How to calculate Saturn's mass from Kepler's third law?

  • #1
sams
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This is not a homework. In Chapter 8: Central-Force Motion, in the Classical Dynamics of Particles and Systems book by Thornton and Marion, Fifth Edition, page 325, Problem 8-19, we are asked to calculate the mass of the planet Saturn. In the instructor's solution manual, the solution for this problem is:

Capture.PNG


ms is the mass of Saturn.
me is the mass of Earth.

I have tried to write ms and me in terms of the semimajor axis and the period and then dividing ms by me, but I did not obtain the above result!

Could anyone explain how could we obtain this ratio? Any help is much appreciated. Many thanks!
 

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  • #2
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Combine it with Newton's law of gravity and the approximation that the distances from sun equal the semi major axis.
 
  • #3
PeroK
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This is not a homework. In Chapter 8: Central-Force Motion, in the Classical Dynamics of Particles and Systems book by Thornton and Marion, Fifth Edition, page 325, Problem 8-19, we are asked to calculate the mass of the planet Saturn. In the instructor's solution manual, the solution for this problem is:

View attachment 234245

ms is the mass of Saturn.
me is the mass of Earth.

I have tried to write ms and me in terms of the semimajor axis and the period and then dividing ms by me, but I did not obtain the above result!

Could anyone explain how could we obtain this ratio? Any help is much appreciated. Many thanks!
The orbits of the Earth and Saturn depend on the Sun's mass, not on the masses of the planets. You could calculate the mass of Saturn from the orbits of its moons.
 
  • #4
RPinPA
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As @PeroK says, this must have involved information about their respective moons. Newton's version of Kepler's Third Law shows that the relationship between ##T^2## and ##R^3## depends on the mass of the central body, and not on the mass of the orbiting body (under the usual approximation that the central body is much heavier than the orbiting body).

Derivation: Centripetal acceleration ##v^2/R = GM/R^2## where ##v## = orbital speed = ##2\pi R/T##.

So ##4\pi^2 R/T^2 = GM/R^2 \iff 4\pi^2R^3 = GMT^2 \iff R^3/T^2 = GM/4\pi^2 ##.

The ratio ##R^3/T^2## is proportional to ##M##, the mass of the central body. If we take that ratio for a moon of Saturn in any units and divide by the same ratio for Earth's moon, the result should be equal to the ratio of their masses.

Earth's moon: T = 27.32 days = 2.36 x ##10^6## s. R = 3.85 x ##10^5## km. In those units, R^3/T^2 = 10250.
Saturn's moon Titan: T = 15.94 days = 1.38 x ##10^6##s. R = 1.22 x ##10^6## km. In those units, R^3/T^2 = 953500.
The ratio between those is 953500/10250 = 93.0

Not sure why I didn't get exactly the same number. Something to do with my rounding perhaps.
 
  • #5
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As @PeroK says, this must have involved information about their respective moons. Newton's version of Kepler's Third Law shows that the relationship between ##T^2## and ##R^3## depends on the mass of the central body, and not on the mass of the orbiting body (under the usual approximation that the central body is much heavier than the orbiting body).

Derivation: Centripetal acceleration ##v^2/R = GM/R^2## where ##v## = orbital speed = ##2\pi R/T##.

So ##4\pi^2 R/T^2 = GM/R^2 \iff 4\pi^2R^3 = GMT^2 \iff R^3/T^2 = GM/4\pi^2 ##.

The ratio ##R^3/T^2## is proportional to ##M##, the mass of the central body. If we take that ratio for a moon of Saturn in any units and divide by the same ratio for Earth's moon, the result should be equal to the ratio of their masses.

Earth's moon: T = 27.32 days = 2.36 x ##10^6## s. R = 3.85 x ##10^5## km. In those units, R^3/T^2 = 10250.
Saturn's moon Titan: T = 15.94 days = 1.38 x ##10^6##s. R = 1.22 x ##10^6## km. In those units, R^3/T^2 = 953500.
The ratio between those is 953500/10250 = 93.0

Not sure why I didn't get exactly the same number. Something to do with my rounding perhaps.
I assume, that the moons are irrelevant here. We can take the pairs (Earth, Sun) and (Saturn, Sun), their orbit periods and their distances approximated once in Newton's law as radius and once in Kepler's law as semi-major axis. Thus you get the quotient ##\frac{m_s}{m_e}## as the sun's mass cancels out.
 
  • #6
RPinPA
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I assume, that the moons are irrelevant here. We can take the pairs (Earth, Sun) and (Saturn, Sun), their orbit periods and their distances approximated once in Newton's law as radius and once in Kepler's law as semi-major axis. Thus you get the quotient ##\frac{m_s}{m_e}## as the sun's mass cancels out.
I'm missing something here. If you are deriving Kepler's Third Law for the planet around the sun, then the derivation for Kepler's Third Law I showed starting from Newton's Law and the expression for centripetal acceleration does not involve the mass of the planet on either side of the equation. In other words, Kepler's Third Law (under the approximation that mass of sun >> mass of planet) tells you that any planet in earth's orbit will have the same period, regardless of mass.

So I'm not quite clear on your description of the derivation that says orbital period depends on planetary mass.
 
  • #7
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I haven't read the details, but as I looked up which of Kepler's laws says which, I found
$$
\left( \frac{T_1}{T_2} \right)^2 = \left(\frac{a_1}{a_2} \right)^3 \frac{M+m_2}{M+m_1}
$$
with the reference to Newton on Wikipedia. I first thought the sun's mass cancels out and didn't realize the small ones would do as well. But this formula was the only way for me to make sense of the hint in your book.
 
  • #8
PeroK
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I haven't read the details, but as I looked up which of Kepler's laws says which, I found
$$
\left( \frac{T_1}{T_2} \right)^2 = \left(\frac{a_1}{a_2} \right)^3 \frac{M+m_2}{M+m_1}
$$
with the reference to Newton on Wikipedia. I first thought the sun's mass cancels out and didn't realize the small ones would do as well. But this formula was the only way for me to make sense of the hint in your book.
I'm not sure how you could use that as the Sun is moving in response to Jupiter's gravity as well. So, trying to include the Sun's motion in a two-body problem with the Earth doesn't look feasible to me.
 
  • #9
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I'm not sure how you could use that as the Sun is moving in response to Jupiter's gravity as well. So, trying to include the Sun's motion in a two-body problem with the Earth doesn't look feasible to me.
You are right, I underestimated the problem. I saw a version of Kepler's 3rd with masses and thought this would do. I don't see another way to combine time and period in a way such that masses are involved. The detour to the sun also looked like the only common factor which combines Saturn and Earth. I tried to make sense out of the hint.
 

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