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(a) Calculate the magnitude of the angular momentum of the earth in a circular orbit around the sun. Is it reasonable to model it as a particle? (b) Calculate the magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles, modeling it as a uniform sphere. Consult Appendix E and the astronomical data in Appendix F.

Respuesta :

Answer:

(a) [tex]L=2.6742\times 10^{40}\, kg.m^2.s^{-1}[/tex]

(b) [tex]L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}[/tex]

Explanation:

For Earth we have:

  • mass of earth, [tex] m_E=5.97\times 10^{24}\, kg[/tex]
  • radius of earth, [tex] R_E=6.38\times 10^6m[/tex]
  • orbital radius, [tex]r=1.5 \times 10^{11}m[/tex]
  • period of rotation, [tex]t_{rot}=24h=86400\, s[/tex]
  • period of revolution, [tex]t_{rev}= 1 yr=3.156\times 10^7 s[/tex]

(a)

Angular momentum, [tex]L=?[/tex]

∵[tex]L=I.\omega[/tex]...............................(1)

For a particle of mass m moving in a circular path at a distance r from the axis,

[tex]I=m.r^2[/tex]

& [tex]v=r.\omega[/tex]

Putting respecstive values in eq. (1)

[tex]L=m_E\times r^{2}\times \omega[/tex]

[tex]L=5.97\times 10^{24}\times (1.5 \times 10^{11})^2\times \frac{2\pi}{3.156\times 10^7}[/tex]

[tex]L=2.6742\times 10^{40}\, kg.m^2.s^{-1}[/tex]

To model earth as a particle is  reasonable because the distance between the sun and the earth is very large s compared to the radius if the earth.

(b)

For a uniform sphere of mass M and radius R and an axis through its center,  [tex]I=\frac{2}{5}M.R^2[/tex]

using eq. (1)

[tex]L_s=(\frac{2}{5}m_E.R_E^2) \omega[/tex]

[tex]L_s=\frac{2}{5} \times 5.97\times 10^{24} \times 6.38\times 10^6\times \frac{2\pi}{86400}[/tex]

[tex]L_s=7.07 \times 10^{23} \, kg.m^2.s^{-1}[/tex]

jayilych4real

The magnitude of the angular momentum of the earth in a circular orbit around the sun is Li= 2.6742 x 10^40 kg.m^2.s^-1

It is reasonable to model the earth as a particle because the distance between the sun and the earth is very large when compared to the radius of the earth.

The magnitude of the angular momentum of the earth due to its rotation around an axis through the north and south poles is L8= 7.07 x 10^23kg.m^2.s^-1

Calculations and Parameters:

For Earth:

  • mass of earth, mE= 5.97 x 10^24kg
  • the radius of the earth, RE= 6.38 x 10^6m
  • orbital radius, r= 1.5 x 10^11m
  • period of rotation, t=24h = 86400s
  • period of revolution t= 1 year = 3.156 x 10^7

To find the angular momentum,

L=mE x r^2 x w

L= 2.6742 x 10^40kg.m^2.s^-1

For a uniform sphere of mass M and radius R and an axis through its center,  

I= 2/5M.R^2

L8= 7.07 x 10^23kg.m^2.s^-1

Read more about angular momentum here:
https://brainly.com/question/7538238

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