What Are the Equations for Calculating Centripetal Acceleration?

AI Thread Summary
Centripetal acceleration can be calculated using the equation ac = v^2/r, where v is the tangential velocity and r is the radius of the circular path. For an object in orbit, the velocity can be expressed as v = 2πr/T, with T representing the orbital period. Substituting this into the centripetal acceleration formula gives ac = 4π^2r/T^2. The radius of Earth's orbit around the sun is approximately 1.5 x 10^11 m, and its mass is about 5.98 x 10^24 kg. Accurate equations are essential for proper calculations in centripetal motion.
SmallPub
Messages
19
Reaction score
0
The radius of the Earth's orbit about the sun is about 1.5x10^11m. The mass of the Earth is 5.98x^10^24kg



Equations:
ac= v^2/r
ac=4∏^2/r
v=2∏r/T


I could not figure out what to do so i could not attempt it
 
Last edited:
Physics news on Phys.org
Centripetal acceleration is ##a_c=\frac{v^2}{R}##.
Since ##v=\frac{2\pi R}{T}##, where ##T## is the period, so
##a_c=\frac{4\pi^2 R}{T^2}##.
 
Your equations, except the first one, are incorrect. Find the correct equations first.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top