Kepler's Laws-What am I doing wrong here?

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In summary, Kepler's Laws--What am I doing wrong here? states that:- Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22 x 10^5km.- From these data, determine the mass of Jupiter.- Homework Equations states that:- T^2=(4(pi)^2/GM)r^3- The Attempt at a Solution states that:- I converted the radius from km to meters and substituted all of the numbers to their variables. However, I still came up with the wrong answer. Please help me check my work, thanks!- Div
  • #1
AznBoi
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Kepler's Laws--What am I doing wrong here?

Homework Statement


Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22 x 10^5km. From these data, determine the mass of Jupiter.


Homework Equations


T^2=(4(pi)^2/GM)r^3


The Attempt at a Solution


I converted the radius from km to meters and substituted all of the numbers to their variables. However, I still came up with the wrong answer. Please help me check my work, thanks!

Divided r^3 from both sides:
(1.77)^2/(4.22x10^8m)^3= 4(pi)^2/(6.673 x 10^-11)M

Multiplied universal gravitation constant with M(mass) on both sides:
(4.1688x10^-26)(6.673x10^-11)M=4(pi)^2

solved for M: M=4(pi)^2/ (2.78184 x 10^-36)

I came up with: 1.4192 x 10^37

The correct answer is: 1.90 x 10^27 kg

What am I doing wrong? Thanks!
 
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  • #2
Don't plug in numbers until you're done with the algebra.

Solve for M:

[itex]T^2 = \frac{4 \pi^2 r^3}{G M}[/itex]

[itex]M = \frac{4 \pi^2 r^3}{G T^2}[/itex]

Convert your units into mks:

[itex]r = 4.22 \cdot 10^8\,m[/itex]

[itex]T = 152928\,s[/itex]

Plug into the equation, and you're done. You appear to have left your T in units of days, which is inconsistent with the SI mks system.

- Warren
 
Last edited:
  • #3
Ah I see.. So is every single variable in Physics in the SI unit? Thanks a lot!
 
  • #4
You're free to use any unit system you want, but you must be consistent. If you want to use G = 6.673 x 10^-11, you must understand its units:

[itex]G = 6.67300 \cdot 10^{-11} \,\textrm{m}^{3} \,\textrm{kg}^{-1} \,\textrm{s}^{-2}[/itex]

That means that everything else in your equation must in terms of meters, kilograms, and seconds: the standard mks SI system.

If you really wanted to use days as your fundamental unit of time, you could, but G would be:

[itex]G = 0.498136781 \,\textrm{m}^{3} \,\textrm{kg}^{-1} \,\textrm{day}^{-2}[/itex]

- Warren
 
  • #5
chroot said:
You're free to use any unit system you want, but you must be consistent. If you want to use G = 6.673 x 10^-11, you must understand its units:

[itex]G = 6.67300 \cdot 10^{-11} \,\textrm{m}^{3} \,\textrm{kg}^{-1} \,\textrm{s}^{-2}[/itex]

That means that everything else in your equation must in terms of meters, kilograms, and seconds: the standard mks SI system.

If you really wanted to use days as your fundamental unit of time, you could, but G would be:

[itex]G = 0.498136781 \,\textrm{m}^{3} \,\textrm{kg}^{-1} \,\textrm{day}^{-2}[/itex]

- Warren

Wow, thanks for all the superfluous replies! Okay, I understand it clearly now. Thanks to you! :smile:
 
  • #6
No problem. Try plugging in your values for r in meters and T in days, using the "version" of G that I provided. :wink:

- Warren
 
  • #7
AznBoi said:
Wow, thanks for all the superfluous replies! Okay, I understand it clearly now. Thanks to you! :smile:

Your adjective "superfluous" is incorrectly applied in this context. Please re-check the definition. Thank you.
 
  • #8
Yeah, sometimes I don't know if the vocabulary I've learned makes sense in what I say/write. Here is the definition: Being beyond what is required or sufficient. How come it doesn't work with chroot's replies? :biggrin:
 
  • #9
Well, the word "superfluous" usually means that something was pointless or needless, not that it was unexpectedly generous.

You might say "Man, this textbook is horrible! It's full of superfluous sidebars," in order to disparage it. You probably wouldn't tell your mom "Thanks for all these superfluous Christmas gifts!"

- Warren
 
  • #10
chroot said:
Well, the word "superfluous" usually means that something was pointless or needless, not that it was unexpectedly generous.

You might say "Man, this textbook is horrible! It's full of superfluous sidebars," in order to disparage it. You probably wouldn't tell your mom "Thanks for all these superfluous Christmas gifts!"

- Warren

Oh I see why berkeman said that to me then. Sorry if you thought of it as the other way. My vocabulary isn't that great and it's still developing. I mean I know the definitions to some words but I just haven't had much experience with them. That's why some of my sentences are weird lol. Alright, thanks for the vocab. lesson xD I'll try not to use the word superfluous as something that is too super. Thanks to both of you.
 

1. What are Kepler's Laws?

Kepler's Laws are a set of three fundamental laws that describe the motion of planets around the sun. They were developed by the German astronomer Johannes Kepler in the early 17th century.

2. What is the first law of Kepler?

The first law, also known as the law of ellipses, states that all planets move in elliptical orbits with the sun at one focus. This means that the distance between the planet and the sun varies throughout its orbit.

3. What is the second law of Kepler?

The second law, also known as the law of equal areas, states that a line drawn from a planet to the sun will sweep out equal areas in equal time intervals. This means that a planet moves faster when it is closer to the sun and slower when it is farther away.

4. What is the third law of Kepler?

The third law, also known as the law of harmonies, states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. In simpler terms, this means that the farther a planet is from the sun, the longer it takes to complete one orbit.

5. What am I doing wrong if Kepler's Laws do not seem to apply?

If Kepler's Laws do not seem to apply, it is likely that the object in question is not orbiting a central body, such as the sun. Kepler's Laws only apply to objects that are orbiting a massive central body, like planets orbiting the sun. If you are observing an object that is not following these laws, it may be due to other factors such as gravitational perturbations from other objects or non-circular orbits.

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