R and r are the radii of earth and moon pe and pm are the densities of rhe earth and moon the ratio of accelearation due to gravity on the surface of the earth to the moon is:
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gmge=16
=Re2MmRm2Me
=Re243πR3mρR2m43πR3eρ
=RmRe
=16
Hence a is the correct answer.
Answered by
5
Hello there,
● Answer-
g/g' = (ρR) / (ρ'R')
● Explanation-
Suppose,
M = mass of earth
M' = mass of moon
ρ = density of earth
ρ' = density of moon
g = gravity at earth surface
g' = gravity at moon surface
Gravitational acceleration on earth surface is-
g = GM/R^2
g = Gρ(4/3)πR^3 / R^2
g = (4/3) GρπR
Similarly, gravitational acceleration on moon is given by-
g' = (4/3) Gρ'πR'
Ratio of respective gravity -
g/g' = (4/3)GρπR / (4/3)Gρ'πR'
g/g' = (ρR) / (ρ'R')
Therefore, ratio of accelearation due to gravity on the surface of the earth to the moon is (ρR)/(ρ'R').
Hope this helps...
● Answer-
g/g' = (ρR) / (ρ'R')
● Explanation-
Suppose,
M = mass of earth
M' = mass of moon
ρ = density of earth
ρ' = density of moon
g = gravity at earth surface
g' = gravity at moon surface
Gravitational acceleration on earth surface is-
g = GM/R^2
g = Gρ(4/3)πR^3 / R^2
g = (4/3) GρπR
Similarly, gravitational acceleration on moon is given by-
g' = (4/3) Gρ'πR'
Ratio of respective gravity -
g/g' = (4/3)GρπR / (4/3)Gρ'πR'
g/g' = (ρR) / (ρ'R')
Therefore, ratio of accelearation due to gravity on the surface of the earth to the moon is (ρR)/(ρ'R').
Hope this helps...
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