Question

Tamsen and Vera imagine visiting another planet, planet X, whose gravitational acceleration, gX, is different from that of Earth's. They envision a pendulum, whose period on Earth is 2.243 s, that is set in motion on planet X, and the period is measured to be 2.000 s. What is the ratio of gX/gEarth? Neglect any effects caused by air resistance.

Answer 1

Answer:

1.27

Explanation:

Period of a pendulum T is

T = 2¶(l/g)^0.5

Where g is acceleration due to gravity

l is lenght of pendulum

For earth, T = 2.243 s

2.243 = 2 x 3.142 x (l/9.81)^0.5

0.36 = (l^0.5)/3.13

1.13 = l^0.5

l = 1.28 m

For planet X of the same lenght

Period T = 2.00 s

2 = 2 x 3.142 (1.28/gx)^0.5

0.32 = 1.13 / gx^0.5

3.53 = gx^0.5

gx^0.5 = 3.53

gx = 12.46 m/s^2

gx/gearth = 12.46/9.81 = 1.27