A body is released from height equal to the radius of earth the velocity of the object when it strikes the surface is? 1) ✓gr 2) ✓2gr 3) 2✓gr

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A body is released from height equal to the radius of earth the velocity of the object when it strikes the surface is? 1) ✓gr 2) ✓2gr 3) 2✓gr

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Answer 1 · 2018-06-12T14:22:33

$\sqrt{g R}$ $sqrt(gR)$

Explanation:

The potential energy of a mass $m$ $m$ at a distance $r$ $r$ from the center of the earth (assumed to be a spherically symmetric object of mass $M$ $M$ and radius $R$ $R$ ) is

$V (r) = - \frac{G M m}{r}$ $V(r) = -(GMm)/r$

Initially the mass is at rest at $r = 2 R$ $r=2R$ . So, its total energy is

$V (2 R) = - \frac{G M m}{2 R}$ $V(2R) = -(GMm)/(2R)$

Thus, from the law of conservation of energy, its speed $v$ $v$ when it hits the earth (at $r = R$ $r=R$ ) is given by

$\frac{1}{2} m v^{2} - \frac{G M m}{R} = - \frac{G M m}{2 R} \Rightarrow$ $1/2 mv^2 -(GMm)/R = -(GMm)/(2R) implies$

$\frac{1}{2} m v^{2} = G M m (\frac{1}{R} - \frac{1}{2 R}) = \frac{G M m}{2 R} \Rightarrow$ $1/2 mv^2 = GMm(1/R-1/(2R)) = (GMm)/(2R) implies$

$v^{2} = G \frac{M}{R}$ $v^2 = GM/R$

Now, the force on the mass at the surface of the earth is given by

$m g = \frac{G M m}{R^{2}} \Rightarrow g = \frac{G M}{R^{2}}$ $mg = (GMm)/R^2 implies g = (GM)/R^2$

Thus

$v^{2} = g R \Rightarrow v = \sqrt{g R}$ $v^2 = gR implies v = sqrt(gR)$

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A body is released from height equal to the radius of earth the velocity of the object when it strikes the surface is? 1) ✓gr 2) ✓2gr 3) 2✓gr

1 Answer

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