Motion In One Dimension Question 7
Question 7 - 30 January - Shift 2
An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth. Its velocity when it strikes the earth’s surface, ignoring air resistance, will be :
(1) $2 \sqrt{gR}$
(2) $\sqrt{g R}$
(3) $\sqrt{\frac{gR}{2}}$
(4) $\sqrt{2 gR}$
Show Answer
Answer: (2)
Solution:
Formula: Work-Energy Theorem
Loss in $PE=$ Gain in $KE$
$(-\frac{GMm}{2 R})-(-\frac{GMm}{R})=\frac{1}{2} mv^{2}$
$\Rightarrow v^{2}=\frac{GM}{R}=gR$
$\Rightarrow v=\sqrt{gR}$
