The Earth is 1.5 times 10^11 m from the Sun and takes a year to make one complet

Question

The Earth is 1.5 times 10^11 m from the Sun and takes a year to make one complete orbit. It rotates on its own axis once per day. It can be treated approximately as a uniform-density sphere of mass 6 times 10^24 kg and radius 6.4 times 10^6 m (actually, its center has higher density than the rest of the planet, and the Earth bulges out a bit at the equator). Using this crude approximation, calculate the following. (a) What is v_CM? (b) What is K_trans? (c) What is omega, the angular speed of rotation around its own axis? (d) What is K_rot? J (e) What is K_tot?

Explanation / Answer

a)
Revolving around sun is Cm
In 1 year, it completes 1 orbit
distance covered , d = 2*pi*r
d = 2*pi*(1.5*10^11 + 6.4*10^6)
= 9.425*10^11 m

time taken,
t = 1 year
=3.154*10^7 s

Vcm = distance covered / time taken
= (9.425*10^11)/(3.154*10^7)
= 2.99*10^4 m/s

b)
Ktrans = 0.5*m*Vcm^2
= 0.5*(6*10^24)*(2.99*10^4)^2
= 2.68*10^33 J

c)
time taken to rotate = 1 day = 86400 s
angle through which it rorate in 1 rotation = 2*pi radians
w = 2*pi/86400
= 7.27*10^-5 rad/s

d)
Moment of inertia of solid sphere, I = (2/5)*m*R^2
= (2/5)*(6*10^24)*(6.4*10^6)^2
= 9.83*10^37 Kg.m^2
K rot = 0.5*I*w^2
= 0.5 *(9.83*10^37)*(7.27*10^-5 )^2
= 2.6*10^29 J

e)
K tot = K rot + K trans =2.6*10^29 + 2.68*10^33 J = 2.68*10^33 J

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.