Earth has mass ' M₁ ' radius ' R₁ ' and for moon mass ' M₂ ' and radius ' R₂ '. Distance between their centres is ' r '. A body of mass ' M ' is placed on the line joining them at a distance r/3 from the centre of the earth. To project a mass ' M ' to escape to infinity, the minimum speed required is (a) [2G/r(M₂+M₁/2)]1/2 (b) [4G/r(M₁+M₂/2)]1/2 (c) [3G/r(M₁+M₂)]1/2 (d) [6G/r(M₁+M₂/2)]1/2

Answer: (d). Point is r/3 from M₁ and 2r/3 from M₂. V = -(3GM₁)/(r) - (3GM₂)/(2r) = -3G/r(M₁ + M₂/2) v² = 2|V| = 6G/r(M₁ + M₂/2) So the answer is (d) [6G/r(M₁+M₂/2)]1/2.

Earth has mass ' M₁ ' radius ' R₁ ' and for moon mass ' M₂ ' and… | MHT CET 2024 (3 May, Shift 1)

Q24MHT CET 2024 · 3 MAY · SHIFT 1Difficult

Escape Velocity

Earth has mass ' $M_1$ ' radius ' $R_1$ ' and for moon mass ' $M_2$ ' and radius ' $R_2$ '. Distance between their centres is ' $r$ '. A body of mass ' $M$ ' is placed on the line joining them at a distance $\frac{r}{3}$ from the centre of the earth. To project a mass ' $M$ ' to escape to infinity, the minimum speed required is

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Answer: (d)

Point is $\dfrac{r}{3}$ from $M_1$ and $\dfrac{2r}{3}$ from $M_2$ .
$V = -\dfrac{3GM_1}{r} - \dfrac{3GM_2}{2r} = -\dfrac{3G}{r}\left(M_1 + \dfrac{M_2}{2}\right)$
$v^2 = 2|V| = \dfrac{6G}{r}\left(M_1 + \dfrac{M_2}{2}\right)$
So the answer is (d) $\left[\dfrac{6G}{r}\left(M_1+\dfrac{M_2}{2}\right)\right]^{1/2}$ .

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Escape Velocity question from MHT CET 2024 (3 May, Shift 1)

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