Measuring Gravity with a Pendulum on Other Planets

Thread starter spherimorph
Start date Nov 18, 2007
Tags

Gravity Pendulum

AI Thread Summary

Measuring gravity on other planets using a pendulum involves understanding the relationship between the pendulum's period and gravitational acceleration. The formula T = 2π√(L/g) indicates that the period T of a pendulum is dependent on its length L and the gravitational acceleration g. By measuring the period of a pendulum on a different planet, one can rearrange the formula to solve for g, thus determining the planet's gravitational strength. This method relies on precise timing and knowledge of the pendulum's length. Overall, using a pendulum provides a practical approach to measuring gravitational differences across celestial bodies.

Nov 18, 2007

spherimorph

How does one go about measuring gravity using a pendulum on other planets? I don't quite understand how and why this works...

Last edited: Nov 18, 2007

Physics news on Phys.org

Nov 18, 2007

jtbell

Staff Emeritus

Science Advisor

Homework Helper

Hint: the period (time for one cycle) of a pendulum with length L is

T = 2 \pi \sqrt{\frac{L}{g}}

See for example

http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html

Thread 'Is calling fictitious forces "not real" just about terminology?'

Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...

View full post »

Thread 'Derivation of Hamilton's Principle: Questions'

I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...

View full post »