Question

We know that the Moon revolves around Earth during a period of 27.3 days. The average distance from the center of Earth to the center of the Moon is 3.84×108 m.
1)What is the magnitude of the acceleration of the Moon due to its motion around Earth?

Answer 1

Answer:

Explanation:

This is a circular motion questions

Where the oscillation is 27.3days

Given radius (r)=3.84×10^8m

Circular motion formulas

V=wr

a=v^2/r

w=θ/t

Now, the moon makes one complete oscillation for 27.3days

Then, one complete oscillation is 2πrad

Therefore, θ=2πrad

Then 27.3 days to secs

1day=24hrs

1hrs=3600sec

Therefore, 1day=24×3600secs

Now, 27.3days= 27.3×24×3600=2358720secs

t=2358720secs

Now,

w=θ/t

w=2π/2358720 rad/secs

Now,

V=wr

V=2π/2358720 ×3.84×10^8

V=1022.9m/s

Then,

a=v^2/r

a=1022.9^2/×3.84×10^8

a=0.0027m/s^2