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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

Expert verified

The free-fall acceleration on the newly discovered planet's surface is12.3m/s2.

Step by step solution

01

Given information.

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

02

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.

03

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(5Me)(2Re}2)2=54(GMeRe2).

04

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(9.80m/s2)=12.3m/s2.

05

Final answer.

The free fall acceleration on the newly discovered planet is 12.3m/s2.

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