Universal gravitation 5- determine which satellite has smaller period

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SUMMARY

The discussion centers on determining which satellite, orbiting Neptune or Earth, has a smaller orbital period using Kepler's 3rd law and Newton's Universal Law of Gravitation. The calculations reveal that Neptune's satellite, with a calculated period of approximately 1265.82 seconds, has a smaller period compared to Earth's satellite, which has a period of approximately 5249.72 seconds. The mass of Neptune is established as 1.02856x1026 kg, and the orbital radius for both satellites is 6.53x106 m. The conclusion is that Neptune's satellite orbits faster due to the greater mass of the planet it orbits.

PREREQUISITES
  • Understanding of Kepler's 3rd law of planetary motion
  • Familiarity with Newton's Universal Law of Gravitation
  • Basic knowledge of circular motion equations
  • Ability to perform calculations involving gravitational constants and orbital periods
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  • Study the implications of Kepler's 3rd law on satellite motion
  • Explore the relationship between mass and orbital period in different planetary systems
  • Investigate the effects of varying orbital radii on satellite periods
  • Learn about the gravitational constant and its applications in astrophysics
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Homework Statement



Neptune is about 17.2 times the mass of Earth. Satellite N orbits Neptune with the same orbital radius as satellite E that orbits the Earth. Determine which satellite has the smaller period. Support your answer with appropriate calculations.

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



mN=17.2x me
=17.2x(5.98x1024kg)=1.02856x1026kg

RN=Re=6.53x106m

For Neptune satellite:

T=2∏√(6.53X106)3/(6.67X10-11)(1.029X1026)
T=1265.82

For Earth satellite:
T=2∏√(6.53x106m)3/(6.67x10-11)(5.98x1024kg)
T=5249.72

∴Neptune's satellite has the smaller period.
If anyone could please verify my work and let me know if/where I went wrong that would be greatly appreciated! Thanks so much for you help! :)
 
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You can just try to find the relation between the mass of planet and the period of its satellite without doing the numerical calculation.
 

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