A spaceship approaches the moon (mass = M,rad | vedclass

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(A) The speed of an object moving in a parabolic path at a distance $R$ from the center of a celestial body is equal to the escape velocity at that po

Vedclass · Accepted Answer

$\sqrt {\frac{{GM}}{R}} \left( {\sqrt 2 - 1} \right)$

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(A) The speed of an object moving in a parabolic path at a distance $R$ from the center of a celestial body is equal to the escape velocity at that point.$V_{p} = V_{e} = \sqrt{\frac{2GM}{R}}$To convert the spaceship into a satellite in a circular orbit of radius $R$,its speed must be reduced to the orbital velocity $V_{0}$ at that distance.$V_{0} = \sqrt{\frac{GM}{R}}$The change in speed required is $\Delta V = V_{p} - V_{0}$.Substituting the values,we get:$\Delta V = \sqrt{\frac{2GM}{R}} - \sqrt{\frac{GM}{R}}$$\Delta V = \sqrt{\frac{GM}{R}} (\sqrt{2} - 1)$

$A$ spaceship approaches the moon (mass $= M$,radius $= R$) along a parabolic path which is almost tangential to its surface. At the moment of maximum approach,the brake rocket is fired to convert the spaceship into a satellite of the moon. Find the change in speed.

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