Question

Luke Skywalker (yes, one of the good guys!) has measured the period of a small angle pendulum on Earth and found that it is 0.650 s.

a) What is the length of this pendulum? cm

b) Luke takes this pendulum such that it is a distance 500km above the surface of the Earth. What will the period of the pendulum be here? s

c) If Luke takes this pendulum to the surface of the planet Hoth where the acceleration of gravity is about 1.45 times that on Earth, what will the period of the pendulum be there? s

Answer 1

Time period on earth surface Te=0.650 s

We know that the time period of a simple pendullum:

; '

(a) length:

Here T=Te=0.65 s and g=9.8 m/s^2, so length l=0.105 m or 10.m

(b) Since the acceleartion due to gravity:

Here G=gravitational constand, Me=mass of the earth and r=total distance of the body from the center of the earth. So, r=R+d, where R=radius of the earth (6400 km) and d=distance of the body from the surface of the earth.

From (2) and (3):

(c) Again from (2):

This will be the time period at planet Hoth.