A body of mass m is released from a height equal to the radius R of the earth. What will be the velocity of the body when it strikes the surface of the earth?

Question

A body of mass m is released from a height equal to the radius R of the earth. What will be the velocity of the body when it strikes the surface of the earth? (A)  √2gR  (B)  √gR  (C)  (gR/√2)  (D)  √3gR

Tardigrade Admin · Accepted Answer

From work-energy theorem, (1/2)mv2=- GMm[ (1/2R)-(1/R) ] ⇒ (1/2)mv2=(- GMm/2R) ⇒ v=√gR ( ∵ g=(GM/R2) )

Q. A body of mass m is released from a height equal to the radius R of the earth. What will be the velocity of the body when it strikes the surface of the earth?

$2 g R$

$g R$

$\frac{g R}{2}$

$3 g R$

From work-energy theorem, $\frac{1}{2} m v^{2} = - GM m [\frac{1}{2 R} - \frac{1}{R}]$ $\Rightarrow$ $\frac{1}{2} m v^{2} = \frac{- GM m}{2 R}$ $\Rightarrow$ $v = g R$ $(∵ g = \frac{GM}{R ^{2}})$

Q. A body of mass m is released from a height equal to the radius R of the earth. What will be the velocity of the body when it strikes the surface of the earth?

2​gR

gR​

2​gR​

3gR​

From work-energy theorem, 21​mv2=−GMm[2R1​−R1​] ⇒ 21​mv2=2R−GMm​ ⇒ v=gR​ (∵g=R2GM​)

$2 g R$

$g R$

$\frac{g R}{2}$

$3 g R$

From work-energy theorem, $\frac{1}{2} m v^{2} = - GM m [\frac{1}{2 R} - \frac{1}{R}]$ $\Rightarrow$ $\frac{1}{2} m v^{2} = \frac{- GM m}{2 R}$ $\Rightarrow$ $v = g R$ $(∵ g = \frac{GM}{R ^{2}})$