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Question

If the density of a small planet is the same as that of the earth while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration on the surface of the planet is _____.


A

0.2g

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B

0.4g

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C

2g

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D

4g

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Solution

The correct option is A

0.2g


Explanation for the correct option:

Step 1: State assumptions and known data

The density of Earth and the small planet is same.
Let the density of earth and the small planet be ρ

Let R be the radius of Earth. Then, 0.2R is the radius of the small planet.

Step 2: Formulas used

Acceleration due to gravity is given as,

g=GMR2 ...1

where G is the gravitational constant, M is the mass of the planet and R is the radius of the planet.

Density of an object, ρ=MV, where V is its volume

Step 3: Derive expression for acceleration due to gravity in term of density of the earth

From the equation relating density and mass,
M=ρV

The shape of a planet can be approximated to be that of a sphere (in fact, shape of the earth is geoid).

The volume of a sphere is V=43πr3, where r is its radius.

Thus, the mass of the sphere is, M=43πr3ρ

Substituting the above expression in 1,

g=43Gπr3ρr2g=43Gπrρ

Step 4: Calculate the acceleration due to gravity on the small planet

The acceleration due to gravity on Earth is,
g=43GπRρ

The acceleration due to gravity on the small planet is,
g'=430.2Rρg'=0.2×43πRρg'=0.2gg=43Rρ

Therefore, if the density of a small planet is the same as that of the earth while the radius of the planet is 0.2 times that of the earth, the gravitational acceleration on the surface of the planet is 0.2g.

Hence, option A is correct.


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