# Oscillation above the Surface of the Earth

1. Dec 4, 2012

### yaylee

1. The problem statement, all variables and given/known data
A pendulum consists of point mass Mo swinging on a massless string of length Lo, with period To = 7.23 s on the surface of the Earth (at RE, the radius of the Earth). Find T1, the period of the same pendulum if it swings on a horizontal platform of height H = 2.23RE (stationary, not orbiting) above the Earth's surface.

2. Relevant equations
T = 2∏√(L/g)
g = G * Mass of Earth/R^2, where R = radius of Earth

3. The attempt at a solution
Squaring both sides of the Period, T, equation, we get,

T on surface of Earth: T^2 = 7.23^2 = 4∏^2(L/g), = 4∏^2(L)*R^2 / (G)(Mearth), and,
T at 2.23R above the Earth: T^2 = 4∏^2(L)*(2.23R)^2 / (G)(Mearth)

Dividing both equations, and solving for T, T = √(7.23^2)(2.23)^2 = 16.1229

Am I doing something wrong here? Many thanks in advance.

2. Dec 4, 2012

### haruspex

In the formula GMe/r2, what exactly is r?

3. Dec 4, 2012

### yaylee

Hi Haruspex,

This r should be (r PLUS 2.23R), or (R + 2.23R) = 3.23R ... Would I be correct here? Thank you for your assistance!

4. Dec 4, 2012

### yaylee

Go figure, it is! Thanks!