A body A of mass m is moving in a circular orbit of radius R about a planet. Another body B of mass ${m \over 2}$ collides with A with a velocity which is half $\left( {{{\overrightarrow v } \over 2}} \right)$ the instantaneous velocity${\overrightarrow v }$ of A. The collision is completely inelastic. Then, the combined body :
Solution
Orbital speed for of A is v = $\sqrt {{{GM} \over R}}$
<br><br>After collision, let the combined mass moves
with speed v<sub>1</sub>
<br><br>$\therefore$ mv + ${m \over 2}{v \over 2}$ = $\left( {{{3m} \over 2}} \right){v_1}$
<br><br>$\Rightarrow$ v<sub>1</sub> = ${{5v} \over 6}$
<br><br>Since after collision, the speed is not equal to
orbital speed at that point. So motion cannot be
circular. Since velocity will remain tangential,
so it cannot fall vertically towards the planet.
Their speed after collision is less than escape
speed $\sqrt 2 v$, so they cannot escape
gravitational field.
So their motion will be elliptical around the
planet.
About this question
Subject: Physics · Chapter: Gravitation · Topic: Satellites and Orbital Motion
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