Satellite X and Y Orbital Period - Answer B: 1.4x10^6s

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SUMMARY

The orbital period of satellite Y, given its orbital radius of 3.0x10^9m, is definitively calculated to be 1.4x10^6 seconds using Kepler's third law of planetary motion. Satellite X has a period of revolution of 3.6x10^5 seconds and an orbital radius of 7.5x10^8m. The relationship established by Kepler's third law, specifically (T1/T2)^2 = (R1/R2)^2, confirms that the periods of revolution and orbital periods are equivalent terms in this context.

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Homework Statement


Two satellites, X and Y, are placed in orbit around a planet. Satellite X has period of revolution of 3.6x10^5s and an orbital radius of 7.5x10^8m. If the orbital radius of satellite Y is 3.0x10^9m, what is the orbital period?

A) 9.1x10^5s
B) 1.4x10^6s
C) 2.9x10^6s
D) 5.2x10^7s

Homework Equations


Kepler's third law equation:
(T1/T2)^2 = (R1/R2)^2


The Attempt at a Solution


For this question what I did was use Kepler's third law and the answer I got was "B". But I was wondering is this correct because satellite X it says it has a "period of revolution of 3.6x10^5s", is this the same as "orbital period"?
 
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Inertialforce said:
For this question what I did was use Kepler's third law and the answer I got was "B". But I was wondering is this correct because satellite X it says it has a "period of revolution of 3.6x10^5s", is this the same as "orbital period"?

they are one and the same.
 

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