If the density of earth Increases by 20% and radius decreases by 20% then the new value of "g" on the surface of earth will be?
Answers
Answered by
1
G = GM / R²
= G × (4/3 × π × R³ × d) / R²
= 4πGdR/3
From above
g ∝ dR
g₁/g₂ = (d₁/d₂) × (R₁/R₂)
g₁/g₂ = (100 / 120) × (100 / 80)
g₁/g₂ = 25/24
g₂ = 24g₁/25
= 0.96 g₁
= 0.96 × 9.8 m/s²
= 9.408 m/s²
Value of g on surface of earth will be 9.408 m/s²
= G × (4/3 × π × R³ × d) / R²
= 4πGdR/3
From above
g ∝ dR
g₁/g₂ = (d₁/d₂) × (R₁/R₂)
g₁/g₂ = (100 / 120) × (100 / 80)
g₁/g₂ = 25/24
g₂ = 24g₁/25
= 0.96 g₁
= 0.96 × 9.8 m/s²
= 9.408 m/s²
Value of g on surface of earth will be 9.408 m/s²
Similar questions