The time period of revolution fo moon around the earth is 28 days and radius of its orbit is 4×10^5 km.If G = 6.67 ×10^-11 Nm^2/Kg then find the mass fo earth.
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# Answer- 6.46×10^24 kg
# Given-
T = 28 days = 2419200 s
r = 4×10^5 km = 4×10^8 m
G = 6.67×10^-11 Nm^2/kg
# Formula-
T = 2π√(r^3/GM)
T^2 = 4π^2×r^3/GM
M = 4π^2×r^3/GT^2
M = 4×3.14^2 × (4×10^8)^3 / (6.67×10^-11×2419200^2)
Solving this you will get
M = 6.46×10^24 kg
Hence mass of earth is 6.46×10^24 kg.
# Given-
T = 28 days = 2419200 s
r = 4×10^5 km = 4×10^8 m
G = 6.67×10^-11 Nm^2/kg
# Formula-
T = 2π√(r^3/GM)
T^2 = 4π^2×r^3/GM
M = 4π^2×r^3/GT^2
M = 4×3.14^2 × (4×10^8)^3 / (6.67×10^-11×2419200^2)
Solving this you will get
M = 6.46×10^24 kg
Hence mass of earth is 6.46×10^24 kg.
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