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Jupiter is a massive planet. We get it. However, we have also heard that, since it has such a huge radius, at different elevations it is possible to experience different levels of gravity. We hear that, at some point above Jupiter’s surface, the gravity is actually Earth-like.

So, at what elevation would this be?

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  • $\begingroup$ Have you tried to find the formula for surface gravity? It's also easy to derive.. $\endgroup$ Commented Mar 2, 2023 at 19:47
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    $\begingroup$ Jupiter has a high rotation rate, so there's a substantial centrifugal force (in a frame rotating with Jupiter). How do you want to deal with that? $\endgroup$
    – PM 2Ring
    Commented Mar 2, 2023 at 19:51
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    $\begingroup$ $$a = \frac{GM}{r^2}$$ $\endgroup$
    – uhoh
    Commented Mar 2, 2023 at 21:46
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    $\begingroup$ The definition of "surface" might be tricky here given we're not entirely sure what the interior structure of Jupiter is. $\endgroup$ Commented Mar 2, 2023 at 22:33

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According to Newton, the gravitational acceleration of a spherical body is given by $$a = \frac{GM}{r^2}$$ where $GM$ is the standard gravitational parameter and $r$ is the radial distance from the centre of the sphere. Jupiter has high rotation, so it's a bit squashed (it's an oblate spheroid), but the above formula is still a good approximation.

It's not easy to determine the internal structure of a giant planet, so the surface of a giant planet is usually defined to be where the the atmospheric pressure is $1$ bar (which is approximately the air pressure at Earth's surface). As I mentioned here, it appears that Jupiter has a diffuse core, but the details are still hazy.

From the JPL Horizons body data page for Jupiter, it has an equatorial radius of $71492\pm4 \,{\rm km}$. Its gravitational parameter is $126686531.900 \,{\rm km^3/s^2}$.

Using those numbers, the surface acceleration is approximately $24.7865 \,{\rm m/s^2}$, which is consistent with the value given by Horizons. Incidentally, the centrifugal acceleration at that radius is approximately $ 2.21 \,{\rm m/s^2}$.

To find the radius where $a$ equals the standard Earth gravitational acceleration, $g=9.80665 \,{\rm m/s^2}$, we can use $$r = \sqrt{\frac{GM}{g}}$$

That gives $r\approx 113659 \,{\rm km}$, which is roughly $42167 \,{\rm km}$ above the nominal surface. The scale height of Jupiter's atmosphere is ~$27 {\rm km}$, so the atmospheric pressure at that altitude is negligible.


As JamesK mentions, the only way to maintain that altitude without using a rocket continuously burning a lot of fuel is to be in orbit, and of course an orbiting body is in freefall, so it doesn't feel the gravitational acceleration.

The speed of a body in a circular orbit at that altitude is ~$33.4 \,{\rm km/s}$, which is a little faster than Earth's orbital speed around the Sun (~$29 \,{\rm km/s}$). The period of that orbit is almost $6$ hours.

The escape velocity at that altitude is ~$47.2 \,{\rm km/s}$, which is much faster than the escape velocity at Earth's surface, ~$11.2 \,{\rm km/s}$.

Orbit speeds can be calculated from the gravitational potential, which is the gravitational potential energy per unit mass. It's given by $$V = -\frac{GM}{r}$$ The circular orbit speed is $$v_c = \sqrt{\frac{GM}{r}}$$ and escape velocity is $$v_e = \sqrt{\frac{2GM}{r}}$$

The radius where Jupiter's gravitational potential equals the gravitational potential on Earth's surface is a little over $2$ million kilometres.

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    $\begingroup$ It follows from the last sentence that the only way to stay at that height, without burning vast amounts of fuel, is to be in orbit. And if you are in orbit, you are in free-fall and don't perceive gravity (just as astronauts don't perceive gravity in the space station but feel weightless.) $\endgroup$
    – James K
    Commented Mar 3, 2023 at 7:07
  • $\begingroup$ An orbital ring could have the outer bit stationary with fast orbiting particles inside. Someone standing on the ring would feel Jupiter's gravity. $\endgroup$ Commented Dec 4, 2023 at 18:45
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Jupiter's gravity can never be fully Earth-like. We can compute a height at which the acceleration matches Earth, but the associated tidal force would be less.

Above a massive spherical body the magnitude of the tidal force may be rendered by

$da/dr=-2a/r,$

where,$a$ is the gravitational acceleration at a distance $r$ from the center of the body. If we match $a$ at Jupiter with the surface gravitational of Earth using we have (from PM 2Ring's answer)

Earth: $a=9.80655\text {m/s}^2, r=6371\text{ km}$

Jupiter: $a=9.80655\text {m/s}^2, r=113659\text{ km}$

meaning the tidal force above Jupiter is only 5.6% of the force at Earth's surface, if we choose an altitude above Jupiter that matches the gravitational acceleration $a$.

Attempting to match tidal forces would put us inside Jupiter's mass causing the above formula to break down. We may not get a match of the tidal force at all. Jupiter's mass density is too low to find such a mass at any point on or above a posited surface.

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