}}\,\text{d }...........\text{(1)} $
By energy conservation
(Kinetic energy $ + $ potential energy), initial $ \text{= } $ (KE $ + $ PE), final
 $ \dfrac{1}{2}\text{m}{{\left( \sqrt{\dfrac{5}{6}}\,{{\text{V}}_{e}} \right)}^{2}}\text{+}\left( \dfrac{-\text{GMm}}{{{\text{R}}_{e}}} \right)=\dfrac{1}{2}\text{m}{{\text{V}}_{2}}^{2}\text{+}\left( \dfrac{-\text{GMm}}{\text{d}} \right).........(2)\, $
From equation $ \text{(1)} $ and $ (2) $
 $ \text{D=5R} $
Hence, option (c) is correct answer.

Additional Information:
Once the escape velocity is reached the rocket moves in circular orbit with increasing radii and finally attains a hyperbola curve and goes into the space with infinity speed and never comes back.

Note :
Escape velocity is the speed required to escape gravity. To move out of the earth’s gravitational pull we need to reach escape velocity otherwise we will be pulled back.