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Mechanics. Gravitation and gravity

  1. May 4, 2015 #1
    1. The problem statement, all variables and given/known data
    The average orbital radii about the Sun of the Earth and Mars are 1.5*11^11 m and 2.3*11^11 m respectively. How many (Earth) years does it take Mars to complete its orbit?

    Answer: 1.9 years.

    2. Relevant equations
    F = G * ((m1*m2) / r^2)
    F = m*(w^2)*r
    T^2 = (4Pi^2 / G*m)*r^3; m = m sun
    g = (4Pi^2*rm^3)/(T^2*re^2); rm = radius Moon, re = radius Earth.

    3. The attempt at a solution
    I get the answer using Kepler's third law.
    http://www.studyphysics.ca/newnotes/20/unit02_circulargravitation/chp08_space/lesson34.htm
    K = T^2 / r^3
    T^2 = K * r^3
    T^2 = 3.95*10^-29 * 2.3*10^11 = ans
    T = ans^1/2
    T = 693.25 days
    -> 1.899 years = 1.9 years.

    But I can't get the required answer using book formulas (A-Level Physics). E.g.: using the last equation I got (4*(pi^2)*((2.3*10^11 m)^3))/((9.8 ms^-1*((1.5*10^11 m)^2))^1/2)=1.02291 × 10^24 and if it is seconds -> 32.436 quadrillion years... Not even close to the 1.9 years answer.

    Any suggestions? Thank you in advance.
     
  2. jcsd
  3. May 4, 2015 #2
    I don't understand your last equation. Please elaborate the cases in which it is used. And yes..... learn a little bit of LaTex. It helps you to put your questions effectively. If not interested then you can always use the symbols given in the template when you click the sigma button. And those ##x^2## and ##x_2## options also help to put superscript and subscript.
     
  4. May 4, 2015 #3
    Possibly you did not apply the equation properly. For one thing, what is that 9.8 m/s value?
     
  5. May 4, 2015 #4
    b756b2647794.jpg
    So I changed rM to my given Mars radius and calculated.

    The formula is derived from G * mmEarth / rEarth^2 = mg. "Newton's test of the inverse square law".
     
  6. May 4, 2015 #5
    Did you also change the ##r_{earth}## to ##r_{sun}##? Since mars is orbiting sun (not earth).
     
  7. May 4, 2015 #6
    I though you will use the formula above this one. It makes more sense. Why bother with g?
    I hope you are aware that in that (last formula) rE is the radius of the Earth (not of some orbit).
    If you want to use this you will need g on the surface of the Sun and radius of the Sun.
    Do yo understand how these formulas were derived?
     
  8. May 5, 2015 #7
    Hm, I recalculated with both Sun and Mars radii, but didn't work out + as nasu pointed out, the previous formula is indeed a better way to solve the problem.

    I used the mentioned formula and got the 1.9 years answer:
    T2 = (4π2 * (2.3*1011)3) / (6.7*10-11 * 1.989*1030)
    T = 60 036 702.33 s
    /60 (minutes) / 60 (hours) / 24 (days) / 365 (years) = 1.9 years

    Though in that case the given Earth's orbit radius is of no use and I looked for the mSun number.

    Yes indeeed the r is the planet radius and not orbit.

    So now I have 2 formulas to solve the problem (the first one is in the first post in the attempts). But how can I solve the question using both orbital radii of Mars and Earth as given in the question?
     
    Last edited: May 5, 2015
  9. May 5, 2015 #8
    You write the formula that gives T^2 for Earth and Mars (separately) and then take the ratio of the two. The mass of the sun will simplify so you don't need to look it up.
    Actually you will end up with Kepler's law.
    Or, on a longer path, but one that some students may find more intuitive, find mass of the Sun from the equation written for Earth (you know period) and then plug in the equation for Mars.
     
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