A ball 'A' of mass m falls to the surface of the earth from infinity. Another ball 'B' of mass 2m falls to the earth from the height equal to six times radius of the earth then ratio of velocities of 'A' and 'B' on reaching the earth is

\sqrt{(} 6 / 5)

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\sqrt{(} 5 / 6)

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\sqrt{(} 7 / 6)

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Solution

The correct option is C $\sqrt{(} 7 / 6)$
$\begin{matrix} v_{A} = \sqrt{\frac{2 G M}{R}} (e s c a p e v e l o c i t y) v_{B} = \sqrt{\frac{2 g h}{1 + \frac{h}{R}}} = \sqrt{\frac{2 g h}{7}} (h = 6 R) v_{B} = \sqrt{\frac{2 g h 6 R}{7}} = \sqrt{\frac{12 G M}{7 R}} \frac{v_{B}}{v_{A}} = \sqrt{\frac{2 G M}{h}} = \sqrt{\frac{7 R}{12 G M}} = \sqrt{\frac{7}{6}} H e n c e, \frac{v_{_{B}}}{v_{A}} = \sqrt{\frac{7}{6}} ∴ o p t i o n D i s c o r r e c t a n s w e r . \end{matrix}$

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A ball 'A' of mass m falls to the surface of the earth from infinity. Another ball 'B' of mass 2m falls to the earth from the height equal to six times radius of the earth then ratio of velocities of 'A' and 'B' on reaching the earth is

The correct option is C √(7/6)vA=√2GMR(escapevelocity)vB= ⎷2gh1+hR=√2gh7(h=6R)vB=√2gh6R7=√12GM7RvBvA=√2GMh=√7R12GM=√76Hence,vBvA=√76∴optionDiscorrectanswer.