Centripetal motion and universal gravitation question: Mars and Sun question


by zeion
Tags: centripetal, gravitation, mars, motion, universal
zeion
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#1
Mar1-09, 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?
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Doc Al
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#2
Mar1-09, 09:12 AM
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Quote Quote by zeion View Post
Orbital speed means centripetal acceleration yes?
No. Speed is distance divided by time, not acceleration.
zeion
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#3
Mar1-09, 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?

Doc Al
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#4
Mar1-09, 09:44 AM
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Centripetal motion and universal gravitation question: Mars and Sun question


Good.
zeion
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#5
Mar1-09, 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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#6
Mar2-09, 06:35 AM
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Quote Quote by zeion View Post
Now can I use this formula to solve for mass of Sun?
V^2 = (G)[m1(sun)] / r
Good.
[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
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


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