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I've recently needed to determine exactly how far Jupiter was from the Sun on April 25th 2018, when a Hubble image of Jupiter was taken.

I'm a beginner in physics and Google searches lead me nowhere. I wouldn’t mind calculating it, but I need data, and a calculation method.

Oh and if anyone also knows which way Jupiter is tilting in 3d space, this is something I'd really need to know, for that same date.

If anyone can help I'd appreciate it. Thank you!

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    $\begingroup$ Do you have a link for that image? Do you know the UTC time that the photo was taken? I can't find a Jupiter image matching that date on hubblesite.org $\endgroup$
    – PM 2Ring
    Commented Sep 4, 2021 at 14:22
  • $\begingroup$ @PM2Ring maybe OP means this image. It wasn't taken in 2018 though (image description says 2016) but it was published on that date. $\endgroup$
    – jng224
    Commented Sep 4, 2021 at 14:36
  • $\begingroup$ @Jonas Maybe. That's the first one I found, too. ;) $\endgroup$
    – PM 2Ring
    Commented Sep 4, 2021 at 14:38

2 Answers 2

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For the position, you can obtain this data (and other solar system data) to high accuracy using the HORIZONS software by NASA.

Use the following settings:

  • Ephemeris Type: VECTORS
  • Target Body: Jupiter [599]
  • Coordinate Origin: Sun (body center) [500@10]

This will generate the position and velocity of Jupiter relative to the Sun at the specified time(s). More settings can be adjusted using Table Settings. The distance is then the magnitude of the position vector.

The orientation of the rotation axes of most major Solar System bodies (as a function of time) can be found here.

On page 7, the orientation of the rotation axis of Jupiter (in ICRF equatorial coordinates; right ascension $\alpha_0$ and declination $\delta_0$) is

$$\alpha_0 = 268.056595 − 0.006499T + 0.000117\sin J_a + 0.000938\sin J_b + 0.001432\sin J_c + 0.000030\sin J_d + 0.002150\sin J_e \\ δ_0 = 64.495303 + 0.002413T + 0.000050 \cos J_a + 0.000404\cos J_b + 0.000617\cos J_c − 0.000013\cos J_d + 0.000926\cos J_e$$

where $$J_a = 99.360714 + 4850.4046T \\ J_b = 175.895369 + 1191.9605T \\ J_c = 300.323162 + 262.5475T \\ J_d = 114.012305 + 6070.2476T \\ J_e = 49.511251 + 64.3000T$$

and $T$ is the duration in Julian centuries since the standard epoch JD 2451545.0 (1 January 2000, 12:00 Temps Dynamique Barycentrique). All angles ($\alpha_0,\delta_0,J_a \text{ to } J_e$) are in degrees.

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  • $\begingroup$ I needed the distance from the sun to Jupiter, and the NASA tool doesn't seem to be able to give that to me, or at least not clearly, it's mostly positions and velocities. But thank you for the tilt information, I'm still finding out what a Julian year is, and how to convert a specific amount of days into a century, but hopefully I can manage! Thank you. $\endgroup$
    – Kieran Le Cam
    Commented Sep 5, 2021 at 1:11
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    $\begingroup$ @KieranLeCam The distance is just $\sqrt{x^2+y^2+z^2}$. To convert between JD and ordinary dates, you can use this tool. A Julian year is exactly 365.25 days of 86400 seconds each. A Julian century is exactly 36525 days of 86400 seconds each. $\endgroup$
    – Vincent Thacker
    Commented Sep 5, 2021 at 1:29
  • $\begingroup$ Thank you Vincent. Very helpful, I found the exact distance for the date I needed, and on my way to calculating the tilt. You guys rock! $\endgroup$
    – Kieran Le Cam
    Commented Sep 5, 2021 at 23:14
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Try Stellarium. It takes a download, but it's free. You can set the location of your telescope to the Sun, and then check the distance to Jupiter.

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  • $\begingroup$ Thank you! It's an approximate distance but usable for now! $\endgroup$
    – Kieran Le Cam
    Commented Sep 5, 2021 at 2:04

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