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

AI Thread Summary
Centripetal acceleration is derived from the difference between the gravitational acceleration at the equator and the observed acceleration of an object dropped, which in this case is 10.00 m/s² minus 9.70 m/s², resulting in 0.3 m/s². This value represents the centripetal acceleration due to the planet's rotation. To understand this conceptually, one can visualize the change in velocity over a small time increment, incorporating a vector triangle that illustrates the relationship between linear and centripetal acceleration. The cosine factor, approximated to 1, simplifies the derivation of the centripetal acceleration formula. Understanding this derivation is crucial for solving similar problems in physics.
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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.
 
Physics news on Phys.org
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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