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23 Nov 2019

What is the speed of Earth around the Sun if 1 AU = 1.50x10^11m,and the Sun's mass is 1.987x10^30kg.

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Casey Durgan
Casey DurganLv2
6 Feb 2019

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Related textbook solutions

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

Related questions

Due to the gravitational interaction between the two stars in a binary system each one moves in a circular orbit around their center of mass. One star has a mass of 15.0x10^30kg; its center is located at x=1.0 AU and y=2.0 AU. Find the center of mass of the system composed of the two stars. (AU stands for astronomical unit. 1 AU= average distance between the Earth and the Sun -1.50x10^8 km.)

Sun
-1.99x10^30kg mass
-Dist. Frm erth=1.5x10^11m
-Avg enrgy perunittime,perunitarea that meets upper atmosph fromsun = 1.34kW/m^2. The sun gives off this enrgy in a compilation offusion rxn’s w/ the result of 4 H nucleuses fusing together, andcreating 1 He nucl. This rxn gives off 25.8MeV. Approximate ofexistence time of sun(years), if sun is mostly made up of H, andburns till ten percent of H is eaten.

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.

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