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Class 9
Chapter 10 Class 9 - Gravitation

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Question 2 Page 138 Why is weight of an object on the moon 1/6 th its weight on the earth? Let mass of an object = m Weight on Earth Let Mass of earth = ME = 5.98 Γ 1024 kg Radius of earth = RE = 6.37 Γ 106 m , Weight on earth = WE = (γπΊπγ_πΈ π)/γπ_πΈγ^2 Weight on Moon Let Mass of moon = MM = 7.36 Γ 1022 kg Radius of moon = RM = 1.74 Γ 106 m Weight on moon = WM = (γπΊπγ_π π)/γπ_πγ^2 Now (ππππβπ‘ ππ ππππ)/(ππππβπ‘ ππ ππππ‘β) = π_π/π_πΈ = (γπΊπγ_π π)/γπ_πγ^2 Γ· (γπΊπγ_πΈ π)/γπ_πΈγ^2 = (γπΊπγ_π π)/γπ_πγ^2 Γ γπ_πΈγ^2/(γπΊπγ_π π) = (π_π Γ γπ_πΈγ^2)/(π_πΈ Γ γπ_πγ^2 ) Putting values = (7.36 Γ γ10γ^22 Γ ( 6.37 Γγ 10γ^6 )^2)/(5.98 Γ γ10γ^(24 )Γ (1.74 Γ γ10γ^6 )^2 ) = (7.36 Γ γ10γ^22 Γ (6.37)^2 Γ γ10γ^12)/(5.98 Γ γ10γ^(24 ) Γ (1.74)^2 Γ γ10γ^12 ) = (7.36 Γ (6.37)^2 )/(5.98 Γ γ10γ^(2 ) Γ (1.74)^2 ) = (7.36 Γ 40.57)/(5.98 Γ 100 Γ 3.02) = 298.59/1805.9 β1/6 Thus, π_π = 1/6 π_πΈ β΄ Weight of an object on the moon is 1/6 of itβs weight on the earth.