What is the derivation of centripetal acceleration on a hypothetical planet?

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SUMMARY

The derivation of centripetal acceleration on a hypothetical planet with an equatorial acceleration of 10.00 m/s² and a radius of 6.2 x 10⁶ m involves understanding the difference between gravitational acceleration and centripetal acceleration. The centripetal acceleration is calculated as the difference between the gravitational acceleration (10.00 m/s²) and the observed acceleration of an object dropped at the equator (9.70 m/s²), resulting in a centripetal acceleration of 0.30 m/s². To derive the formula for centripetal acceleration in terms of velocity, one can utilize a vector triangle that illustrates the change in velocity over a small time increment, incorporating a cosine factor that approximates to 1.

PREREQUISITES
  • Understanding of gravitational acceleration and centripetal acceleration concepts
  • Familiarity with vector analysis in physics
  • Knowledge of basic kinematics and motion equations
  • Ability to interpret and manipulate trigonometric functions
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  • Study the derivation of centripetal acceleration formulas in physics
  • Learn about vector triangles and their applications in motion analysis
  • Explore the relationship between velocity, radius, and centripetal acceleration
  • Investigate the effects of planetary rotation on gravitational forces
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Students studying physics, educators teaching kinematics, and anyone interested in the dynamics of planetary motion and acceleration concepts.

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Homework Statement


Imagine another planet with an acceleration of 10.00 m/s^2 at its equator when ignoring the rotation of the planet. The radius is 6.2 x10^6 m. An object dropped at the equator yields an acceleration of 9.70 m/s^2. Determine the length of 1 day on this planet.

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The Attempt at a Solution


My teacher taught me how to solve this question, but there is just one step I don't understand
10.00m/s^2 - 9.70m/s^2 = 0.3m/s^2 (centripetal acceleration)
why does subtracting these two values give us the centripetal acceleration??
On a test, if I get a similar question I know I have to subtract the two values, but I really want to learn the reason behind it.
 
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Do you know how to derive the formula for centripetal acceleration in terms of velocity, using a vector triangle showing the change in velocity over a small increment of time ##\delta t##? A similar construction can be used to justify the above formula. There's a ##\cos\delta\theta## factor in there that is approximated by 1 to get the above formula.
 

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