How to Calculate the Mass of Jupiter Using Orbital Data

In summary: From what I can see, the formula you want is \frac{2πr}{T}=\sqrt{\frac{GM}{r}}. With M being the mass of Jupiter, G being the gravitational constant, r being the radius from the centre of Jupiter, and T being the period. So, T^2=GM^2. So, T^2=GM^2.
  • #1
bona0002
15
0
Hi guys,

I'm not sure if I'm going about this the correct way, but it seems to be the only one that makes sense right now. The problem reads: Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22E5 km. From these data, determine the mass of Jupiter.

So, with that in mind, the equation that pops out at me is T^2 = (4*pi^2)(a^3)/(G*M_big_). Now, assuming that M_big_ is the size of jupiter, one can solve for M_big_: M_big_ = (4*pi^2)(a^3)/(T^2*G). Before substituition, I convered 1.77 days into 1.53E5 seconds. Then, I substituted: M_big_ = ((4*pi^2)*(4.22E5km)^3)/((1.53E5)^2)*(6.67E-11) = 1.07E7 kg.

So, is my process right and I'm simply punching it in wrong, or is it that my logic is flawed?

Thanks!
 
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  • #2
bona0002 said:
Hi guys,

I'm not sure if I'm going about this the correct way, but it seems to be the only one that makes sense right now. The problem reads: Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22E5 km. From these data, determine the mass of Jupiter.

So, with that in mind, the equation that pops out at me is T^2 = (4*pi^2)(a^3)/(G*M_big_). Now, assuming that M_big_ is the size of jupiter, one can solve for M_big_: M_big_ = (4*pi^2)(a^3)/(T^2*G). Before substituition, I convered 1.77 days into 1.53E5 seconds. Then, I substituted: M_big_ = ((4*pi^2)*(4.22E5km)^3)/((1.53E5)^2)*(6.67E-11) = 1.07E7 kg.

So, is my process right and I'm simply punching it in wrong, or is it that my logic is flawed?

Thanks!

Change km to m.
 
  • #3
rude man said:
Change km to m.

I did. The answer I get then is 1.07E10, which is supposedly incorrect.
 
  • #4
bona0002 said:
I did. The answer I get then is 1.07E10, which is supposedly incorrect.

From what I can see, the formula you want is [itex]\frac{2πr}{T}=\sqrt{\frac{GM}{r}}[/itex]

With M being the mass of Jupiter, G being the gravitational constant, r being the radius from the centre of Jupiter, and T being the period.

Good luck :)
 
  • #5
Alright, figured it out. Turns out I had the process down just fine, but just that when punching in the numbers on my calculator, I forgot to cube a. Thanks for the help guys!
 

1. How does Newton's law of universal gravitation work?

Newton's law of universal gravitation states that any two objects in the universe are attracted to each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This means that the larger the masses of the objects and the closer they are to each other, the stronger the gravitational force between them.

2. What is the formula for calculating gravitational force?

The formula for calculating gravitational force is F = G(m1m2)/r^2, where F is the gravitational force, G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between them.

3. How does the distance between two objects affect the gravitational force between them?

The distance between two objects has an inverse square relationship with the gravitational force between them. This means that as the distance between two objects increases, the force of gravity decreases exponentially.

4. What is the difference between mass and weight in relation to universal gravitation?

Mass is a measurement of the amount of matter in an object, while weight is a measurement of the force of gravity acting on an object. In relation to universal gravitation, mass is used to calculate the gravitational force between two objects, while weight is the result of that force acting on an object.

5. Can universal gravitation be applied to objects on Earth?

Yes, universal gravitation can be applied to objects on Earth. This is because the law of universal gravitation applies to all objects in the universe, including those on Earth. The gravitational force between two objects on Earth is simply much smaller compared to objects in the universe due to the larger masses and distances involved.

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