Tangential speed of moon around earth

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SUMMARY

The discussion focuses on calculating the tangential speed of the Moon as it orbits the Earth, utilizing the formula for tangential speed: \( v = \frac{2\pi r}{T} \), where \( r \) is the radius of rotation and \( T \) is the period of revolution. Participants clarify that while the value of \( \pi \) is irrational, the final speed should be rounded to a reasonable number of significant figures based on the precision of the input measurements. The term "radius of rotation" refers to the distance from the Moon to the Earth's center, which is essential for accurate calculations.

PREREQUISITES
  • Understanding of tangential speed and linear velocity
  • Familiarity with the concept of significant figures in measurements
  • Basic knowledge of circular motion and orbital mechanics
  • Ability to perform calculations involving \( \pi \) and scientific notation
NEXT STEPS
  • Learn about significant figures and their importance in scientific calculations
  • Study the principles of circular motion in physics
  • Explore the concept of orbital mechanics and gravitational forces
  • Practice calculating tangential speed using different celestial bodies
USEFUL FOR

Students in physics, particularly those studying orbital mechanics, as well as educators and anyone interested in understanding the calculations involved in celestial motion.

rkrk
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I have this problem for my AP Physics class that discusses the moon circling around the Earth. I am given the mass of the moon, time it takes for one revolution, and the moon's distance from the Earth (the radius of rotation). I am supposed to find out the tangential speed of the Moon traveling around the Earth, and put it into scientific notation. Easy enough, because tangential speed is simply the distance the moon is from the Earth times two times pi divided by the time it takes for one full revolution. However, no matter what the distance is, when ever doing any operation involving pi you will end up with an irrational number. How are you supposed to write a neverending number in scientific notation? Also what exactly does it mean by radius of rotation? Am I doing anything wrong or is my formula for tangential speed incorrent?
 
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rkrk said:
However, no matter what the distance is, when ever doing any operation involving pi you will end up with an irrational number. How are you supposed to write a neverending number in scientific notation?
While the pure number \pi might be irrational and "neverending" (which is why we represent it by a symbol), your calculation of the speed will not be a pure number. You have to roundoff your final answer to a reasonable number of significant figures. (Based on the accuracy of your values for time and distance.)


Also what exactly does it mean by radius of rotation?
Just what you thought it meant when you produced your formula for tangential speed. The moon travels in a circle around the Earth's center; you are given the radius of that circle, I presume.
Am I doing anything wrong or is my formula for tangential speed incorrent?
Nothing wrong with your formula.
 
Since the distance from the moon to the Earth (radius of rotation) is a measured quantity, it is not exact and is given to some number of "significant figures". G and M are also "measured" and so are given with some number of significant figures. Your answer should have the number of significant figures equal to the smallest of these. (Your calculation can't be more accurate than the least accurate measurement.)
 
Tangential speed is also known as linear velocity. It is tangential speed because the velocity component is tangential to the acceleration component, which is towards Earth. The answer is simple: You know the period and radius of rotation, velocity is simply 2(PI)radius / period.
 
Doc Al said:
The moon travels in a circle around the Earth's center; you are given the radius of that circle, I presume.

Nothing wrong with your formula.
The moon does not travel in a circle around the Earth's center.
 
MeJennifer said:
The moon does not travel in a circle around the Earth's center.
True, but close enough for this problem. :wink:
 

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