To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.
Our values are given as
The angular velocity of a body can be described as a function of frequency as
PART A) The expression for the maximum angular velocity is given by the amplitude so that
PART B) The maximum acceleration on your part would be given by the expression
Answer:
3.85*10^8m
Explanation:
We know that
Speed v = distance/time t
So distance = v x t
So = 3*10^8 x 2.56s
= distatance = 7.7*10^8m
However this is double the distance from the earth to the moon, so to get distance from earth to the moon we divide by 2
7.7*10^8/2= 3.85*10^8m
A force sets an object in motion when the force is multiplied by the time of its application.
Thank you for your question, what you say is true, the gravitational force exerted by the Earth on the Moon has to be equal to the centripetal force.
An interesting application of this principle is that it allows you to determine a relation between the period of an orbit and its size. Let us assume for simplicity the Moon's orbit as circular (it is not, but this is a good approximation for our purposes).
The gravitational acceleration that the Moon experience due to the gravitational attraction from the Earth is given by:
ag=G(MEarth+MMoon)/r2
Where G is the gravitational constant, M stands for mass, and r is the radius of the orbit. The centripetal acceleration is given by:
acentr=(4 pi2 r)/T2
Where T is the period. Since the two accelerations have to be equal, we obtain:
(4 pi2 r) /T2=G(MEarth+MMoon)/r2
Which implies:
r3/T2=G(MEarth+MMoon)/4 pi2=const.
This is the so-called third Kepler law, that states that the cube of the radius of the orbit is proportional to the square of the period.
This has interesting applications. In the Solar System, for example, if you know the period and the radius of one planet orbit, by knowing another planet's period you can determine its orbit radius. I hope that this answers your question.
