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)
Show Answer
Answer:
Correct Answer: 11. (c)
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
$$ \begin{aligned} \therefore \quad \frac{G M _e m}{2 R} & =\frac{-G M _e m}{(R+h)}-\frac{-G M _e m}{R} \\ \frac{1}{2 R} & =\frac{1}{R}-\frac{1}{R+h} \\ \frac{1}{2 R} & =\frac{1}{R+h} \\ h+R & =2 R \\ h & =R \end{aligned} $$
