Will someone check my application of kepler's laws?

  • Thread starter Thread starter imsoconfused
  • Start date Start date
  • Tags Tags
    Application Laws
Click For Summary
SUMMARY

The discussion centers on the application of Kepler's laws, specifically Kepler's Third Law, to calculate the length of the major axis of Earth's orbit and the altitude required for a geostationary satellite. The period of Earth's orbit is approximately 365.25 days, and using the mass of the sun (1.99x10^30 kg) and the gravitational constant (6.67x10^-11 Nm^2/kg^2), the calculations must convert days to seconds for accuracy. Additionally, the Earth's mass (5.98x10^24 kg) and radius (6.37x10^6 m) are critical for determining the satellite's altitude, which requires precise arithmetic for correct results.

PREREQUISITES
  • Understanding of Kepler's laws, particularly Kepler's Third Law
  • Knowledge of gravitational constants and their applications
  • Ability to perform unit conversions, specifically from days to seconds
  • Familiarity with basic orbital mechanics and satellite positioning
NEXT STEPS
  • Study the derivation and applications of Kepler's Third Law in celestial mechanics
  • Learn about gravitational force calculations using Newton's law of gravitation
  • Research methods for calculating geostationary satellite altitudes
  • Explore the implications of orbital mechanics on satellite communication systems
USEFUL FOR

Astronomy students, physics enthusiasts, and aerospace engineers interested in orbital mechanics and satellite technology will benefit from this discussion.

imsoconfused
Messages
50
Reaction score
0
1. The period of the Earth's orbit is approximately 365.25 days. Use this fact and Kepler's Third Law to find the length of the major (not semi-major) axis of the Earth's orbit. You will need the mass of the sun, M = 1.99x10^30 kg, and the gravitational constant, G = 6.67x10^-11 Nm^2/kg^2.


2. It's possible to place a satellite into orbit about the Earth so that it remains fixed above a given location on the equator. Compute the altitude that is needed for such a satellite. The Earth's mass is 5.98x10^24 kg; its radius is 6.37x10^6 m.

Sorry, I couldn't figure out how to get the equations from microsoft equation editor into the question, so you'll have to open the attachment (you may have to save it, I apologize). I've got full solutions worked out, so all I need is for someone to double check me.
Thanks in advance!
 

Attachments

Last edited:
Physics news on Phys.org
First one you have the wrong answer. You forgot to convert days to seconds. Also your maths in the various stages is a little off. I'll let you see if you can spot what it is.

Second one is slightly off as well. I think that's just arithmetic.
 

Similar threads

Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K