Centripetal motion and universal gravitation question: Mars and Sun questionby zeion Tags: centripetal, gravitation, mars, motion, universal 

#1
Mar109, 09:07 AM

P: 467

1. The problem statement, all variables and given/known data
Mars travels around the Sun in 1.88 (Earth) years, in an approximately circular orbit with a radius of 2.28 x 10^8 km Determine a) the orbital speed of Mars (relative to the Sun) b) the mass of the Sun 2. Relevant equations acceleration centripetal = 4(pi^2)(r) / (T^2) universal attraction = (G)(m1)(m2) / (radius)^2 3. The attempt at a solution Given: T = 1.88 years = 59287680s = 5.9x10^7s r = 2.28 x 10^8km = 2.28 x 10^11m Orbital speed means centripetal acceleration yes? Then, acceleration centripetal = 4(pi^2)(r) / (T^2) acceleration centripetal = 4(9.869604401)(2.28x10^11m) / (34.81x10^14s) acceleration centripetal = 90.01x10^11m / 34.81x10^14s acceleration centripetal = 2.59x10^3m/s = 0.00259m/s How come its so slow? 



#3
Mar109, 09:36 AM

P: 467

Ok so,
Mars travels around the Sun once every 1.88 years. So I have to find the distance of that circle and divide it by the time to find the speed. I have the radius, so I have to find the perimeter. Perimeter of circle is 2(pi)(r) So the distance traveled is 14.33x10^11m In 1.88 years = 5.9x10^7s So the speed is 14.33x10^11m / 5.9x10^7s = 2.43x10^4m/s yes? 



#4
Mar109, 09:44 AM

Mentor
P: 40,877

Centripetal motion and universal gravitation question: Mars and Sun question
Good.




#5
Mar109, 06:33 PM

P: 467

Now can I use this formula to solve for mass of Sun?
V^2 = (G)[m1(sun)] / r [m1(sun)] = (V^2)(r) / (G) [m1(sun)] = (2.43x10^4m/s)^2(2.28 x 10^11m) / [6.67x10^11(N)(m^2)/kg^2] [m1(sun)] = (5.9x10^8m^2/s^2)(2.28 x 10^11m) / [6.67x10^11(N)(m^2)/kg^2] [m1(sun)] = (13.45x19^19m^3/s^2) / [6.67x10^11(N)(m^2)/kg^2] [m1(sun)] = 2.02x10^8(N)m/kg^2 Why are the units all weird? 


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