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How can you estimate Saturn's mass using data from Cassini's final moments in September 2017 (apoapsis on September 12 at 1:27 a.m. EDT Saturn time at a distance of about $1.3*10^6$ km from Saturn, last signal was recorded at 6:32 am EDT September 15?

My ideas: Using Kepler's Laws and using the following information I come to $ M_s = \frac{4π^2 a^3}{G T^2} $ and by plotting for $T$ the difference between the signals and for a the distance I get a value which is much more higher than Saturn's effective Mass. Any hint?

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  • $\begingroup$ I think you are one the right track, but suspect you used the wrong value for $T$. T has to be a period of the orbit. It isn't clear from the info given if the duration in question was a whole period, multiple periods, or just one period. $\endgroup$
    – R. Romero
    Commented Feb 6 at 15:53
  • $\begingroup$ In September 2017, the Cassini1 mission neared its end after 20 years due to a lack of fuel. NASA's spacecraft reached its apoapsis on September 12 at 1:27 a.m. EDT (Saturn time) at a distance of about $1.3*10^6$ km from Saturn. There the probe came to a standstill and then plunged into Saturn. The last signal from Cassini was recorded at 6:32 a.m. EDT on September 15th - shortly after its transmission, the probe presumably burned up in the gas giant's atmosphere. Using the given data, estimate the mass of Saturn and compare it with the literature value. $\endgroup$
    – Enkt Enktson
    Commented Feb 6 at 18:37

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I think I see what they were getting at. The Apoapsis is the farthest part of the craft's orbit. It burns up at the opposite point, the nearest part of it orbit ,approximately half way through its total orbit. The time lapse between those times given is is 3 days, five hours, and 5 minutes or $277,500s$. Double that to get the full period: $555,000s$

$G=6.6743e^{-11}m^3/(kgs^2)$

Assuming a circular orbit:

$a=1.3\cdot 10^9m$

$\frac{4\pi^2a^3}{GT^2}\implies M_s=4.2188\cdot 10^{27} Kg$

Assuming as per Michael Seifert's suggestion $a=(1/2)1.3\cdot 10^9m $ then implies $M_s=5.27\cdot 10^{26}$

https://science.nasa.gov/mission/cassini/grand-finale/grand-finale-orbit-guide/ gives the orbital period for Cassini the last several weeks of its adventure is 6.5 days or $561500$ seconds.

5.683 × 10^26 kg is the mass of Saturn from Google.

$(5.683-42.188)/5.683*100\%=-642.354\%$ : per cent error for circular orbit.

$(5.683-5.27)/5.683*100\%=7.26\%$: per cent error of highly elliptical orbit.

Not as accurate as I'd like.

I could see sources of error, assuming I got the arithmetic right this time.

The period estimate I used comes from assuming a circular orbit so not a highly eccentric one. That alone might account for much of the error as mentioned by Michael Seifert in the comments. I'm also assuming a two body problem. Saturn has dozens of moons that might effect the period. Jupiter might be near enough and massive enough to change things up. Then how long was Cassini experiencing drag before contact was lost? Drag would throw off the period estimate.

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    $\begingroup$ Was it necessary to list 50 digits? And what was the point of converting it to $10^{27.625}$, which is, by the way, not equal to $4.21\times 10^{26}$ but rather $4.21\times 10^{27}$. So you're off by a factor of 12.5, not 25% $\endgroup$
    – AXensen
    Commented Feb 6 at 22:07
  • $\begingroup$ ah, thanks for catchign that. $\endgroup$
    – R. Romero
    Commented Feb 6 at 22:08
  • $\begingroup$ Thanks a lot for the solution, the problem is I can not understand why Cassini estimated a mass 12,5 the official one: it is kind of a big mistake to be made by such advanced technology. What do you think on that? $\endgroup$
    – Enkt Enktson
    Commented Feb 6 at 22:14
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    $\begingroup$ If the probe is really a "plunge orbit", i.e., a very very thin ellipse with periapsis ≈ 0 km, then the semi-major axis a of the orbit is half of the apoapsis. This brings your estimated mass down by a factor of 8, which actually makes it a fairly reasonable estimate given its crudeness. $\endgroup$
    – Michael Seifert
    Commented Feb 6 at 22:26
  • $\begingroup$ Regarding assuming I got the arithmetic right this time. You didn't get the arithmetic right this time; see the comment by @MichaelSeifert. You edited the answer after Michael left his comment, and yet did not incorporate that comment into your answer. The semi-major axis length is the average of the apoapsis and periapsis distances. $\endgroup$
    – David Hammen
    Commented Feb 10 at 11:50

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