Kepler's Questions: Saturn's Orbit & Satellite Radius Calculations

  • Thread starter mcdowellmg
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In summary: But I think you have to convert 2.35 hours to seconds, not years. And, since you are using 2.35 hours, you should convert the units for the gravitational constant from N-m^2/kg^2 to N-m^2/s^2. So the mass of the planet should be 3.22*10^24 kg and the radius should be 4.89*10^6 meters.In summary, the first question involves calculating the radius of an orbit around Saturn using the given equations and values. After correcting some calculation errors, the correct answer is 52,732 km. The second question involves finding the radius of a planet using the law of gravitation and given values.
  • #1
mcdowellmg
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Question 1:

Homework Statement



https://www.physicsforums.com/showthread.php?t=44239 except the planet is Saturn.


Homework Equations



[tex]r = (t / (2pi)) ^ {2 / 3} * (G * M) ^ {1 / 3}[/tex]


The Attempt at a Solution



I tried many things, so I thought I would give tony873004's solution attempt from the thread I linked to a shot.

"where r is the radius of your orbit (distance from the center of Saturn), t is the period of your orbit, which is the same as the rotational period of Saturn (38,745 seconds), G is the gravitational constant [tex]6.67 * 10^{-11}[/tex], M is the mass of Saturn ([tex]5.6851 * 10^{26} kg[/tex]).

Subtract from this the radius of Saturn (60,268 km) to get your altitude."

I calculated from all of that 456,708 km, which is apparently wrong.
Am I doing something wrong here with finding the right rotational period or mass of Saturn?



The second question is:

Homework Statement



A 25.0-kg satellite has a circular orbit with a period of 2.35 h and a radius of 7.30×10^6 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 9.10 m/s2, what is the radius of the planet?


The Attempt at a Solution



GMm/r^2 = m*v^2/r
v = 2*pi*r/T
so GM/r^2 = (4*pi^2*r)/T^2

So the mass of the planet M = 4*pi^2*r^3/(G*T^2) = (4*pi^2*(7.30*10^6)^3)/(6.67x10^-11*(2.35*3600)^2)= 3.22*10^24 kg

b) Now Using the law of gravitation we get GMm/R^2 =m*g at the surface we get

GM/R^2 = g So R = sqrt(GM/g) = sqrt(6.67x10^-11*3.22*10^24/9.10) = 4.89*10^6

That is somehow also wrong. Are the formulas set-up correctly?
 
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  • #2
For Question 1 your method looks okay, but your answer is not correct. Probably a calculation slip-up. Try running the numbers again. If you don't make headway, post some partial results that can be checked.

For Question 2, is your result to be specified in meters or in kilometers?
 
  • #3
The second question is to be answered in meters.I tried the first question again, and got 112,732 km, which is much different! Is that more in line with what you were calculating? It is difficult dealing with all of this scientific notation.

Here is how I calculated. [tex]r = (t / (2pi)) ^ {2 / 3} = (38,745 seconds / 6.2832...) = 6166.44... ^ {2/3} = 336.27[/tex]
[tex](G * M) ^ {1 / 3} = (6.67 * 10 ^ {-11} (m ^ {3})/(kgs ^ {2} * 5.6851 * 10^{26} kg = (37919617000000000) ^ {1/3} = 335960.32[/tex]
[tex]336.27 * 335960.32 = 1.13*10 ^ {8}[/tex], which is in meters, so it would be 113,000 km.
[tex]113,000 km - 60,268 km (the radius of Saturn) = 52,732 km.[/tex]

Well, now I have gotten 52,732 km! I swear that I have the worst ability at using calculators/Wolfram Alpha. I need to go back to doing everything manually, if only I could remember how!
 
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  • #4
mcdowellmg said:
The second question is to be answered in meters.


I tried the first question again, and got 112,732 km, which is much different! Is that more in line with what you were calculating? It is difficult dealing with all of this scientific notation.

I obtained a smaller value (very roughly half) for the first question.

For the second question, the value I obtained using the same numbers that you did differed in the second decimal place. So, maybe run through the numbers again.
 
  • #5
I might as well try the calculation again for question #2 on here, step-by-step.

[tex](G*M*m)/(r ^ {2}) = (m*v ^ {2})/r[/tex]
[tex]v = (2*\pi*r)/T[/tex]
Therefore, [tex](G*M)/r ^ {2} = (4*\pi ^ {2}*r)/T ^ {2}[/tex]
So, the mass of the planet, [tex]M=(4*\pi ^ {2}*(7.30*10 ^ {6}) ^ {3})/(6.67*10 ^ {-11}*(2.35 years * 3600 seconds) ^ {2}) = (1.54*10 ^ {22})/0.00477382572 = 3.23*10 ^ {24}[/tex]

Now Using the law of gravitation we get [tex]G*M*m/Radius of planet ^ {2} =m*g[/tex] at the surface we get

[tex]GM/R^2 = g[/tex] So, [tex]R = \sqrt{GM/g}= \sqrt{(6.67*10 ^ {-11}*3.23*10 ^ {24})/9.10)} = 333931335560000000/9.10 = 4.87*10 ^ {6}[/tex] meters

Is that right?
 
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  • #6
The numbers look good this time.
 

Related to Kepler's Questions: Saturn's Orbit & Satellite Radius Calculations

1. What is Kepler's first law?

Kepler's first law states that the orbit of a planet around the sun is an ellipse with the sun at one of the two foci.

2. How is Kepler's second law related to the speed of a planet in its orbit?

Kepler's second law, also known as the law of equal areas, states that a line connecting a planet to the sun sweeps out equal areas in equal time intervals. This means that a planet will travel faster when it is closer to the sun and slower when it is farther away.

3. What is the significance of using Saturn's orbit to study planetary motion?

Saturn's orbit is important in studying planetary motion because it is one of the most stable and predictable planetary orbits in our solar system. This allows scientists to make accurate calculations and predictions based on Kepler's laws.

4. How is the satellite radius of Saturn calculated?

The satellite radius of Saturn is calculated by using the formula a = ∛(GMp²/4π²), where a is the radius of the orbit, G is the gravitational constant, M is the mass of Saturn, and p is the orbital period of the satellite.

5. What are some real-world applications of studying Kepler's laws?

Studying Kepler's laws has many practical applications, including predicting the positions and movements of planets, satellites, and other celestial bodies. It also helps in the design and operation of spacecraft and satellites, as well as in understanding the dynamics of our solar system and other planetary systems in the universe.

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