You have been visiting a distant planet. your measurements have determined that the planet's mass is three times that of earth but the free-fall acceleration at the surface is only one-fourth as large.

Question

Jan Wiley

Physics

13 January, 03:34

You have been visiting a distant planet. your measurements have determined that the planet's mass is three times that of earth but the free-fall acceleration at the surface is only one-fourth as large.

+1

Answers (1)

Know the Answer?

Victor Dunlap · Answer 1

The free-fall acceleration is proportional to the mass of the planet and inversely proportional to the square of the radius of the planet.

g = (G*m) * M / (RE^2), where G and m are constants. G is Universal Gravitational Constant and m is your mass, which is the same everywhere.

Then, if ME is the mass of the planet Earth, RE is the radius of planet Earth, MP is the mass of the distant planet and RE is the radius of the distant planet you can write:

g on Earth / g on distant planet = [ME / (RE^2) ] / [MP / (RP^2) ] = 4

And MP = 3ME

Then you can obtain the ratio of RP/RE

[ME / (RE^2) ] / [3ME / (RP) ^2 ] = 4

=> (RP) ^2 / (RE) ^2 = 4*3 = 12

=> (RP) ^2 = 12 * (RE) ^2

=> RP = (√12) RE

This is the radius of the distant planet is √12 or about 3.46 times the radius of the Earth.