Im having trouble starting out this problem:: The Earth’s acceleration due to gr
ID: 1663227 • Letter: I
Question
Im having trouble starting out this problem:: The Earth’s acceleration due to gravity variesfrom 9.78 m/s2 at the equator to 9.83 m/s2 at the poles, because the Earth is not a perfect sphere. A pendulum whoselength is precisely 1.000 m can be used to measure g. Such a device is called a gravimeter. a) How long do 100 oscillations take at the equator? b) How long do 100 oscillations take at the poles? c) Is the difference between your answers to parts (a) and (b)measurable? What kind of instrument could you use to measure the difference? d) Suppose you take your gravimeter to the top of a highmountain peak near the equator. There you find that 100 oscillations take 201.0 s. What is g on themountaintop? The Earth’s acceleration due to gravity variesfrom 9.78 m/s2 at the equator to 9.83 m/s2 at the poles, because the Earth is not a perfect sphere. A pendulum whoselength is precisely 1.000 m can be used to measure g. Such a device is called a gravimeter. a) How long do 100 oscillations take at the equator? b) How long do 100 oscillations take at the poles? c) Is the difference between your answers to parts (a) and (b)measurable? What kind of instrument could you use to measure the difference? d) Suppose you take your gravimeter to the top of a highmountain peak near the equator. There you find that 100 oscillations take 201.0 s. What is g on themountaintop? because the Earth is not a perfect sphere. A pendulum whoselength is precisely 1.000 m can be used to measure g. Such a device is called a gravimeter. a) How long do 100 oscillations take at the equator? b) How long do 100 oscillations take at the poles? c) Is the difference between your answers to parts (a) and (b)measurable? What kind of instrument could you use to measure the difference? d) Suppose you take your gravimeter to the top of a highmountain peak near the equator. There you find that 100 oscillations take 201.0 s. What is g on themountaintop?Explanation / Answer
a) T = 2(L/g) time for 100 oscillations = 100T = 200(1/9.78) =200.914 s b) at the poles, time = 200(1/9.83) = 200.402 s c) the difference, about half a second is quite measurable(e.g., bystroboscopic camera). d) T = 201 = 200(L/g) = 200/g g = (200/201)^2 = 9.772 m/s^2