At what point above Jupiter is the gravity Earth-like?

Question

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?

Answer 1

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.

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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.

Answer 2

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.