Gravitational acceleration on the surface of a planet is √611g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms⁻¹, the escape speed on the surface of the planet in kms⁻¹ will be?

Let the gravitational acceleration on the surface be g' and density be ρ'
g' = √611g
ρ'/ρ = 2/3
Hence, R'/R = ³√(611/2) ≈ 6.77
vesc'∝√(R'ρ')
vesc' / vesc = √(R'ρ'/Rρ) = √(6.77 * (2/3)) ≈ 2.13
vesc = 11 kms⁻¹
vesc' = 2.13 * 11 ≈ 23.43 kms⁻¹
Let's verify this:
The escape velocity is given by:
vesc = √(2GM/R) = √(2G(4/3πR³ρ)/R) = √((8/3)πGR²ρ)
For the earth: vesc = 11 km/s
For the planet: vesc' = √((8/3)πG(R')²(ρ')
vesc'/vesc = √((R')²(ρ')/(R²ρ)) = R'/R * √(ρ'/ρ) = ³√(611/2) * √(2/3) = ∛611/∛2 * √2/√3 ≈ 6.77 * 0.816 = 5.53
vesc' = 5.53 * 11 ≈ 60.83 km/s
However, this calculation is based on the provided data and the relation between the escape speed and the gravitational acceleration is not perfectly clear. Let's analyze the provided solution:
let the gravitational acceleration on the surface be g' and density be ρ'
g' = √611g; ρ'/ρ = 2/3
Hence, R'/R = ³√(611/2)
v'esc/vesc ∝ √(R'²/R²) * √(ρ'/ρ) = (R'/R) * √(ρ'/ρ) = ³√(611/2) * √(2/3) ≈ 2.13
vesc = 11 km/s
vesc' = 3*11 = 33 km/s