PHYS 201 Fall 2017 17-Circular Motion 7. Assume the earth travels in uniform cir
ID: 1791458 • Letter: P
Question
PHYS 201 Fall 2017 17-Circular Motion 7. Assume the earth travels in uniform circular motion around the sun.Mo a. What is the period of the earth's orbit (first in years, then in seconds)? su 2.0y 070 b. What is the frequency of the earth's orbit (in 1/s)? What is the angular velocity of the earth? (You will need to know that the circle the earth goes around the sun has a radius of r-1.50 × e. 1011 m) d. What is the speed of the earth? e. What is the acceleration of the earth? What is the force on the earth? (You will need to know that the mass of the earth is m - 5.97 × 1024 kg) f.Explanation / Answer
a)
For circular motion, force of gravity will provide the necessary centripetal force
So, GMm/R^2 = mv^2/R
here , G = gravitational constant = 6.67*10^-11
M = mass of sun = 1.99*10^30 kg
m = mass of earth
R = distance between earth and sun = 1.496*10^11 m
So, v = sqrt(GM/R) = 2*pi*R/T
So, T = 2*pi*R^(3/2)/sqrt(GM)
= 2*pi*(1.496*10^11)^1.5/sqrt(6.67*10^-11*1.99*10^30)
= 3.1556*10^7 s <------------ answer
= 3.15564*10^7/(3600*24*365) years
= 1 year <----------- answer
b)
frequency, f = 1/T = 1/(3.1556*10^7)
= 3.17*10^-8 s^-1
c)
angular velocity , W= 2*pi*f = 2*pi*3.17*10^-8
= 1.99*10^-7 rad/s
d)
speed of earth, v = W*R = 1.99*10^-7*(1.496*10^11)
= 2.98*10^4 m/s
e)
acceleration = v^2/R = (2.98*10^4)^2/(1.496*10^11)
= 0.00594 m/s2
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