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Did i do this question right?

  • Thread starter lanvin
  • Start date
17
0
Mars travels around the sun in 1.88 {Earth} yrs,in an approximately circular orbit with a radius of 2.28 * 10^8km.Determine
{a}The orbital speed of mars {relative to the sun}


Answer {a}:
period {T} = 1.88 * 365 *24 * 3600 sec , R = 2.28 * 10^8 * 10^2 = 2.28 * 10^11
d = 2 pie R ------> 2 * 3.14 * 2.28 * 10^11
V=d / T -------> { 2 * 3.14 * 2.28 * 10^11 } / { 1.88 * 365 *24 * 3600 }
=2.4 * 10^4m/s


Is that correct?


or do I use

acceleration of centripetal motion = (4)(pi^2)(r)(f^2)

then use the acceleration and find velocity in

V = √(acceleration x radius)

which method is correct? if any...
 
Last edited:

Chi Meson

Science Advisor
Homework Helper
1,767
10
You did this correctly, though longishly.

And the second way is the same thing just with with extra steps. Notice that your initial solution was to find v=d/t, where d is the circumference, 2 pi r , and t is the period, T.

so v=(2 pi r)/T (all done right there)

and since f= 1/T , v=2 pi r f

Your second solution takes the square of this, divides by radius, then multiplies by radius, then square roots it.
 

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