Editor Staff asked 1 year ago
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
Interview Questions and Answers, Database Interview Questions and Answers for Freshers and Experience
Editor Staff answered 1 year ago
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.
