Gravitational acceleration on the surface of a planet is , where is the gravitational acceleration on the surface of the Earth. The average mass density of the planet is times that of the Earth. If the escape speed on the surface of the Earth is taken t

Question

Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11} g$ , where $g$ is the gravitational acceleration on the surface of the Earth. The average mass density of the planet is $\frac{2}{3}$ times that of the Earth. If the escape speed on the surface of the Earth is taken to be $11 {km s}^{- 1}$ , the escape speed on the surface of the planet in ${km s}^{- 1}$ will be [2010]

Smart Achievers · Accepted Answer

(3)

We know that escape speed $v = \sqrt{2 g R}$

$∴$ $\frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{p}}{R}} \dots (i)$

$Given \frac{g_{p}}{g_{e}} = \frac{\sqrt{6}}{11} \dots (i i)$

$Also g = \frac{4}{3} π G ρ R$ $∴$ $\frac{g_{p}}{g_{e}} = \frac{ρ_{p}}{ρ} \times \frac{R_{p}}{R}$

$∴$ $\frac{\sqrt{6}}{11} = \frac{2}{3} \times \frac{R_{p}}{R}$ $[∵ \frac{ρ_{p}}{ρ} = \frac{2}{3} (given)]$

$∴$ $\frac{R_{p}}{R} = \frac{3 \sqrt{6}}{22} \dots (i i i)$

$From (i), (ii) & (iii)$

$\frac{v_{p}}{v_{e}} = \sqrt{\frac{\sqrt{6}}{11} \times \frac{3 \sqrt{6}}{22}} = \sqrt{\frac{3 \times 6}{11 \times 22}} = \frac{3}{11}$

$∴$ $v_{p} = \frac{3}{11} \times v_{e} = \frac{3}{11} \times 11 km/s = 3 km/s$

Solution

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Solution

Ans. (3) We know that escape speed v=2gR ∴ vpve=gpge×RpR ⋯(i) Given gpge=611 ⋯(ii) Also g=43πGρR ∴ gpge=ρpρ×RpR ∴ 611=23×RpR [∵ρpρ=23 (given)] ∴ RpR=3622 ⋯(iii) From (i), (ii) & (iii) vpve=611×3622=3×611×22=311 ∴ vp=311×ve=311×11 km/s=3 km/s

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