formula for planet mass according to Kepler's harmonic law

L

lcortina

New member
Joined
Nov 13, 2008
Messages
8
Problem: A result of Kepler’s harmonic law is that the mass of a planet (M) with a satellite is directly proportional to the cube of the mean distance (d) from the satellite to the planet and inversely proportional to the square of the period of revolution (p). Early astronomers estimated the mass of the earth to be kg and observed that the moon orbited the earth with a period of 27.322 days at a mean distance of km.

(1) Write a formula for the mass of a planet according to Kepler.

(2) Find the proportionality constant using the observations and estimates for the earth and moon.

(3) Extend your knowledge across the solar system to find the mass of Mars based on observations that Phobos orbits Mars in 7.65 hours at a mean distance of 9330 km.

(4) What is the approximate ratio of the mass of the earth to the mass of Mars?
 
lcortina said:
Problem: A result of Kepler’s harmonic law is that the mass of a planet (M) with a satellite is directly proportional to the cube of the mean distance (d) from the satellite to the planet and inversely proportional to the square of the period of revolution (p). Early astronomers estimated the mass of the earth to be [you forgot to copy-n-past this value kg and observed that the moon orbited the earth with a period of 27.322 days at a mean distance of km.
Click to expand...
What are your thoughts? After reviewing the section in your book and the pages in your class notes on variation equations, what have you tried? How far did you get? Where are you stuck? For instance:

1) You reviewed the definitions for "directly proportional", "cube", "inversely proportional", and "square"; created a formula incorporating these definitions and the provided relationship, plugged in the given variables, and... then what?

2) You plugged the given values into the formula, leaving only the variable for the constant, and... then what?

3) You plugged these new values into the same formula, leaving only the variable for the mass, and... then what?

4) You created the division and... then what?

Please be complete. Thank you! :D

Eliz.
 
Start answering Stapel's questions - you will be on your way.....
 
lcortina said:
I can't figure out how to write the formula according to kepler.
Click to expand...


Why not?

You have not shown any initiative; so, perhaps, the answer to my question is that you have not even tried.

(Kepler is a proper noun, by the way, because it's the man's name.)

 
lcortina said:
I wasn't even able to start because I did not understand any of it.
Click to expand...
I'm sorry to hear that you have been mistakenly placed in a course for which you are entirely unprepared (otherwise, you would have at least recognized terms, like "kilometers","planet", "variable", and the basic concept of equations, from previous years of coursework). How very frustrating for you! :shock:

Please conference with your academic advisor at your earliest convenience, so that you can be properly evaluated and then be allowed to drop back and re-start your studies with material of which you have at least some passing familiarity -- a basic requirement in math studies, since new material builds on previous concepts and methods. When you demonstrate how inappropriate this present course is to your current abilities, you may be able even to get your money back for this course; it's certainly worth asking.

Our best wishes to you in your future studies! :D

Eliz.
 
stapel said:
I'm sorry to hear that you have been mistakenly placed in a course for which you are entirely unprepared (otherwise, you would have at least recognized terms, like "kilometers","planet", "variable", and the basic concept of equations, from previous years of coursework). How very frustrating for you!

Our best wishes to you in your future studies! :D

Eliz.
Click to expand...

lcortina: Hi. Are you trying to understand how these words "translate" into an equation? You haven't shown any effort here. Whenever a word problem asks you to find an equation, the first thing you need to do is translate what it says in words into a mathematical representation.

For example, if the problem says "the cube of the mean distance," that translates, mathematically, into \(\displaystyle d^3\). So, you just write that in every place in the equation where "cube of the mean distance" occurs in words. Equations are sort of like mathematical "sentences," but instead of translating into French or something, you're translating into mathematical terms.

Can you at least translate a "part" of this? Any part? Everybody here is very nice and willing to help, even if you have no choice but to be in this class, in a school that didn't prepare you adequately for this course? (Or didn't you try to start in those classes either?)

I'm sorry to be so mean, but it's not going to do you any good if I just tell you what the answer is. But I'll give you a hint: You're going to have to understand what a constant of proportionality is (usually represented by k). I'll make you a promise: If you write back with something more than a restatement of the problem and a statement that you don't know how to solve it, I'll help you all I can. I'm sure others will do the same. But if all you ask is "What do I do?" the answer is "Do some research and try to get started." We're all friends here, and if you make a mistake, we'll just tell you where you messed up and we can go from there. No harm in that: it's just a school assignment, not like anyone's life depends on it. But if you don't tackle these problems (and understand what you need to understand), they only get harder as you go through life.
 
(1) Write a formula for the mass of a planet according to Kepler.

(d x p)= M

M= 384.4 x 10^3

This is what i understood.

Letty
 
Problem: A result of Kepler’s harmonic law is that the mass of a planet (M) with a satellite is directly proportional to the cube of the mean distance (d) from the satellite to the planet and inversely proportional to the square of the period of revolution (p).
Click to expand...

