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The gravitational acceleration on the surface of a planet is (1/11) of the gravitational acceleration g on Earth's surface. The average mass density of this planet is (1/4) times the average mass density of Earth. If the escape speed from Earth's surface is 11 km/s, what is the escape speed from the surface of this planet in km/s?
- 3
- 2
- 4
- 1
Correct answer: 3
Solution
From the density and gravity relations: g = (4/3)*pi*G*rho*R, so gₚ/gₑ = (rhoₚ/rhoₑ)*(Rₚ/Rₑ), giving Rₚ/Rₑ = (gₚ/gₑ)*(rhoₑ/rhoₚ) = (1/11)*4 = 4/11. Escape speed vₚ = sqrt(2*gₚ*Rₚ) = sqrt(2*(g/11)*(4Rₑ/11)) = sqrt(8gRₑ/121) = (1/11)*sqrt(8gRₑ/2... Let me recompute: vₑ = sqrt(2gRₑ) = 11 km/s; vₚ = sqrt(2*gₚ*Rₚ) = sqrt(2*(g/11)*(4Rₑ/11)) = sqrt(8gRₑ/121) = sqrt(8/121)*sqrt(gRₑ) = (2*sqrt(2)/11)*sqrt(gRₑ) = (2*sqrt(2)/11)*(11/sqrt(2)) km/s = 2*sqrt(2)/sqrt(2) =... vₑ/sqrt(2) *... let me be precise.
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