Solve Kepler's Law: Find Satellite's Period

  • Thread starter Thread starter keeholee
  • Start date Start date
  • Tags Tags
    Kepler's law Law
Click For Summary

Homework Help Overview

The discussion revolves around applying Kepler's Law to find the period of an artificial satellite based on its distance from the Earth compared to that of the Moon. The subject area is celestial mechanics and orbital dynamics.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Kepler's Law and the relationship between the periods and distances of celestial bodies. Questions arise regarding the validity of the calculated period and the implications of the distances involved.

Discussion Status

Some participants express uncertainty about the calculated period of the artificial satellite, questioning whether it is reasonable given the distances involved. There is an exploration of potential conversion issues and the need for clarification on the units used.

Contextual Notes

Participants note the differences in distances between the Moon and the artificial satellite, highlighting the need to consider unit conversions and the implications of the results in the context of orbital mechanics.

keeholee
Messages
3
Reaction score
0
Kepler's Law help!

Use this equation: T^2 of A / R^3 of A = T^2 of B / R^3 of B

The moon has a period of 27.3 days and has a mean distance of 3.9*10^5 km from the center of the earth. Find the period of an artifical satellite that is 7.5*10^3 km from the center of the earth.

How do i solve this?
 
Physics news on Phys.org


Kepler's law states that the period of a satellite squared over its radius cubed is always constant.

Therefore, you can just plug the values into the equation and solve for T^2 of (the period of the artificial satellite).
 


yes i got .075 for the period. but isn't that way too small considering the distance from center of the Earth to artifical satellite is bigger than the mean distance from the center of the Earth to the moon?
 


The distance to the artificial satellite from the Earth is a whole 10^2 smaller, according to the question you posted. The moon is farther away. .075 days for the period however, seems off.
 


oh yea Oo.


welll yea it is really off but is that possible? maybe its in Earth's i have to convert it or something? which i don't know how
 

Similar threads

Undergrad Kepler orbits for planets of similar masses
  • · Replies 2 ·
Replies
2
Views
1K
High School Proof of inverse square law for gravitation?
  • · Replies 18 ·
Replies
18
Views
3K
Find the location knowing the resonance using Kepler's third law
  • · Replies 2 ·
Replies
2
Views
1K
What is the relationship between tension and mass for an artificial satellite?
  • · Replies 10 ·
Replies
10
Views
2K
Third kepler's law error on sun mass calculation
  • · Replies 1 ·
Replies
1
Views
3K
High School Solving Kepler's 1st law as a function of time
  • · Replies 23 ·
Replies
23
Views
6K
Kepler's Third Law vs Conservation of angular momentum
  • · Replies 8 ·
Replies
8
Views
2K
Completing satellite's state vector by given info
  • · Replies 1 ·
Replies
1
Views
1K
Calculating the period of a low Earth satelite using Kepler
  • · Replies 6 ·
Replies
6
Views
2K
Kepler's Third Law and Motion of Two Point Masses
  • · Replies 1 ·
Replies
1
Views
2K