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The free-fall acceleration on the surface of the Moon is about one sixth that on the surface of the Earth. The radius of the Moon is about 0.250 RE (RE = Earth's radius = 6.4 106 m). Find the ratio of their average densities, rhoMoon/rhoEarth.

Respuesta :

On the surface of the Moon, the free-fall acceleration is roughly one-sixth that of the Earth. Their average densities are 0.667 apart.

Given the radius of the Moon is (Rm) = 0.250 RE

The radius of the earth (RE) = 6.4 x 10^6 m

Let freefall acceleration of the moon = gm

Let freefall acceleration of the earth = ge

Given gm = ge/6

We know that g = Gm/r^2 where G is the gravitational constant

Let mass of moon = m1 and mass of earth = m2

Gm1/Rm^2 = Gm2/6xRE^2

mass = densityxvolume

let density of moon = d1 and density of earth = d2

d1 x (4/3Rm^3)/Rm^2  =  d2x (4/3RE^3)/6xRE^2

d1xRm = d2xRE/6

d1/d2 = RE/6X0.250RE

d1/d2 = 0.667

Hence the ratio of densities of moon and earth is 0.667

To learn more about freefall acceleration click here https://brainly.com/question/28863645

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