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23 Nov 2019

If on September fifteenth the north pole of the planet Earth istilted 15 degrees from the Sun.Rotation period of sun T= 27 daysand its axis points up. a)What would be the magnitude and directionof angular momentum vector (L) for this system at that exact day?b)How to find the moment of inertia (I in kg*m^2) of the Earthrotating on its axis at that day. c)How to find the ang. momentumfor Earth rotating on its axis (L in (kg*m^2)/s). d) What would bethe angular momentum vector of Sun rotating on its axis?. e) Whatwould be the total ang. momentum of the system?

I calculated the moment of inertia of earth and got I =9.72*10^37 kgm^2

Angular momentum of earth L=7.07*10^33 (kgm^2)/s

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

If on New Years Eve the North pole of the planet Earth is tilted 23degrees away from sun. The mass of the planet Earth is 5.97*10^24kg . Radius of the Earth is 6.38*10^6 m. The mass of sun is1.99*10^30 kg. The radius of the sun is 6.96*10^8 m.ç a) What wouldbe the moment of inertia of earth rotating around its axis, earthrotating around sun axis. B) What would be the angular momentum ofearth rotating on its axis, earth rotating around the sun, sunrotating around its axis. C ) What would be the total angularmomentum of the system?

I solved :
Moment of inertia Erath around its axis is I = 2/5MR^2 =2/5(5.97*10^24 kg)*(6.38*10^6m)^2=9.72*10^37 kgm^2
Angular momentum of the earth around its axis is L=I*omega =(9.72*10^37kgm^2)8(2pi/86400 sec/day) = 7.07*10^33 (kgm^2)/s
Moment of inertia earth around sun is I = MR^2 = (5.97*10^24kg)*(1.50*10^11m) = 4.48*10^52 kgm^2
Angular momentum Earth around sun is L=I*omega =2/5(1.99*10^30kg)(6.96*10^8m)^2(2.42*10^-6rad/s)=1.34*10^33kgm^2/s
Total momentum of the system =8.91*10^45 kgm^2/sß

Did I solve this problem correctly? Please help.

calculate the angular momentum of the earth in its orbit around the sun?

 

 

m   5.97 10 kg 6 E R   6.38 10 m Orbital radius 11 r   1.50 10 m Period of rotation Prot = 24 h 86,400 s  Period of revolution Prev = 7 1 y 3.156 10 s   Identify: L I z z   Set Up: For a particle of mass m moving in a circular path at a distance r from the axis, 2 I mr  and v r   . For a uniform sphere of mass M and radius R and an axis through its center, 2 2 5 I MR  . The earth has mass 24 E m   5.97 10 kg , radius 6 E R   6.38 10 m and orbit radius 11 r   1.50 10 m . The earth completes one rotation on its axis in 24 h 86,400 s  and one orbit in 7 1 y 3.156 10 s   . Execute: (a) 2 24 11 2 40 2 7 2 rad (5.97 10 kg)(1.50 10 m) 2.67 10 kg m /s 3.156 10 s L I mr z z z                   . The radius of the earth is much less than its orbit radius, so it is very reasonable to model it as a particle for this calculation. (b)   2 2 2 24 6 2 33 2 5 5 2 rad (5.97 10 kg)(6.38 10 m) 7.07 10 kg m /s 86,400 s L I MR z z                  Evaluate: The angular momentum associated with each of these motions is very large. Note that the orbital angular momentum is over 7 orders of magnitude greater than its rotational angular momentum. In fact, most of the total angular momentum of the solar system comes from the sum of the orbital angular momenta of its planets.

Earth is rotating about the Sun in our solar system. The distance between the Earth and Sun
is 1.5 × 1011m. Consider the impact of an asteroid of mass 8.9 ×1015 kg and a speed of 90,000
km/h with respect to the Earth’s surface. The asteroid strikes the Earth at the equator as shown
in Figure. Consider the mass and radius of the Earth are ME = 5.97 × 1024 kg and RE = 6.38
×106 m, respectively. Assume that the axial and orbital axis of rotations are parallel to each
other.

(a) Calculate the axial and total angular momentum of the Earth. (4 marks)
(b) What is the change in the angular velocity of the Earth due to the collision? (6 marks)
(c) What is the maximum possible change in the length of a day due to the collision?
(3 marks)
(d) How much energy is needed to stop the Earth’s rotation about its axis after the collision?
(4 marks)
(e) What would be the moment of inertia of the Earth if the axis of rotation is parallelly
shifted so that it touches the equator?

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