What would be the magnitude of the acceleration due to gravity on the surface of the earth if the density of the earth increased by 3 times and the radius remained the same?

(a) 9.81 m/s^2

(b) 12.26 m/s^2

(c) 15.33 m/s^2

(d) 29.43 m/s^2

The question was posed to me in an internship interview.

Query is from Acceleration due to Gravity of the Earth in portion Gravitation of Physics – Class 11

Select the correct answer from above options

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The correct choice is (d) 29.43 m/s^2

The explanation is: g = (G*M1)/R^2;

M1 = Mass of the earth

R = Radius of the earth

Since the radius is the same, the volume would remain constant.

Density = mass/volume

Since density is increased 3 times and volume is the same, this implies that mass is increased 3 times.

We know; g = 9.81 m/s^2

New mass = 3 x M1

Therefore, new acceleration;

3 x (G*M1)/R^2 = 3x g

= 3 x 9.81

= 29.43 m/s^2.