# Pendulum swinging above earth's surface

## Homework Statement

Assume all oscillators are frictionless.
a) A pendulum consists of point mass Mo swinging on a massless string of length Lo, with period To = 9.42 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 = 2.96RE (stationary, not orbiting) above the Earth's surface.

## Homework Equations

T = 2∏√(L/g) , g = GM/R2 , Radius of earth = 6.37e6m, mass of earth = 5.98e24kg

## The Attempt at a Solution

g(T/2∏)2 = L
9.81m/s2√(9.42s/2∏)2 = 22.05012604m

g = GM/R2
(6.67e-11 * 5.98e24kg)/(2.96 * 6.37e6)2 = g1
3.98866e14/3.55518567e14 = 1.121927339m/s2 = g1

T1 = 2∏√(L/g1)
T1 = 2∏√(22.05012604/1.121927339)
T1 = 27.85499219s

The answer i calculated is wrong, any help with figuring out why it's wrong would be greatly appreciated

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PeterO
Homework Helper

## Homework Statement

Assume all oscillators are frictionless.
a) A pendulum consists of point mass Mo swinging on a massless string of length Lo, with period To = 9.42 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 = 2.96RE (stationary, not orbiting) above the Earth's surface.

## Homework Equations

T = 2∏√(L/g) , g = GM/R2 , Radius of earth = 6.37e6m, mass of earth = 5.98e24kg

## The Attempt at a Solution

g(T/2∏)2 = L
9.81m/s2√(9.42s/2∏)2 = 22.05012604m

g = GM/R2
(6.67e-11 * 5.98e24kg)/(2.96 * 6.37e6)2 = g1
3.98866e14/3.55518567e14 = 1.121927339m/s2 = g1

T1 = 2∏√(L/g1)
T1 = 2∏√(22.05012604/1.121927339)
T1 = 27.85499219s

The answer i calculated is wrong, any help with figuring out why it's wrong would be greatly appreciated
The R in the formula g = GM/R2 is the distance from the centre of the Earth.
This pendulum is set up 2.96RE above the surface, so 3.96RE from the centre.

Thank you for the response Peter. Are you saying that the R in the equation should be RE + 2.96RE?

PeterO
Homework Helper
Thank you for the response Peter. Are you saying that the R in the equation should be RE + 2.96RE?
I think yes. After you have worked it out, I will show you how I would have done the question - in a very simply way.

Ok so I redid the problem with the correction of the R. g = (6.67e-11 * 5.98e24kg)/((2.96*6.37e6)+6.37e6)2 and got a g = .626841558. Then T = 2∏√(22.05012604/.626841558) and got T = 37.26s. I think that period should be right, I'm interested in your alternate method to solving this type of problem

PeterO
Homework Helper
Ok so I redid the problem with the correction of the R. g = (6.67e-11 * 5.98e24kg)/((2.96*6.37e6)+6.37e6)2 and got a g = .626841558. Then T = 2∏√(22.05012604/.626841558) and got T = 37.26s. I think that period should be right, I'm interested in your alternate method to solving this type of problem
OK it amounts to using variation.

T = 2∏√L/g and g = GM/R2

Substitute the expression for g into the first expression and we get

T = 2∏√(LR2/GM)

most of those terms don't change, so replacing 2∏√(L/GM) with something simple like k we get

T = kR

Now the new R value is 3.96 times the R value at the surface, so simple arithmetic →

T = 9.42 x 3.96 → 37.3072s or 37.3 s when respecting the significant figures in supplied data.

A couple of things to note:

1: This type of comparative approach is very powerful and leads to simple calculation.

2: Giving the distance above the Earth Surface is a common "tester" used in problems.
It is always fun to see people calculate that gravity 600km above the surface is way stronger than at the surface, because someone calculates with R = 600 000 m rather than 600 000 m MORE than the Radius of the Earth.

Another example:
Compare the centripetal acceleration of a satellite orbiting a distance RE above the surface of the earth, to one orbiting at a distance 2RE above the surface.

The centripetal acceleration is simple g at the point.

Knowing the formula g = GM/R2 leads some to assume an answer 4:1, but of course it is actually 9:4 since the R values to use in the formula are not RE & 2RE but 2RE & 3RE

PeterO
Homework Helper
Ok so I redid the problem with the correction of the R. g = (6.67e-11 * 5.98e24kg)/((2.96*6.37e6)+6.37e6)2 and got a g = .626841558. Then T = 2∏√(22.05012604/.626841558) and got T = 37.26s. I think that period should be right, I'm interested in your alternate method to solving this type of problem
Note: I think your answer of 37.26 varies from my 37.3026 due to the rounded off values of Earth mass and Radius you have used.