Universal gravitation 10- Determine the rate

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SUMMARY

The discussion focuses on determining the rate at which an imaginary line between the Earth and the Sun sweeps out area as the Earth orbits the Sun. The relevant parameters include the Earth's orbital period of 365 days and a mean orbital radius of 1.50 x 1011 meters. The key equation to solve this problem is the area swept out per unit time, which can be derived from the formula for the area of a circle, A = πR2, divided by the period, T. The solution emphasizes that the problem is fundamentally geometric rather than reliant on complex gravitational equations.

PREREQUISITES
  • Understanding of Kepler's laws of planetary motion
  • Basic geometry, specifically area calculations
  • Familiarity with Newton's Universal Law of Gravitation
  • Knowledge of orbital mechanics and period calculations
NEXT STEPS
  • Study the derivation of area swept out by a radius vector in circular motion
  • Learn about Kepler's laws, particularly Kepler's 2nd law regarding area and time
  • Explore the relationship between orbital radius and period using T2 = R3/K
  • Investigate the application of Newton's laws in celestial mechanics
USEFUL FOR

This discussion is beneficial for students studying physics, particularly those focusing on celestial mechanics, orbital dynamics, and gravitational theory. It is also useful for educators seeking to clarify concepts related to universal gravitation and geometric interpretations of orbital motion.

dani123
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Homework Statement



As the Earth revolves around the sun, if not only travels a certain distance every second, it also causes an imaginary line between the Earth and the sun to pass through a certain area every second. During one complete trip around the sun, the total area would be approximately equal to ∏R2. The time it would take to do this would be the period, T. Determine the rate at which the area is swept by an imaginary line joining the sun and the Earth as the Earth orbits the sun. Use the Earth's period of 365 days and the mean orbital radius of 1.50x1011.

Homework Equations


I have made a list of equations that are relevant for this entire module on universal gravitation. So although there are many of them does not mean that they all apply in this circumstance. The ones relevant to this question will be placed in bold.

Kepler's 3rd law: (Ta/Tb)2=(Ra/Rb)3

motion of planets must conform to circular motion equation: Fc=4∏2mR/T2

From Kepler's 3rd law: R3/T2=K or T2=R3/K

Gravitational force of attraction between the sun and its orbiting planets: F=(4∏2Ks)*m/R2=Gmsm/R2

Gravitational force of attraction between the Earth and its orbiting satelittes: F=(4∏2Ke)m/R2=Gmem/R2

Newton's Universal Law of Gravitation: F=Gm1m2/d2

value of universal gravitation constant is: G=6.67x10-11N*m2/kg2

weight of object on or near Earth: weight=Fg=mog, where g=9.8 N/kg

Fg=Gmome/Re2

g=Gme/(Re)2

determine the mass of the Earth: me=g(Re)2/G

speed of satellite as it orbits the Earth: v=√GMe/R, where R=Re+h

period of the Earth-orbiting satellite: T=2∏√R3/GMe

Field strength in units N/kg: g=F/m

Determine mass of planet when given orbital period and mean orbital radius: Mp=4∏2Rp3/GTp2


The Attempt at a Solution



I highlighted the above equation because I think I am supposed to use this equation for this problem. But I honestly don't understand what this question is actually looking for, so if someone could let me know where I should even begin it would be greatly appreciated! Thank you so much in advance!
 
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hi dani123! :smile:
dani123 said:
As the Earth revolves around the sun, if not only travels a certain distance every second, it also causes an imaginary line between the Earth and the sun to pass through a certain area every second. During one complete trip around the sun, the total area would be approximately equal to ∏R2. The time it would take to do this would be the period, T. Determine the rate at which the area is swept by an imaginary line joining the sun and the Earth as the Earth orbits the sun. Use the Earth's period of 365 days and the mean orbital radius of 1.50x1011.

sorry, but all the equations you have listed are irrelevant :redface:

this is just geometry …

you're given the values of R and T, and you're asked to find the area per time

start with an easy case … what is the area after time T?​
 

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