Solving Problem 9: Smaller than Any of the Choices?

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In summary, The problem involves finding the mass of a planet based on the orbital period and radius of its moons. It is not necessary to assume a circular orbit. Kepler's Third Law can be used, where the ratio of the radius cubed to the period squared must be constant for all objects orbiting the planet. This ratio can be found using data from one moon and then applied to the period of another moon to solve for the planet's mass.
  • #1
jimbo71
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Homework Statement


see problem 9 attachment


Homework Equations


see problem 9 attempt attachment


The Attempt at a Solution


I come up with a solution that is much smaller than any of the choices. I do not know if I am to assume the orbital is circular. I do not know if there is another error besides that in my work.
 

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  • #2
Hello,

You need not assume that the orbit is circular. Use one of Kepler's Laws (which applies for both circular and elliptical orbits).
 
  • #3
ok if I use kepler's third law would I start by using moon II orbital period and radius to solve for the mass of the planet?
 
  • #4
You don't need to solve for the planet's mass. Remember that R^3 is proportional to T^2, and that must hold for all objects orbiting a body. Hence R^3/T^2 = constant.
 
  • #5
ok so I can find that ratio from moon II data and then multiply that ratio by T^2 of moon III?
 
  • #6
yup, that is right
 
  • #7
thank you for your help
 

1. What is the purpose of "Solving Problem 9: Smaller than Any of the Choices?"

The purpose of this problem is to find the smallest number among a given set of choices.

2. How do I approach solving this problem?

One way to approach this problem is by comparing each choice to the others and keeping track of the smallest number. Another approach is to use a loop to iterate through the choices and update the smallest number as you go.

3. Can I use any specific mathematical formulas or concepts to solve this problem?

Yes, you can use the concept of minimum and maximum values to solve this problem. You can also use the concept of arrays and loops to efficiently compare and find the smallest number.

4. What if all the choices are negative numbers?

If all the choices are negative numbers, the approach to solving the problem would be the same. You would just need to make sure to account for negative numbers in your comparisons and updates.

5. Is there a more efficient way to solve this problem?

Yes, you can use a sorting algorithm to sort the choices in ascending order and then simply take the first number as the smallest. This would be more efficient for larger sets of choices, but may not be necessary for smaller sets.

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