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7. gravitation — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics7. gravitation1 Mark
MCQ
A particle is projected from the mid-point of the line joining two fixed particles each of mass $m.$ If the distance of separation between the fixed particles is $l,$ the minimum velocity of projection of the particle so as to escape is equal to
  • A
    $\sqrt {\frac{{GM}}{l}} $
  • B
    $\sqrt {\frac{{GM}}{2l}} $
  • C
    $\sqrt {\frac{{2GM}}{l}} $
  • ✓
    $2 \, \sqrt {\frac{{2GM}}{l}} $
✓

Answer

Correct option: D.
$2 \, \sqrt {\frac{{2GM}}{l}} $
d
The gravitational potential at the mid-point $P$

$V=V_{1}+V_{2}=\frac{-G m}{(l / 2)}-\frac{G m}{(l / 2)}=-\frac{4 G m}{l}$

$\Rightarrow$ The gravitational potential energy

$=U=-\frac{4 G m m_{0}}{l}, m_{0}=$ mass of particle

When it is projected with a speed $v$, it just escapes to infinity, and the potential and kinetic energy will become zero.

$\Rightarrow \triangle K E+\triangle P E=0$

$\Rightarrow\left(0-\frac{1}{2} m_{0} v^{2}\right)+\left\{0-\left(-\frac{4 G m m_{0}}{l}\right)\right\}=0 \Rightarrow v=2 \sqrt{\frac{2 G m}{l}}$

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