What is the Time for an Object to Orbit the Sun at a Radius of 3*10^6 m?

Click For Summary
SUMMARY

The discussion centers on calculating the orbital period of an object at a radius of 3*10^6 meters from the Sun using the equation T^2/R^3 = 4π^2/GM. Participants clarify that the radius in the formula refers to the total distance from the center of the Sun, not just the given radius. Since 3*10^6 meters is less than the Sun's radius of 6.96*10^8 meters, it is concluded that the question likely intended for the radius to be measured from the Sun's surface, necessitating the addition of the Sun's radius to the orbital radius for accurate calculations.

PREREQUISITES
  • Understanding of Kepler's Third Law of Planetary Motion
  • Familiarity with gravitational constant (G) and mass of the Sun (M)
  • Basic knowledge of orbital mechanics
  • Ability to perform calculations involving exponents and units of measurement
NEXT STEPS
  • Research the implications of Kepler's Laws on orbital dynamics
  • Learn how to apply the gravitational constant in orbital calculations
  • Study the effects of altitude on orbital mechanics
  • Explore advanced topics in celestial mechanics and perturbation theory
USEFUL FOR

Astronomy students, physicists, and anyone interested in orbital mechanics and gravitational calculations will benefit from this discussion.

avb203796
Messages
63
Reaction score
0
If an object were discovered oriting at a radius of 3*10^6 m, what would be the time that it would take to complete one orbit around the sun?
So I know I need to use the following equation:

T^2/R^3 = 4pie^2/GM

but do I use the radius they gave me or do I have to add that radius to the the radius of the sun to get R?
 
Physics news on Phys.org
check your radius ... it seems to be smaller then the radius of the sun!
 
The radius in the formula is the radius of the circle of the orbit. However, the Sun's radius is quite a bit more than 3 million meters (The Earth's radius is 6 million meters)!

If this is exactly how the question is stated, the the question is flawed. Perhaps the problem was intended to be 3 million meters above the surface of the sun. In that case you would add 3*10^6 to the Sun's radius of 6.96*10^8 meters, which would be 6.99*10^8 meters. This is almost insignificant, but it would be my assumption as to what the intent of the question was.
 
Last edited:

Similar threads

  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K