What is Kepler's Formula and How is it Used in Circular Motion and Gravitation?

Click For Summary
SUMMARY

Kepler's Formula is essential for understanding the relationship between the orbital period of a satellite and its distance from the central body in circular motion and gravitation. The discussion highlighted a common error in calculating the number of seconds in a day, emphasizing the importance of accuracy in calculations. A specific example mentioned was the confusion surrounding the value of ##4 \times 10^{-7}m## for satellite orbits, which was deemed incorrect. The participants provided valuable insights into the correct application of Kepler's Formula and the significance of precise calculations in orbital mechanics.

PREREQUISITES
  • Understanding of Kepler's Laws of Planetary Motion
  • Basic knowledge of circular motion principles
  • Familiarity with gravitational force calculations
  • Proficiency in unit conversions and time calculations
NEXT STEPS
  • Study Kepler's Third Law of Planetary Motion in detail
  • Learn about gravitational force equations and their applications
  • Explore the concept of orbital mechanics and satellite dynamics
  • Practice unit conversion techniques, especially for time and distance
USEFUL FOR

Astronomy students, physics enthusiasts, and professionals in aerospace engineering who seek to deepen their understanding of orbital mechanics and gravitational calculations.

Mongster
Messages
20
Reaction score
7
Homework Statement
Find the radius R of the orbit of a geosynchronous satellite that circles the Earth. (Note that R is measured from the center of the Earth, not the surface of the Earth.)

The universal gravitational constant G is 6.67×10−11Nm2/kg2.
The mass of the earth is 5.98×1024kg.
The radius of the earth is 6.38×106m.
Relevant Equations
r^3 = (G*Mass of Earth*T^2)/(4pi^2)
Using Kepler's Formula, I tried to solve for the answer but was told that it's incorrect.
 

Attachments

  • IMG_2295 (1).jpg
    IMG_2295 (1).jpg
    59 KB · Views: 167
  • Capture 1.PNG
    Capture 1.PNG
    22.2 KB · Views: 182
Physics news on Phys.org
Doesn't ##4 \times 10^{-7}m## seem just a little small for a satellite orbit?
 
  • Like
  • Haha
Likes   Reactions: mpresic3, berkeman, epenguin and 2 others
But that's what the calculator said!

Also, 24 x 60 is not 86400. There are 86400 seconds in a day, but that's not 24 x 60.
 
  • Like
Likes   Reactions: Mongster
Ahh I apologise. It's a silly mistake I've made, shouldn't have rushed through things. Thank you all for the valuable inputs!
 
  • Like
Likes   Reactions: berkeman

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
2K
Replies
8
Views
1K
  • · Replies 12 ·
Replies
12
Views
990
Replies
19
Views
4K
Replies
55
Views
3K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
4
Views
2K
Replies
19
Views
3K