Question

A body is released from a point distance r from the center of earth. If R is the earth r > R, then the velocity of the body at the time of striking the earth will be:

• A. $\sqrt{gR}$
• B. $\sqrt{2gR}$
• C. $\sqrt{\frac{2gR}{r-R}}$
• D. $\sqrt{\frac{2gR(r-R)}{r}}$

Answer 1

The correct option is D $\sqrt{\frac{2gR(r-R)}{r}}$

${\left(PE\right)}_{height}={\left(PE\right)}_{surface}+{\left(KE\right)}_{surface}-\frac{GMm}{r}=-\frac{GMm}{R}+\frac{1}{2}m{v}^2\therefore \frac{1}{2}{v}^2=GM(\frac{1}{R}-\frac{1}{r})\Rightarrow v=\sqrt{\frac{2GM}{R}\left(\frac{r-R}{r}\right)}\frac{R}{R}\Rightarrow \sqrt{\frac{2GMR}{R}\left(\frac{r-R}{r}\right)}\Rightarrow \sqrt{\frac{2gR(r-R)}{r}}$