Q115 / 200 · NEET 2005 · Physics

For a satellite in an orbit around the earth, the ratio of kinetic energy to potential energy is:

  • Correct answer
    A
    $\frac{1}{2}$
  • B
    $\frac{1}{\sqrt{2}}$
  • C
    2
  • D
    $\sqrt{2}$

Solution

$-\frac{\mathrm{GMm}}{R^2}+m \omega^2 R=0$ $$ \begin{aligned} \therefore \quad \frac{\mathrm{GMm}}{R^2} & =m \omega^2 R \\ \text { K.E. }=\frac{1}{2} I \omega^2 & =\frac{1}{2} m R^2 \omega^2 \\ & =\frac{\mathrm{GMm}}{2 R} \end{aligned} $$ $$ \begin{aligned} & P . E .=-\frac{\mathrm{GMm}}{R} \therefore K . E=\frac{|P \cdot E|}{2} \\ & \therefore \quad \frac{K \cdot E}{P \cdot E}=\frac{1}{2} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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