How Fast Does Mercury Travel Around the Sun?

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SUMMARY

The orbital velocity of Mercury is calculated using the formula v = (2πr)/T, where r is the orbit radius (5.79 x 1010 m) and T is the orbital period (88.0 days). The correct conversion of the period into seconds is essential for accurate calculations. The acceleration of Mercury can be derived from the formula a = v²/r, where v is the orbital velocity. Both calculations require precise unit conversions to yield correct results.

PREREQUISITES
  • Understanding of orbital mechanics
  • Familiarity with unit conversions (days to seconds)
  • Knowledge of basic physics equations for velocity and acceleration
  • Ability to manipulate mathematical formulas
NEXT STEPS
  • Research the derivation of orbital velocity formulas in celestial mechanics
  • Learn about unit conversion techniques, especially for time
  • Study the relationship between velocity and acceleration in circular motion
  • Explore examples of calculating orbital parameters for other planets
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Astronomy students, physics enthusiasts, and anyone interested in celestial mechanics and orbital dynamics.

trajan22
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What is the magnitude of the orbital velocity of the planet Mercury (orbit radius =5.79 * 10^{7} {km}, orbital period = 88.0 days)?

Ive been trying to solve this using this equation v=(2pi(57.9*10^9m))/(7.6*10^3)s
but it is not coming out right...the numbers in the equation are the units in meters and seconds because that must be the form of the answer. I also have to solve for accelereation and have used this equation
a=((4pi^2(57.9*10^9))/(7.6*10^6)^2
this has also not worked out and its the only equations that could even fit the problem from this particular section in the book. please help!
 
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The period is correct in the acceleration equation, but not in the velocity equation.

The acceleration is given by the velocity squared divided by the radius.
 

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