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

AI Thread Summary
To determine the orbital period of an object at a radius of 3*10^6 m around the Sun, the correct formula to use is T^2/R^3 = 4π^2/GM. The radius in the equation refers specifically to the orbital radius, not the radius of the Sun. Since 3 million meters is less than the Sun's radius of approximately 6.96*10^8 meters, the question may be flawed. It is suggested that the intended radius might be 3 million meters above the Sun's surface, requiring the addition of the Sun's radius to this value. Clarifying this point is essential for accurate calculations.
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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?
 
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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.
 
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