Circular Motion of the Earth around the sun

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SUMMARY

The period of Earth's nearly circular motion around the Sun is definitively 1 year. This conclusion is based on the formula for circular motion, where the period T can be derived from the equation (2*pi*r)/T. The gravitational force equation Fc=G(mMe/r^2) = mv^2/r indicates that a satellite, or in this case, Earth, maintains a specific speed to remain in orbit. The discussion confirms that no additional calculations are necessary beyond recognizing that 1 year represents the time taken for one complete revolution around the Sun.

PREREQUISITES
  • Understanding of circular motion and orbital mechanics
  • Familiarity with gravitational force equations
  • Basic knowledge of the concept of a year in relation to Earth's orbit
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Explore the derivation of Kepler's laws of planetary motion
  • Learn about the implications of gravitational force on satellite orbits
  • Study the relationship between orbital radius and period in circular motion
  • Investigate the effects of eccentricity on planetary orbits
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and orbital dynamics, as well as educators seeking to clarify concepts related to Earth's motion around the Sun.

layla2525
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Homework Statement




What is the period of Earth's (nearly) circular motion around the Sun?


(2*pi*r)/T

1 year

(1 year)/(2*pi)

(2*pi)/(1 year)

2*pi*(1 year)

Homework Equations



There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius.

Fc=G(mMe/r^2) =mv^2/r

The Attempt at a Solution



I am leaning towards 1 year...
 
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layla2525 said:

Homework Statement




What is the period of Earth's (nearly) circular motion around the Sun?


(2*pi*r)/T

1 year

(1 year)/(2*pi)

(2*pi)/(1 year)

2*pi*(1 year)

Homework Equations



There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius.

Fc=G(mMe/r^2) =mv^2/r

The Attempt at a Solution



I am leaning towards 1 year...

Well you should, as 1 year is the time take for the Earth to make one revolution around the Earth! So you didn't need to do anything with the 2*pi.
 
layla2525 said:

Homework Statement




What is the period of Earth's (nearly) circular motion around the Sun?


(2*pi*r)/T

1 year

(1 year)/(2*pi)

(2*pi)/(1 year)

2*pi*(1 year)

Homework Equations



There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius.

Fc=G(mMe/r^2) =mv^2/r

The Attempt at a Solution



I am leaning towards 1 year...
Welcome to Physics Forums :smile:

I am a little confused. We already know that the period is 1 year, from the basic definition of a year. Was there any more instruction given with the question? Are you supposed to use the radius of Earth's orbit, or the speed of Earth in its orbit, or something along those lines?
 

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