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For its orbital motion around the sun, find the average daily angular displacement of earth?
Earth has an average distance from the sun of approximately 1.5 x 10^8 km. For its orbital motion around the sun, find the following:
a) average daily angular displacement
b) the average daily distance traveled (arc length)
if you need to know:
there are 365 1/4 days in a year.
earth has an equatorial radius of approximately 6380 km and rotates 360 degrees every 24 hours.
Help!! My physics teacher doesn't teach us anything.
2 Answers
- PearlsawmeLv 71 decade agoFavorite Answer
In 365.25 days it covers an angle of 360 degrees.
In one day it will cover = 360 /365.25 = 0.9856.
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Long ago it was thought that in 360 days, the sun comes to the same position in the celestial sphere. That is the reason why there are 360 degrees in a circle. Every day the sun moves towards East aproximately by 1 degree.
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The circumference of the orbit = 2*π*r = 2* π *1.5*10^8 = 9.42*10^8 km.
The distance covered in one day = 9.42*10^8 km /365.25 =
2.579055*10^6 km
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- 1 decade ago
Suppose we assume the orbit is circular , then the circumference is 2*pi*r = 2*3.14*1.5*10^8 = 9.42*10^8 km
So this distance is covered in 365.25 days ....so in one day it travels 2.579055*10^6 km daily ..... so this is the answer for (b). But ofcourse the orbit is elliptical so it varies a lot . If u can calculate the distance of elliptical path u can find out the exact answer....
