Asked by Brenda
I found the average acceleration of earths orbit around the sun over a period of 5 days by calculating delta theta and dividing by radius and 1/T.
I got the right answer but as I am studying for my test, I am seeing the units do not work...does this make sense? Can someone derive that equation for me?
So here is what I have:
theta = 0.086 rads
time = 432,000 seconds (5 days)
radius = 6.371x10^6
(0.086 x 432,000)/6.371x10^6 = 0.006
I got the right answer but as I am studying for my test, I am seeing the units do not work...does this make sense? Can someone derive that equation for me?
So here is what I have:
theta = 0.086 rads
time = 432,000 seconds (5 days)
radius = 6.371x10^6
(0.086 x 432,000)/6.371x10^6 = 0.006
Answers
Answered by
Elena
You accidentally got the right answer.
Correect solution is following:
The Earth makes one revolution (2π) during 1 year => the angular displacement for 5 days is
θ= 2π•5/365 = 0.086 rad.
The angular velocity of the Earth is
ω= θ/t = 0.086/5•24•3600 =2•10⁻⁷ rad/s
If R = 1.496•10¹¹ m, the centripetal acceleration of the Earth is
a=ω²•R =(2•10⁻⁷)²•1.496•10¹¹ = 0.006 m/s²
Correect solution is following:
The Earth makes one revolution (2π) during 1 year => the angular displacement for 5 days is
θ= 2π•5/365 = 0.086 rad.
The angular velocity of the Earth is
ω= θ/t = 0.086/5•24•3600 =2•10⁻⁷ rad/s
If R = 1.496•10¹¹ m, the centripetal acceleration of the Earth is
a=ω²•R =(2•10⁻⁷)²•1.496•10¹¹ = 0.006 m/s²
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