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

Answer: (b). 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 (b) [6G/r(M₁+M₂/2)]1/2.

Earth has mass M₁ and radius R₁. Moon has mass M₂ and radius R₂… | MHT CET 2020 (19 Oct, Shift 2)

Q60MHT CET 2020 · 19 OCT · SHIFT 2Difficult

Escape Velocity

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

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Answer: (b)

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 (b) $\left[\dfrac{6G}{r}\left(M_1+\dfrac{M_2}{2}\right)\right]^{1/2}$ .

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Escape Velocity question from MHT CET 2020 (19 Oct, Shift 2)

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