Question

A newly discovered planet has a radius twice as large as earth’s and a mass five times as large. What is the free-fall acceleration on its surface?

Short answer

The free-fall acceleration on the newly discovered planet's surface is$12.3m/s^2.$

Step-by-step solution

Given information.

Free-fall acceleration is the acceleration experienced by a freely falling body towards a body with gravitational attraction.

Calculation.

The free fall acceleration is given by : $g=GMR2.$

Here,

Mass of the planet is 5 times greater than earth: $Mp=5Me.$

Radius is twice as large as earth: $Rp=2Re.$

Here the mass and radius of earth is Me and Re respectively; and the mass and radius of newly discovered planet is Mp and Rp respectively.

Continuation of calculation.

Hence the free-fall acceleration on earth is : $ge=GMeRe2.$

and the free-fall acceleration on the newly discovered planet is : $gp=GMpRp2.$

Now replace Mp=5M and Rp=2Re at above equation

$gp=G\left(5Me\right)\left(2Re\}2\right)2=54\left(GMeRe2\right)$.

Continuation of calculation.

Now substitute the value of ge : $gp=54ge.$

Substitute the known value of ge to above equation to calculate gp.

localid="1648481965361" $gp=54(9.80\mathrm{m}/{\mathrm{s}}^2)=12.3\mathrm{m}/{\mathrm{s}}^2.$

Final answer.

The free fall acceleration on the newly discovered planet is $12.3\mathrm{m}/{\mathrm{s}}^2.$