Part A What is the magnitude of the orbital velocity, in m/s, of the earth aroun
ID: 1362096 • Letter: P
Question
Part A
What is the magnitude of the orbital velocity, in m/s, of the earth around the sun? The orbit radius of the earth is 1.50×1011m.
Answer: v= 2.98 x 10^4 m/s
Part B
What is the radial acceleration, in m/s2, of the earth toward the sun?
Answer: 5.93 x 10^-3 m/s^2
Part C
What is the magnitude of the orbital velocity, in m/s, of Mercury around the sun? The orbit radius of the Mercury is 5.79×1010m, the orbit period of Mercury is 88.0 d.
Part D
What is the radial acceleration, in m/s2, of Mercury toward the sun?
Explanation / Answer
Each planet moves in a circular orbit and therefore has acceleration a(rad) = v^2/R.
The radius of the earth’s orbit is r =1.50*10^11 m and its orbital period is
T = 365 days = 3.16*10^7 sec.
For Mercury, r = 5.79*10^10 m and T = 88.0 days = 7.60*10^6 sec
A. v = 2*pi*r/T = 2*3.14*1.5*10^11/(3.16*10^7) = 2.98*10^4 m/s
B. a = v^2/r = (2.98*10^4)^2/(1.5*10^11) = 5.92*10^-3 m/sec^2
C v = 2*3.14*5.79*10^10/(7.6*10^6) = 4.78*10^4 m/s
D. a = (4.78*10^4)^2/(5.79*10^10) = 3.96*10^-2 m/sec^2