Centripetal motion and universal gravitation question: Mars and Sun question

zeion

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?

Doc Al

No. Speed is distance divided by time, not acceleration.

zeion

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?

Doc Al

Good.

zeion

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?

Doc Al

Good.
In your last step you didn't divide the units properly. You should have gotten:
[m^3/s^2]/[(N)(m^2)/kg^2] = [m^3/s^2]*[kg^2/(N)(m^2)] = [m kg^2]/[N s^2]

To simplify further, express Newtons in terms of more fundamental units:
N = kg-m/s^2