Gravitation 2 Question 8
11. A particle is projected vertically upwards from the surface of Earth (radius $R$ ) with a kinetic energy equal to half of the minimum value needed for it to escape. The height to which it rises above the surface of Earth is
(1997, 2M)
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Answer:
Correct Answer: 11. h = R
Solution:
- Kinetic energy needed to escape $=\frac{G M_{e} m}{R}$
Therefore, energy given to the particle $=\frac{G M_{e} m}{2 R}$
Now, from conservation of mechanical energy. Kinetic energy at the surface of earth $=$ Difference in potential energy at a height $h$ and on the surface of earth
$$ \begin{aligned} \therefore \quad \frac{G M_{e} m}{2 R} & =\frac{-G M_{e} m}{(R+h)}-\frac{-G M_{e} m}{R} \\ \frac{1}{2 R} & =\frac{1}{R}-\frac{1}{R+h} \\ \frac{1}{2 R} & =\frac{1}{R+h} \\ h+R & =2 R \\ h & =R \end{aligned} $$
