JEE PYQ: Gravitation Question 15
Question 15 - 2021 (25 Feb Shift 2)
The initial velocity $v_i$ required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity $v_e$ such that $v_i = \sqrt{\frac{x}{y}} \times v_e$. The value of $x$ will be:
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Answer: 10
Solution
$\frac{1}{2}mv_i^2 - \frac{GMm}{R} = -\frac{GMm}{11R}$. $\frac{1}{2}v_i^2 = \frac{GM}{R}\left(1 - \frac{1}{11}\right) = \frac{10GM}{11R} = \frac{10}{11} \times \frac{v_e^2}{2}$. So $v_i = \sqrt{\frac{10}{11}}v_e$. Hence $x = 10$.
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