# If Earth was a moon of Jupiter orbiting at the same distance as Metis (Jupiter's innermost moon), then how much lighter would a 100 kg person feel?

Metis is Jupiter's innermost moon, orbiting at a semi-major axis of 128,000 km. If Earth were to magically replace Metis, then how much lighter would a human weighing 100 kg be?

There's no need to ponder over how Earth would interact with other Jovian moons, this is purely a hypothetical.

I assume that tidal forces are the factor at play here, but they're out of my knowledge and that's why I'm asking this question.

• You know the Earth's gravitational strength. Now calculate Jupiter's strength using $G\frac{m_1m_2}{R}$ and add the two. Apr 25 at 11:16

### TL; DR

Around $$8$$ kg.

You can calculate the tidal force using Newton's law of gravitation, $$F = G\frac{m_1m_2}{r^2}$$ where $$G$$ is the universal gravitational constant, $$m_1$$ and $$m_2$$ are the masses of the two bodies, and $$r$$ is the distance between the body centres.

It is often convenient to work with gravitational acceleration. Now $$F=ma$$, so the gravitational acceleration due to a body of mass $$M$$ is simply $$a = \frac{GM}{r^2}$$ As Wikipedia mentions, it's hard to measure $$G$$, but it's easy to obtain quite precise values of $$GM$$ for bodies with satellites. According to JPL Horizons, $$GM$$ for Jupiter is $$126686531.9\,\rm{km^3/s^2}$$.

The tidal force is a differential force. For this scenario with Earth orbiting Jupiter, we need to calculate the force due to Jupiter at the centre of the Earth and subtract that from the force due to Jupiter at the surface of the Earth. So the tidal acceleration at the surface of the Earth is $$a = GM\left(\frac1{(R+\Delta r)^2} - \frac1{R^2}\right)$$ where $$R$$ is the orbital radius ($$128000$$ km), and $$\Delta r$$ is the radius of the Earth ($$6378.1$$ km).

Wikipedia gives an approximation equation, which is ok when $$\Delta r$$ is much smaller than $$R$$: $$a = 2\Delta r GM / R^3$$

We're using units of km, so we need to multiply the results by $$1000$$ to convert the acceleration to $$\rm{m/s^2}$$. And we can divide by $$g = 9.80665\,\rm{m/s^2}$$ to express the result in terms of standard Earth gravity.

Using that formula, I get $$a \approx 0.0785780\,\rm{m/s^2} = 0.0785780\,g$$ Thus a $$100$$ kg person would feel roughly $$7.86$$ kg lighter when Jupiter is directly above or below them.

Using the more accurate equation, $$a = 0.83226 \,\rm{m/s^2} = 0.084867\,g$$ when Jupiter is at the zenith, and $$a = 0.71659 \,\rm{m/s^2} = 0.073072\,g$$ when Jupiter is at the nadir.

These calculations are only approximate. If Earth were in that orbit, its shape would change due to the tidal force.

And of course, it's not safe for humans to be near Jupiter. It's too cold, and the radiation is too intense. Jupiter's radiation belts are huge, and far more energetic than Earth's. See The in-situ exploration of Jupiter’s radiation belts for details.

• I don't normally answer impossible hypothetical questions, but these equations can be used calculate the tidal acceleration on real moons. Apr 25 at 14:29
• I was questioning whether this question will fall under list of hypothetical questions but you wrote a fantastic answer. +1 Apr 26 at 3:16
• Late reply, but if said person was standing on a location where Jupiter is on the horizon (i.e. 90° from where Jupiter is at zenith/nadir), then would their weight increase or still decrease but by half as much as in the original scenario? Apr 28 at 10:50