Acceleration due to gravity on the moon is - 1 - 6 -th of the acceleration due to gravity on the earth- If the ratio of densities of the earth and the moon is - 5 - 3 - then radius on the moon in terms of radius of earth will be

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We know g-4-x3C0-G-x3C1-r3-K-x3C1-rGiven- gm-16ge-x21D2-xA0-gmge-x3C1-mrm-x3C1-ere-16Also-xA0-x3C1-m-35-x3C1-eSo-xA0- 3rm5re-16-x21D2-rm-5re18-xA0-

Acceleration due to gravity on the moon is $1 / 6 t h$ of the acceleration due to gravity on the earth. If the ratio of densities of the earth and the moon is $5 / 3$ , then radius on the moon in terms of radius of earth will be

We know $g = \frac{4 π G ρ r}{3} = K ρ r$

Given, $g_{m} = \frac{1}{6} g_{e}$

$\Rightarrow$ $\frac{g_{m}}{g_{e}} = \frac{ρ_{m} r_{m}}{ρ_{e} r_{e}} = \frac{1}{6}$

Also $ρ_{m} = \frac{3}{5} ρ_{e}$

So, $\frac{3 r_{m}}{5 r_{e}} = \frac{1}{6}$

$\Rightarrow r_{m} = \frac{5 r_{e}}{18}$

Acceleration due to gravity on the moon is 1/6th of the acceleration due to gravity on the earth. If the ratio of densities of the earth and the moon is 5/3, then radius on the moon in terms of radius of earth will be

We know g=4πGρr3=KρrGiven, gm=16ge⇒ gmge=ρmrmρere=16Also ρm=35ρeSo, 3rm5re=16⇒rm=5re18

Acceleration due to gravity on the moon is $1 / 6 t h$ of the acceleration due to gravity on the earth. If the ratio of densities of the earth and the moon is $5 / 3$ , then radius on the moon in terms of radius of earth will be

We know $g = \frac{4 π G ρ r}{3} = K ρ r$

Given, $g_{m} = \frac{1}{6} g_{e}$

$\Rightarrow$ $\frac{g_{m}}{g_{e}} = \frac{ρ_{m} r_{m}}{ρ_{e} r_{e}} = \frac{1}{6}$

Also $ρ_{m} = \frac{3}{5} ρ_{e}$

So, $\frac{3 r_{m}}{5 r_{e}} = \frac{1}{6}$

$\Rightarrow r_{m} = \frac{5 r_{e}}{18}$