Kepler's Third Law satellite problem

Click For Summary
SUMMARY

The discussion focuses on calculating the altitude of a synchronous orbit for a satellite over Pluto using Kepler's Third Law. The formula used is r = [(G*M*T^2)/4pi^2]^(1/3), where G is the gravitational constant (6.67e^-11), M is Pluto's mass (1.31e^22 kg), and T is the orbital period (551852 s). The calculated radius is 1.89e7 m, leading to an altitude of 1.77e4 km after subtracting Pluto's radius (1,153,000 m). Despite these calculations, the user reports that the system indicates the answer is incorrect, prompting a request for further insights.

PREREQUISITES
  • Understanding of Kepler's Third Law of planetary motion
  • Familiarity with gravitational constant and its application
  • Knowledge of orbital mechanics and synchronous orbits
  • Basic proficiency in unit conversions (meters to kilometers)
NEXT STEPS
  • Review the application of Kepler's Third Law in different celestial contexts
  • Investigate potential errors in gravitational calculations involving Pluto
  • Explore the significance of synchronous orbits in satellite technology
  • Learn about the effects of planetary radius on orbital calculations
USEFUL FOR

Astronomy students, astrophysicists, and anyone involved in satellite design or orbital mechanics will benefit from this discussion.

bonekrushur
Messages
3
Reaction score
0

Homework Statement



An orbiting satellite stays over a certain spot on the equator of (rotating) Pluto. What is the altitude (in km) of the orbit (called a "synchronous orbit")?


Homework Equations


r = [(G*M*T^2)/4pi^2]^(1/3)
h= r-radius of pluto


The Attempt at a Solution


I have plugged in the numbers (6.67e^-11) for G, (1.31e^22 kg) for M, and (551852 s) for T. I get (1.89e^7 m) for the radius. Then I subtract 1,153,000 meters from my answer to account for the radius of Pluto. My final answer in meters is (1.77e^7 m). My final answer in km is (1.77e^4 km). However, when I input this, the system shows this is wrong. I've tried several variations of the answer, but nothing works. Does anyone know if I've overlooked something in the equation? Thanks.
 
Physics news on Phys.org
Seems right.
 
I get the same answer.
 

Similar threads

Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
2K