Oscillation above the Surface of the Earth

[yaylee]

Homework Statement

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.

Homework Equations

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

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.

 

[haruspex]

height H = 2.23RE (stationary, not orbiting) above the Earth's surface.

T at 2.23R above the Earth: T^2 = 4∏^2(L)*(2.23R)^2 / (G)(Mearth)

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

 

[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!

 

[yaylee]

Go figure, it is! Thanks!

 

[Nickie]

Your approach is correct. However, there seems to be an error in your calculation of T for the pendulum at 2.23R above the Earth's surface. It should be T = √(7.23^2)(2.23)^2 = 11.0756 s. This makes sense, as the period of a pendulum is proportional to the square root of its length, so increasing the length by a factor of 2.23 would result in a longer period. Additionally, your final answer should include the units of time (s), so the final answer would be T = 11.0756 s.