as 1 astronomical unit (au). The mean radius of the Moon orbit around the Earth is 0.002570 au. The mean radius of the Earth at the equator is 6378 km. The distance from the Earth of the Moon is equivalent of how many trips around the circumference of the Earth of the equator? thanks!
2007-07-04 19:21:31 · 4 answers · asked by help_with_physics 1 in Science & Mathematics ➔ Mathematics
the circumference of the Earth of the equator
= 2 pi r= 2 pi 6378 = 40 000 km
The mean radius of the Moon orbit around the Earth is =0.002570 au = 0.002570*149 milion km = 382930 km
The answer is
382930 / 40000= 9.57325 times
2007-07-04 19:38:10 · answer #1 · answered by zohair 2 · 0⤊ 1⤋
Now i'm doing all of this p/p no calc all brain so lets hope its right:
if radius of Earth is 6378km, then circumference is 2(pi)6378, or about 39,603.84 km...
Since the mean distance is .00257 AU, that is the same as .00257 times 1.496x10^8 km, or about 3.845x10^5 km, or 384,500 km, so i'd say about 10 trips, maybe 9.9?
2007-07-04 19:34:38 · answer #2 · answered by Kirk 1 · 0⤊ 1⤋
1.496X10*8 * .00257 = 384,472 km
Circumference of Earth = 6.28*6378=40,054 km
384472/40054=9.6 = trips arond earth's equator.
2007-07-04 19:40:46 · answer #3 · answered by ironduke8159 7 · 0⤊ 1⤋
2π(6,378) ≈ 40,074 km
(1.496*10^8 km)(0.002570) ≈ 384,472 km
(384,472 - 6,378)/40,074 ≈ 9.4349 trips (to Moon orbit from Earth's surface)
2007-07-04 20:23:20 · answer #4 · answered by Helmut 7 · 1⤊ 0⤋