Astrophysical background: The celestial equator and celestial pole are the line

Question

Astrophysical background: The celestial equator and celestial pole are the line and points respectively in the sky above the Earth's equator and north and south poles. The ecliptic is the plane of Earth's orbit around the Sun. (a) Model the Earth as a uniform sphere. Calculate the angular momentum of the Earth due to its spinning motion about its axis to three significant figures.

1-Magnitude

2- direction


(b) Calculate the angular momentum of the Earth due to its orbital motion about the Sun to three significant figures.
1- Mgnitude
2- direction


(c) Compare the two quantities of angular momentum.
Which is larger in magnitude?

5 ---Select--- rotational or orbital

6- By what factor?

Explanation / Answer

P.S : small {x} means multiplication & capital {X} means cross product !

m = 5.98 x 10^24 Kg (mass of Earth)

M = 1.991 x 10^30 Kg (mass of Sun)

R = 6.37 x 10^6 m (radius of Earth)

r = 1.496 x 10^11 m (distance between Sun & Earth)

G = 6.672 x 10^-11 N.m^2/Kg^2 (Universal gravitational constant)

I = (2/5)xmxR^2 (where (I) is the moment of inertia of a solid sphere ; like Earth)

I = 9.706x10^37 Kg.m^2

F = GxMxm/r^2 (where (F) is the gravitaional force between two masses ; like Earth & Sun)

from Newton second law ---> F=mxa

F=ma (where (a) is the centripetal acceleration given by a = v^2/r , where (v) is the linear speed)

GxMxm/r^2 = mxv^2/r (after some algebra)

v = (GxM/r)

v = 29798.74 m/s

now = v/r (where () is the angular speed)

= 1.992x10^-7 rad/s

L = Ix (where L is the angular momentum)

L = 1.933x10^31 Kg.m^2/s (it's direction is along the spin axis whether up or down)

B) Assuming Earth orbit is circular

L = rXp (another (general) form of angular momentum where p is the linear momentum)

p = mxv

L = rxmxvxsin() (where () is the angle between (r) & (p) , & =90 according to our assumption)

L = rxmxv

L = 2.666x10^40 Kg.m^2/s

P.s : I left part(C) for you !

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