Orbital period questions continued

Click For Summary
SUMMARY

The discussion focuses on calculating the orbital period of a satellite with a mass of 9.0x103 kg and an orbital radius of 3.20x107 m. The key equations used include Kepler's third law and the formula for centripetal acceleration, Ac = 4π2r/T2. The calculated orbital period is 15.79 hours, derived from the velocity of the satellite, which is determined using the gravitational constant G = 6.67x10-11 m3kg-1s-2.

PREREQUISITES
  • Understanding of Kepler's third law of planetary motion
  • Familiarity with gravitational force calculations
  • Knowledge of centripetal acceleration formulas
  • Basic proficiency in algebra for manipulating equations
NEXT STEPS
  • Study the derivation and applications of Kepler's third law
  • Learn how to calculate gravitational force using Newton's law of universal gravitation
  • Explore the relationship between velocity and centripetal acceleration in orbital mechanics
  • Investigate the effects of altitude on satellite motion and orbital period
USEFUL FOR

Students in physics, aerospace engineers, and anyone interested in understanding satellite dynamics and orbital mechanics.

Inertialforce
Messages
67
Reaction score
2

Homework Statement


A 9.0x10^3kg satellite with an orbital radius of 3.20x10^7m orbits the Earth at an altitude of 2.56x10^7m. What is the orbital period?

Homework Equations


ΣFc = mac
Kepler's third law equation

The Attempt at a Solution


For this question I know that it is asking me to find the orbital period, but it has also given me an altitude and mass for the satellite. Are they there just to through me off because you are supposed to use Kepler's third law equation to find the orbital period?

Or are they important and I am supposed to use the equation:

Fg = 4(pie)^2r/T^2

Basically what I am wondering is which equation do we use?
 
Last edited:
Physics news on Phys.org
What does Kepler's third law give you? In relation to what you are looking for? What are the inputs required to use Kepler's Law?

~Lyuokdea
 
Keplers third law: "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." i don't think it is necessary to use keplars law to solve this question, this is how i would figure it out;

ok so you can solve how fast the satellite is moving in relation the Earth: v=(GMe/r)^.5
(G=6.67x10^-11 gravitational constant) so plugging the numbers in gets you a velocity of 3536.4176...ms^-1, next we use the formula Ac=4[tex]\pi[/tex]^2r/T^2 like you had before but you can solve the acceleration by using Ac=v^2/r now that you have the velocity. This gives you 0.39082m/s^2 now put this back into your original formula which gives you a period of 15.79 hours

hope this helps
 

Similar threads

Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
6K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
1K