Early skeptics of the idea of a rotating earth said that the fast spin of Earth would throw people at the equator
into space. The radius of Earth is about 6400 km. Show why this objection is wrong by calculating.
(a) Calculate the speed of a 95 kg person at the equator.
(b) Calculate the force needed to accelerate the person in the circle.
(c) Calculate the weight of the person.
(d) Calculate the normal force of Earth on the person, that is, the person's apparent weight.
1 Answer | Add Yours
a) When the earth rotates an object(man) on the equator undergoes distance equal to the circuference of earth at equator in 24 hours. So, its velocity = distance undergone /seconds in 24 hrs = 2pi*64000*1000 metrs/(24*60*60) =465.4,211339meters/second is speed of any object on the earth due to the earth's rotation.
The person along the earth moves with a constant speed along the earth's circumference. In other words his constant relative speed along the earth wrt the earth is zero. So to accelerate the 95 kg man by 1meter/s^2 you require 95kg*1m/s^2 newtons =95N
c)The weight of the person is due to the inherent gravitational attraction between the person and the earth. So gravitational force on the person =GMm/R^2 where G is the gravitational constant and its value = 6.67258*10^-11Nm^2/kg^2, M is the mass of the earth = 5.978*10^24 kg and m is the mass of the person =95 kg , given. So the weight due to gravitation = 6.67258*10^-11)(5.978*10^24)(95)/(6400000)^2 = 925.1526 N
d)The apparent weight of the person, here in this question due to earths rotation about its axis alone, is reduced by mv^2/R = 95*(465.4211339)^2/(6400*1000) = 95*(0.033846379) =3.2154N which is due to the rotation of the earth. So the apparent weight of the person = weight due to gravitation - force on the person due to the earth's rotation = (925.1526-3.2154)N = 921.9372 N.
We’ve answered 317,404 questions. We can answer yours, too.Ask a question