of circular motions of planets) find the mass of Object X.
Calculations
Step 1
Kepler’s Law: M=(v^2 r)/G
The M stands for the mass of the object, so that is what we are working on to find. v^2 Is the velocity of the object through its radius. The value of velocity is 2πr/P, so 2πr determines the distance from both Object X and the moon, so it is divided by P which is the amount it takes to finish one orbit around Planet X. A light body that orbits the object (Planet X’s moon) is signified by r. G is basically 6.673*〖10〗^(-11) m^3 kg^(-1) s^(-2)(the gravitational calculation of force between two bodies) which m is meters, kg is kilograms, and s is seconds. This is in scientific notation and we use it because it lowers the amount of zeros in 6.673 but it is still the same number. The small numbers on the right upper corner is an exponent which we use to simplify the multiplication writing like 5^3 is 5*5*5. When we plug in 2πr into v, the exponent of (2πr/P)2 distributes to the variables (the letters in the equation) becoming (4π^2 r^3 /P^2 ). The equation turns to:
Kepler’s Law: M=(v^2 r)/G →M =((4π^2 r^3 ))/(GP^2 )
Step 2
Now that we have our equation ready to plug in our numbers, first we need to change the units from kilometers to meters for r and days to seconds for P. The evaluations are:
r=57,783km (convert to meters)
G = 6.673*〖10〗^(-11) m^3 kg^(-1) s^(-2) (ready to plug into the equation)
P = 32.17 days (change to seconds)
It is required to convert to smaller units because it gives us bigger numbers, thus giving us accurate results when finding the solution.
57,783km=1000m/km → m = 57,783,000 or 5.778 * 107
One kilometer equals 1000 meters so we have 1000m/km.
Km on both sides cancel out because we are dividing the same unit which leaves 57,783 multiplying to 1000m. This gives us 57,783,000m, but we have to turn it into scientific notation to make calculations easier. There is a decimal point in 57,783,000. and we move it 7 times to the left which automatically becomes 5.778 * 107. The exponent is 7 because we moved the decimal 7 times to the left producing a smaller number and is multiplied by 10 because it is a base (system of decimals). Now we have to convert 32.17 days to seconds.
(32.17days/24hours/ 60minutes /60seconds)/(days/hours/minutes) → s = 2,779,488s or 2.779 * 106
There are twenty four hours in a day, sixty minutes in an hours, and sixty seconds in a minute. We divide to cancel the units leaving us seconds and multiplying all the numbers. The answer in seconds is 2,779,488 seconds, and scientific notation of that is 2.779 * 106. Now that we have our numbers in meters and seconds, we can plug the number in the equation to evaluate the mass of Planet X.
Step 3
M =((4π^2 r^3 ))/(GP^2 ) G = 6.673*〖10〗^(-11) m^3 kg^(-1) s^(-2) r = 5.778*107 P = 2.779 * …show more content…
106
M is the formula we need to find the mass of Planet X and G, r, and P will be utilized in the equation. First we will solve the numerator (the top part of the fraction) so the calculations will not be a disorder.
(4π^2)(r^3) → (4π^2)((5.778 * 107m)3
Everything is plugged in the numerator of the equation and I will now solve for 4π2. The sign π has an exponent of 2, so you multiply π twice. The answer for that is 9.8596, and then you multiply that with 4. That answer is 39.489
(4π^2 )→ 39.489
For r we plug in 5.778 * 107 from our calculations. Our answer for that is 192.900 * 1021.
(5.778 * 107)3 → 192.900 * 1021
Since the exponent 3 is outside the parenthesis, then that exponent is distributed to the whole equation. 5.778 is multiplied 3 times so it becomes 192.900. The 107 becomes 1021 because you are multiplying those two exponents, so 7 * 10 is 21. Now we will multiply our two answers for the numerator, but first have both numbers in scientific notation
39.489 * 192.900 * 1021
We must turn our decimal to scientific notation in order to get an accurate answer.
39.489→3.949 * 101
We rounded our number to the nearest decimal and we moved our decimal to the left resulting in a positive exponent 1. We will have our 192.900 * 1021 as 1.929 * 1023 because it simplifies our scientific notation. Now we are ready to solve the tope portion of the fraction.
3.949 * 101 * 1.929 * 1023 → 7.618 * 1024
We found the top portion of the fraction is 7.618 * 1024 and now we can save that number for later because we will solve the bottom portion next. GP^2 → 6.673 * 10-11m3 kg-1 s-2 * (2.779 * 106s) 2
Distribute the 2 to the whole scientific notation turning it into 7.723 * 1012s2. The negative exponents on G can automatically be turned around as a fraction. The 6.673 * 10-11m3 kg-1 s-2 has an invisible 1 under that number and you can flip it as□(m^3/(6.673*〖10〗^11 kg^1 s^2 )). The m3 stays on top because positive numbers cannot be moved down and the when negative numbers are flipped, they turn positive. Now we multiply and cancel out units because they are multiplying. □(m^3/(6.673*〖10〗^11 kg^1 s^2 ))*7.723*〖10〗^12 s^2 → 51.536 *101 m3 kg-1 =
5.154 * 102 m3 kg-1
We simplified the number after we canceled out and crossed out the exponents and units. Now we can solve the whole equation.
Step 4 (7.618*〖10〗^24 m^3)/(5.154*〖10〗^2 m^3 kg^(-1) ) = (7.618*〖10〗^24)/(5.154*〖10〗^2 kg^(-1) ) →1.478 * 1022 kg
This answer is the mass of Planet X.
We still have to see if Planet X’s mass is enough to be considered a planet, so we have to divide the mass of planet x and the mass of earth which is 5.980 * 1024.
(1.478*〖10〗^22 kg)/(5.980*〖10〗^24 ) = .2472 * 10-2 kg
We divide the numbers and we get .2472 * 10-2. Now we simplify the scientific notation so we end with:
2.472 * 10-3
We multiply this scientific notation to 100 so we can find a percentage of the mass. The Scientific notation will convert to an actual decimal in order to find an accurate percentage of the mass of Planet X.
.002472 * 100 =
.247%
Conclusion
The mass of Planet X is .247% meaning it is less than 1% of mass. This is an unusually small number for a planet. Planet X has similar qualities to a planet, but its mass is too small. Even if Planet X is not a planet, there are many undiscovered planets waiting to be observed. It may seem a waste of time, but we learn new things through every calculation we make for every existent body in the galaxy.
Reference "International Astronomical Union | IAU." International Astronomical Union | IAU. Web. 21 Sept.
2015.