Radical Expressions and Functions

Question

In the pendulum formula T = 2 (pi) sqrt (L/g), we use g = 9.8 m/s^2 for the acceleration due to gravity on Earth. But what about Mars? If an astronaut on the surface of mars swings a 1-meter long pendulum and it has a period of 3.27 seconds, what is the acceleration due to gravity, g, on Mars?

Answer 1

On Mars,

 

T = 2π√(L/g') ⇒ 

 

g' = (2π/T)²L = (2π/3.27)²(1) m/s² = 3.692 m/s²