Question

A particle is projected vertically upwards the surface of the earth (radius $R_e$) with a speed equal to one fourth of escape velocity. What is the maximum height attained by it ?

• A. $\frac{16}{15}{R}_e$
• B. $\frac{R_e}{15}$
• C. $\frac{4}{15}{R}_e$
• D. $None$

Answer 1

Answer: (B)

$\frac{R_e}{15}$

From conservation of mechanical energy $\frac{1}{2}m{v}^2=\frac{GMm}{R_e}-\frac{GMm}{R}$

where R=maximum distance from centre of the earth

Also $v=\frac{1}{4}{v}_e=\frac{1}{4}\sqrt{\frac{2GM}{R_e}}$

$\Rightarrow \frac{1}{2}m\times \frac{1}{16}\times \frac{2GM}{R_e}=\frac{GMm}{R_e}-\frac{GMm}{R}$

$\Rightarrow R=\frac{16}{15}{R}_e$

$h=R-R_e=\frac{R_e}{15}$