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

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• sams
In summary: Anyway, I'm not sure how you could use that as the Sun is moving in its own orbit around the barycenter, but I do see how it would work if you calculated T^2/R^3 for two different orbits of the same planet around the sun. Assuming the sun's mass is the same in both cases, you would then have$$\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3}$$and then you could use that combined with Newton's Law to get the expression I used.In summary, the mass of the planet Saturn can be calculated by using Newton's Law of Gravity and the approximation that the distances from the
sams
Gold Member
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:

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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Combine it with Newton's law of gravity and the approximation that the distances from sun equal the semi major axis.

sams said:
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.

sophiecentaur
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.

RPinPA said:
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.

fresh_42 said:
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.

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.

fresh_42 said:
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.

PeroK said:
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.

## 1. How is Kepler's third law used to calculate Saturn's mass?

Kepler's third law states that the square of a planet's orbital period is proportional to the cube of its semi-major axis. By knowing Saturn's orbital period and distance from the sun, we can calculate its mass using this equation: M = 4π²a³/GT², where M is the mass, a is the semi-major axis, G is the gravitational constant, and T is the orbital period.

## 2. What is the value of Saturn's semi-major axis?

Saturn's semi-major axis, or the distance from its center to the sun, is approximately 9.5 astronomical units (AU) or 1.43 x 10^12 meters.

## 3. How do we measure Saturn's orbital period?

Saturn's orbital period can be measured by observing its position in the sky over time. This can be done using telescopes or spacecraft, and the average length of time it takes for Saturn to complete one orbit around the sun is approximately 29.5 Earth years.

## 4. What is the value of the gravitational constant (G) used in the calculation?

The gravitational constant, denoted as G, is a fundamental constant in physics and its value is approximately 6.674 x 10^-11 cubic meters per kilogram per second squared. It is used in the calculation of Saturn's mass from Kepler's third law.

## 5. Can the mass of Saturn be accurately calculated using Kepler's third law?

Yes, Kepler's third law has been proven to accurately calculate the mass of planets, including Saturn. However, other factors such as the presence of other celestial bodies and gravitational interactions may affect the precision of the calculation.

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