A simple pendulum has a period T on Earth. If it were used on Planet X and found to have a period of only 71% of T, what would be the free fall acceleration due to gravity on this planet X?

gₓ = 1.4 g = 13.8 m/s²

Explanation:

The time period of simple pendulum on earth is given as:

T = 2π√(L/g)

where,

L = Length of pendulum

g = acceleration due to gravity on the surface of earth = 9.8 m/s²

T = Time Period on Earth

but, on the other planets, it becomes:

Tₓ = 2π√(L/gₓ)

where,

Tₓ = Time Period on Surface of Other Planet

gₓ = acceleration due to gravity on other planet  = ?

It is given that:

Tₓ = 71% of T

Tₓ = 0.71 T

using values:

2π√(L/gₓ) = (0.71) 2π√(L/g)

√(1/gₓ) = (0.71)√(1/g)

squaring both sides:

1/gₓ = 0.71/g

gₓ = g/0.71

gₓ = 1.4 g = 13.8 m/s²