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Can someone please help me with a planetary acceleration problem?

 
 
Reply Wed 28 Jul, 2010 02:54 am
Someone posted this question on Yahoo! Answers and I want to answer but I'm not sure how:

What is the acceleration of Jupiter produced by the gravitational force ?
When 'g' between jupiter and the sun is 4.19 * 10 ^23 N, and Jupiter's orbital radius is 778.3 x 10^9 m.
If possible use the same information to calculate the acceleration of the Sun.

How do I set this up so I can solve this? Please help me out-thank you.
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fresco
 
  1  
Reply Wed 28 Jul, 2010 03:06 am
@JGoldman10,
I would have thought you needed information on mass to do this calculation.
JGoldman10
 
  1  
Reply Wed 28 Jul, 2010 04:44 am
@fresco,
Hi, is this solution correct?:

F (force of attraction) = G (M(s) m/ r^2 = mv^2/ r = mg

where m is Jupiter's mass and M(s) is the Sun's (~ 2 x 10^30 kg)
m = 4.19 * 10 ^23 N = (4.19 * 10 ^23/9.81) kg = 4.27115189 × 10^22 kg
G = 6.67428 * 10^(-11) N(m/kg)^2

G[M(s)]m/(r^2) = 6.67428 * 10^(-11) N(m/kg)^2 * 2 x 10^30 kg * 4.27115189 × 10^22 kg/(778.3 x 10^9 m)^2 = 9.41207487 × 10^18 newtons

Jupiter's acceleration: 9.41207487 × 10^18 newtons/(4.27115189 × 10^22 kg) = 0.000220363853 m / s^2

Sun's acceleration: 9.41207487 × 10^18 newtons/(2 x 10^30 kg) = 0.000220363853 m / s^2
raprap
 
  1  
Reply Wed 28 Jul, 2010 05:03 am
@JGoldman10,
The formula is Correct, but the math is wrong.

Use a little knowledge to sanity check your answers. The acceleration of gravity of the Earth is 9.8 m/s^2 and the Earth has a smaller acceleration than the more massive bodies of the sun and Jupiter.

Ex

F=G*ms*mj/d^2 where G is the gravitational constant, ms & mj are the mass of the sun and Jupiter, and d is the distance between centers.

Hint--cancel units algebraically to check answer and convert to Nt

Then use F=ms*a to calculate the gravitational acceleration of the sun.

Finally compare the calculated solar acceleration to 9.8m/s^2 to see if the answer makes sense.

Rap

JGoldman10
 
  1  
Reply Wed 28 Jul, 2010 10:52 am
@raprap,
That math was wrong:

Jupiter's acceleration: (9.41207487 × 10^18 newtons)/(4.27115189 × 10^22 kg) = 0.000220363853 m / s^2

Sun's acceleration:
(9.41207487 × ((10^18) newtons)) / (2 x 10^30 kg)) = 4.70603743 × 10^(-12) m / s^2
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JGoldman10
 
  1  
Reply Wed 28 Jul, 2010 01:07 pm
@raprap,
Hi raprap, ignore that last comment. This is the correct math,

F (force of attraction) = G (M(s) m/ r^2 = mv^2/ r = mg

where m is Jupiter's mass and M(s) is the Sun's (~ 2 x 10^30 kg)

First we have to find the mass of Jupiter m:

F= G[M(s)]m/(r^2)

m = F(r^2)/[GM(s)] = (4.19 * 10 ^23 N)(778.3 x 10^9 m)^2/([6.67428 * 10^(-11) N(m/kg)^2][2 x 10^30 kg]) = 1.90140077 × 10^27 kilograms

Acceleration of Jupiter: (4.19 * 10 ^23 N)/(1.90140077 × 10^27 kilograms) = 0.000220363853 m / s^2

Sun's acceleration:
(4.19 * 10 ^23 N)/(2 x 10^30 kg) = 2.09500 × 10^(-7) m / s^2
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