Pendulum swinging above earth's surface

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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:smile:
 

Answers and Replies

  • #2
PeterO
Homework Helper
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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:smile:
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.
 
  • #3
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Thank you for the response Peter. Are you saying that the R in the equation should be RE + 2.96RE?
 
  • #4
PeterO
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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.
 
  • #5
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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:smile:
 
  • #6
PeterO
Homework Helper
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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:smile:
OK it amounts to using variation.

You already knew

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
 
  • #7
PeterO
Homework Helper
2,426
48
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:smile:
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.

had you expressed your answer to 3 figures you would be giving the same answer as me.
 
  • #8
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Thank you for your help. I have a habit of not rounding, I'm glad my professor isn't too harsh about it. I found that your alternate way of doing this calculation to be much more efficient. I think the extra example you mentioned is cleverly worded and is very possible for my professor to put something similar on the next exam. I appreciate your time and patience spent helping me.
 

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