What is the linear speed of Mercury in its orbit about the sun? Give answer in meters per second. (Consider the orbit of Mercury to be a circle of radius 5.79×1010 m and that it takes 0.24 yr for one orbit around the sun.)
2007-02-13 09:17:56 · 2 answers · asked by avemaria 2 in Science & Mathematics ➔ Astronomy & Space
The radius is 5.79 × 10^10? The the circumference would be about 36.38 × 10^10.
An Earth-year is about 365.25 × 24 × 60 × 60 seconds long (= 31557600 sec), but a Mercury year is 24% of this, i.e., 7573824 sec
So the answer is to be given as m/s, so just divide:
363800000000 ÷ 7573824,
which is about 4.8 × 10^4.
Feel free to email me for further clarifications.
PS, omg, finally, someone else who uses the correct '×' symbol for multiplication! (as opposed to x, X, *, and so on...)
2007-02-13 09:55:15 · answer #1 · answered by Bog-man 4 · 0⤊ 0⤋
C = 2 pi (5.79E+10 m) = 3.64E+11 m
T = 0.24 years (31556926 sec/year) = 7573662 sec
V = C/T = 48061 m/s
2007-02-13 10:19:35 · answer #2 · answered by Anonymous · 0⤊ 0⤋