Solving Orbital Mechanics: 24 Hours to Seconds Conversion

  • Thread starter Thread starter triden
  • Start date Start date
  • Tags Tags
    Orbit Satellite
Click For Summary
SUMMARY

The discussion focuses on converting 24 hours into seconds within the context of orbital mechanics. Participants emphasize the importance of circular motion and the relationship between angular velocity (omega) and gravitational force using the equations r*(omega)^2 = GM/r^2 and omega^2 = GM/r. The correct approach involves considering the radius of the Earth, approximately 6371 km, when calculating gravitational forces. The final answer for the conversion and related calculations is approximately 24,000 miles.

PREREQUISITES
  • Understanding of gravitational force equations, specifically Fg = -gmm/r
  • Familiarity with circular motion concepts in physics
  • Knowledge of angular velocity and its relation to gravitational parameters
  • Basic understanding of the Earth's radius and its significance in orbital mechanics
NEXT STEPS
  • Study the derivation of the equation omega^2 = GM/r^3
  • Learn about the implications of circular motion in orbital mechanics
  • Research the effects of Earth's radius on gravitational calculations
  • Explore advanced gravitational force equations and their applications in astrophysics
USEFUL FOR

Students studying physics, particularly those focusing on orbital mechanics, as well as educators and anyone interested in understanding the conversion of time into orbital parameters.

triden
Messages
172
Reaction score
0
http://upload.cybermart.ca/2004/hepmesad.gif


I can't decide what to do with this question. I am studying for a final. I converted 24 hours to seconds and tried using the Fg = -gmm/r but I don't think that's going to work...

any leads? Thanks guys
 
Last edited by a moderator:
Physics news on Phys.org
You're on the right track. Think about circular motion, too.
 
theFuture said:
You're on the right track. Think about circular motion, too.

ok I think I am getting it.

r*(omega)^2 = GM/r^2

then i can cancel the R's to get

Omega^2 = GM/r ?
 
triden said:
ok I think I am getting it.

r*(omega)^2 = GM/r^2

then i can cancel the R's to get

Omega^2 = GM/r ?

Yes,it's correct.To check your answer,though,u should be gettin round about 35000km.
 
you mean omega^2 = GM/r^3
 
That's one of my favorite problems.

I think the answer is about 24,000mi. Don't forget to subtract the radius of the earth. I always forget to do that!
 
i Think radius should be (R+r) and use F=GMm/(R+r)^2
 
saltrock said:
i Think radius should be (R+r) and use F=GMm/(R+r)^2

Why complicate?Use "r" as your length variable (the radius of the trajectory)and then,once u got the result,subtract the mean radius of the Earth which is round about 6371km.
 

Similar threads

Problem about orbital mechanics
  • · Replies 5 ·
Replies
5
Views
2K
A satellite in orbit in a system with one planet and two moons
  • · Replies 1 ·
Replies
1
Views
1K
Solving Satellite Orbit Problem: Finding Time in Hours
  • · Replies 3 ·
Replies
3
Views
2K
Satellite Orbiting: Speed, Period, Altitude Calc.
  • · Replies 2 ·
Replies
2
Views
3K
Finding the radius of the satellite's circular orbit
  • · Replies 3 ·
Replies
3
Views
2K
Satellite in orbit losing speed
  • · Replies 4 ·
Replies
4
Views
1K
Additional Velocity Required for a Satellite Already in Orbit to Escape
  • · Replies 4 ·
Replies
4
Views
2K
Calculating Height of a Twice-Daily Orbit Above Earth's Surface
  • · Replies 38 ·
2
Replies
38
Views
4K
Geostationary satellite orbit radius as a multiple of Earth radius.
  • · Replies 13 ·
Replies
13
Views
3K
The difference in mechanical energy of a satellite
  • · Replies 5 ·
Replies
5
Views
2K