\(\displaystyle M = k \cdot \frac{d^3}{p^2}\)

now use your given data to find the proportionality constant \(\displaystyle k\)
 
For those with an interest in finer detail of Kepler’s laws, there is a nice discussion/explanation here:

http://homepages.wmich.edu/~korista/Newton-Kepler.html

“Kepler's 3rd Law: p^2(yrs) = a^3(AU), meaning larger orbits (a) have longer orbital periods (p), and the average orbital speeds are slower for planets with larger orbits.
After deriving gravitational orbital motion from the laws of motion, law of gravity and quite a bit of calculus, Newton found that Kepler's 3rd Law should actually be this: p^2(yrs) = a^3(AU) / (M1 + M2); where the masses are measured in units of our Sun's mass. Newton also showed that this equation may also be expressed in physical units as
p^2 = 4(pi^2)(a^3)/ G(M1 + M2)
for p measured in seconds, a in meters, and M in kg (and pi = 3.141592653...).”

And, of course, there’s always wikipedia:

“Planets distant from the sun have longer orbital periods than close planets. Kepler's third law describes this fact quantitatively.
"The square of the orbital period of a planet is directly proportional to the third power of the semi-major axis of its orbit. Moreover, the constant of proportionality has the same value for all planets."
For example, suppose planet A is four times as far from the sun as planet B. Then planet A must traverse four times the distance of Planet B each orbit, and moreover it turns out that planet A travels at half the speed of planet B. In total it takes 4×2=8 times as long for planet A to travel an orbit, in agreement with the law (8^2=4^3).
Symbolically:

P^2 is proportional to a^3

where P is the orbital period of planet and a is the semi-major axis of the orbit.”

http://en.wikipedia.org/wiki/Kepler%27s ... ary_motion

I mention this also as an example for students who may be trying to understand/research some “famous” equations that are often used in math classes. Even if you don’t understand all the physics, the equations can be easily found. “The truth is out there…”
 
lcortina said:
(1) Write a formula for the mass of a planet according to Kepler.

(d x p)= M

M= 384.4 x 10^3

This is what i understood.

Letty
Click to expand...

Thanks. This is how we can help... You got the "mass is" part; that translates into "M =".

The problem is I don't think you understand the difference between directly and inversely proportional. If one thing is directly proportional to something else, that means when one goes up, the other one goes up too. An example is "The mass of a newspaper is directly proportional to the number of pages in the newspaper." As the number of pages goes up, the mass of the newspaper also goes up, by an amount based on the number of pages. In this tiny example, we don't know how much each page weighs, but we could say, "The mass of the newspaper is some number (the mass of each page) times the number of pages. If we use variables for these things, we can write an equation:

\(\displaystyle m = k \cdot p \quad\) where m = mass of newspaper, k = mass of each page, p = number of pages.

On the other hand, if the problem says something is "inversely" proportional to something else, that means as one goes up, the other goes down. Another totally contrived example of this would be "The price of a pair of jeans is inversely proportional to the number of tears in the material." That is, as the material gets torn a lot more, the price goes down. (That might not be the case for jeans that are torn on purpose, but I hope you get the idea.) Let's say we let t represent the number of tears in a pair of jeans and p represent the price of those jeans. We could write this inverse proportionality like this:

\(\displaystyle p = k \cdot \frac{1}{t} \quad\) where k = an unknown value lost with each imperfection in the material.

I realize you have planets here, but the principle is the same. When you increase the denominator, the value of the whole fraction goes down, and when you increase the numerator, the value of the whole fraction goes up.

In your problem, it says, "the mass of a planet (M) with a satellite is directly proportional to the cube of the mean distance (d) from the satellite to the planet and inversely proportional to the square of the period of revolution (p)".

Since we have "mass ... is", we write M =.

Since we have "directly proportional", we write that quantity in the numerator (the cube of the mean distance): \(\displaystyle d^3\).

Since we have "inversely proportional", we write that quantity in the denominator (the square of p): \(\displaystyle \frac{1}{p^2}\).

Now, we need a constant of proportionality, like the mass of each page in the newspaper. We don't know "how much" the mass increases when distance goes up or when period goes down; we just know that it does increase. (That's the second part of the problem.) So, we just use a temporary variable k to represent the constant of proportionality. We'll solve for it when we can plug values into this formula.

\(\displaystyle M = k \cdot d^3 \cdot \frac{1}{p^2} = k \cdot \frac{d^3}{p^2}\)

Now, part 2 is just a sort of plug-and-chug. Put in the values and find k.

Then write back and remind me of where we stand with the whole proportionality thing.
 
Thank you all for the help. I finally understood. The answer would be

The formula is m ? d^3 / p^2 ? m = k d^3 /p^2, where k = Proportionality constant
k = m p^2 /d^3

I think this is correct.

Letty
 
You must log in or register to reply here.
Top