A particle of mass 3.6 x 10\(^{-6}\)kg revolving around the earth has a radial acceleration of 4.3 x 10\(^7\)ms\(^{-2}\). Calculate the centripetal force of the particle.
A
1.55 x 10\(^2\)N
B
2.00 x 10\(^{-2}\)N
C
6.25 x 10\(^{-14}\)N
D
1.00 x 10\(^2\)N
Verified Correct Answer: A.
Physical Principles & Numerical Solution
F = mass x acceleration due to gravity
F\(_c\) = m x a\(_c\)
a\(_c\) = a\(_r\) = 4.3 x 10\(^7\)ms\(^{-2}\), m = 3.6 x 10\(^{-6}\)kg
F\(_c\) = m x a\(_c\) = 3.6 x 10\(^{-6}\) x 4.3 x 10\(^7\) = 154.8N ≈ 155N ⇒ 1.55 x 10\(^2\)N
Source: WAEC / JAMB / NECO Classroom Archive (Record #72051)