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

Question

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 (A) √g R (B) 2 √g R (C) √2 g R (D) √(g R/2)

Tardigrade Admin · Accepted Answer

Loss in PE = Gain in KE (-( GMm /2 R ))-(-( GMm / R ))=(1/2) mv 2 ⇒ v 2=( GM / R )= gR ⇒ v =√ gR

Q. 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

$g R$

$2 g R$

$2 g R$

$\frac{g R}{2}$

Loss in PE = Gain in KE
$(- \frac{GM m}{2 R}) - (- \frac{GM m}{R}) = \frac{1}{2} m v^{2}$
$\Rightarrow v^{2} = \frac{GM}{R} = g R$
$\Rightarrow v = g R$

Q. 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

gR​

2gR​

2gR​

2gR​​

Loss in PE = Gain in KE (−2RGMm​)−(−RGMm​)=21​mv2 ⇒v2=RGM​=gR ⇒v=gR​

Q. 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

$g R$

$2 g R$

$2 g R$

$\frac{g R}{2}$

Loss in PE = Gain in KE
$(- \frac{GM m}{2 R}) - (- \frac{GM m}{R}) = \frac{1}{2} m v^{2}$
$\Rightarrow v^{2} = \frac{GM}{R} = g R$
$\Rightarrow v = g R$