What will the escape speed from planet having radius thrice that of earth and the same mean density as taht of earth
Answers
M = mass of earth
R = radius of earth
V = Volume of earth = 4πR³/3
density is given as
ρ = M/V = M/(4πR³/3)
M = 4πρR³/3 eq-1
escape speed is given as
v = sqrt(2GM/R)
using eq-1
v = sqrt(2G(4πρR³/3)/R)
v = sqrt(8GπρR²/3) eq-2
v₁ =escape speed on earth
v₂ =escape speed on planet
R₁ = radius of earth
R₂ = radius of planet = 3 R₁
using eq-2 , for earth
v₁ = sqrt(8GπρR₁²/3) eq-3
using eq-2 , for planet
v₂ = sqrt(8GπρR₂²/3) eq-4
dividing eq-4 by eq-3
v₂/v₁ = sqrt(8GπρR₂²/3) / sqrt(8GπρR₁²/3)
v₂/v₁ = R₂/R₁
v₂/v₁ = 3 R₁/R₁
v₂ = 3 v₁
hence the escape speed on the planet is thrice that on earth
Answer:
Answer is 33.6km/sec i.e escape Speed of that planet is thrice of the escape speed of earth
Explanation:
just see the uploaded photo
