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Keplers Third Law Essay

Calculation: Use Kepler’s Third Law (M=(4Tt^2 r^3)/(GP^2 )) to calculate the mass of “Object X” from the orbit of its moon- whose orbital radius (r) is 64,780 kilometers and whose orbital period (P) is 38. 200 days-and knowing that the gravitational constant G= 6. 673? [10] ^(-11) m^3 [kg] ^(-1) s^(-2)(where m= meters, kg= kilograms, and s=seconds). Express the answer in units of Earth Mass (M_Earth), knowing that M_Earth= 5. 980x [10] ^24 kg. (Hint: first express r and p in standard units: meters and seconds. ) Given: M=(4Tt^2 r^3)/(GP^2 ) is known as Kepler’s Third Law where P is the orbital period, r is he orbital radius, and G is the gravitational constant.

When given this type of problem, it is important to first understand what it is asking. Identifying what it is asking helps prevent doing more work than necessary and helps in getting the correct answer. First, the question asks us to calculate the mass of Object X. The question also asks to express the answer in units of Earth mass (M_Earth). Therefore, we will have two answers.

Next we identify the variables given in the question. The variables given are: 64,780 kilometers (km) which is the orbital radius (r), 38. 200 days which is the orbital radius, and . 673x [10] ^(-11) m^3 [kg] ^(-1) s^(-2) which is the gravitational constant. Now that we know the variables given, we can identify if any of the given variables need to be converted into proper units. In order to do this, we look at the units of the gravitational constant since it is a natural law and the units are standard. A natural law simply means that it is constant throughout time, meaning the gravitational constant will still be 6. 673x [10] ^(-11) m^3 [kg] ^(-1) s^(-2) in two thousand years into the future or past. Furthermore, by comparing the unit of the gravitational onstant to the variables given, we see that we must convert kilometers to meters and days into seconds.

First lets convert days to seconds (you can start with converting kilometers into meters first as well, the order does not matter at this point). Remember to always write down you units in order to keep take of them. Step 1 38. 200daysx((24 hours)/(1 day))? ((60 minutes)/(1 hour))? ((60 seconds)/(1 minute))=3300480 sec The whole point of the above calculation was to get rid of the days and convert them into seconds. That is why we multiplied by 24 hours over days, since there is 24 hours in one day (a similar process was used to onvert hours into minutes and minutes into seconds.

It is best to leave the answer as 3300480 sec, since rounding will cause the final answer to be less accurate. Now we can convert kilometers to meters. Step 2 64,780 kmx(1000 m)/(1 km)=64780000 or [6. 478? 10] ^7 m This times the answer can be in exponential form in order to make keeping track of numbers easier which will help in arriving to the correct answer. It is important to remember that the exponent on the 10 should not be a negative in this case. Having a negative 7 instead of a positive 7 will, will make the value smaller.

For instance if the exponent was negative 7, this means the number would actually be 0. 000000648 instead of 64780000. Also the reason that the exponent is 7, is because the decimal place was moved to the left sever times. Now that we have the proper units, we can plug in the values we obtained (the values that have a box around them) into Kepler’s Third Law, which again is M=(4Tt^2 r^3)/(GP^2 ). We know that 3300480 sec should be the P and that [6. 478? 10] ^7 m is the r for the equation since it was said earlier in the question.

When all the variables are inserted into the equation, we should get he following: Step 3 M=(4TT^2 ( [6. 478? 10] ^7 m)^3)/((6. 673? [10 ^(-11) m^3 [kg] ^(-1) s^(-2) ) (3300480 sec)^2) Again, it is important to write down all the units so that you know what units can cancel out. From the calculations above (Step 3), we notice that in (3300480 sec)^2, the seconds can be turned in (sec) ^2 when distributing the 2. By doing the same process, we can also turn the meters in ( [6. 478? 10] ^7 m)^3 into m^3. By distributing the exponents, we can cancel out units.

Therefore, the following should be obtained: Step 4 M=(4TT^2 ( [6. 478? 10] ^7 )^3)/((6. 673? [10] ^(-11) (kg] ^(-1)) 3300480)^2 ) We can now begin to solve for M. However, plugging this into the calculator or doing it by hand will be easier if you break Step 4 into separate parts. In other words, solve for the numerator first, then solve the denominator and finally put the two values together to solve for M. When solving for the numerator, start off by distributing the exponent 3 to [6. 478×10] ^7. The value obtained by this distribution should be 2. 6830x [10] ^24. We can then multiple 2. 6830? [10) ^24 with T^2, then you can multiply by 4. The number obtained after multiplying should be [1. 073205? 10] ^25.

The reason for multiplying in this order is due to the order of operations rule. First you multiply what is in the parenthesis, then multiply your exponents and then multiplying the rest of the numbers. The order used to multiply also helps in keeping track of your values and allows for more accuracy. (Note: you can simplify your answer to [1. 073? 10] ^25. However, having more numbers will allow for a more accurate final answer). When solving for the denominator, we should start off by multiplying 3300480 with the exponent 2. By multiplying the exponent, you should obtain [1. 0893? 10] ^13.

We can then multiple [1. 0893×10] ^13 with 6. 673x [10] ^(-11) [kg) ^(-1) (remember to also distribute the units). You should then get a value of 726. 901116 kg. The same order of operations used for solving the numerator, is used here as well. Now that we have solved for both numerator and denominator, we can now plug the values we obtained (the circled values) back into the equation which should give the following: M=(1. 073205? [10] ^25)/(726. 901116 kg) M= [1. 476×10] ^22 kg The answer can now be rounded to the thousands place since the gravitational constant number was also given in the housands place-this is also allows consistency within a problem.

Now, that we have solved for the first part of the question, we solve for the second part of the question. In order to express the answer in units of Earth mass (M_Earth), we must divide our answer with the given M_Earthwhich is 5. 980x [10] ^24 kg. Therefore, the following should be obtained: M_Earth=(1. 476x [10] ^22 kg)/ (5. 980x [10] ^24 kg) M_Earth=2. 468? [10] ^(-3) or 0. 25% Notice that the units cancel out. Furthermore, the answer can be expressed exponentially or in decimal form (depending on what your instructor specifically wants).

In order to convert the decimal for into a percentage, you would multiply by 100. Also, notice that the decimal form is rounded to the tenths place (this is your choice or the choice of your instructor). Congratulations, you have now solved the entire problem! A final note: throughout the problem I have assumed you understood how to distribute exponents. However, if you did not understand how to distribute exponents see table 1. 1 for further explanation. Finally, while it is important to find the correct solution to a problem, it is also important to take a minute and consider what you have done.

For instance, you have just calculate the mass of Object X using three values: the gravitational constant, the orbital radius and the orbital period. That’s pretty amazing, especially considering the fact that there aren’t many ways to calculate the mass of an object that is billions of light years away. For example, you cannot simply find a balance or electric scale and find the mass. Furthermore, knowing something, such as the mass, about an object in outer space that we don’t fully understand is in itself quite amazing. So, do consider what you have done, instead of just blindly working through the problem.

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