<h3>TL; DR</h3>
<p>Around <span class="math-container">$8$</span> kg.</p>
<hr />
<p>You can calculate the <a href="https://en.wikipedia.org/wiki/Tidal_force" rel="noreferrer">tidal force</a> using <a href="https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation" rel="noreferrer">Newton's law of gravitation</a>,
<span class="math-container">$$F = G\frac{m_1m_2}{r^2}$$</span>
where <span class="math-container">$G$</span> is the <a href="https://en.wikipedia.org/wiki/Gravitational_constant" rel="noreferrer">universal gravitational constant</a>, <span class="math-container">$m_1$</span> and <span class="math-container">$m_2$</span> are the masses of the two bodies, and <span class="math-container">$r$</span> is the distance between the body centres.</p>
<p>It is often convenient to work with gravitational acceleration. Now <span class="math-container">$F=ma$</span>, so the gravitational acceleration due to a body of mass <span class="math-container">$M$</span> is simply
<span class="math-container">$$a = \frac{GM}{r^2}$$</span>
As Wikipedia mentions, it's hard to measure <span class="math-container">$G$</span>, but it's easy to obtain quite precise values of <span class="math-container">$GM$</span> for bodies with satellites. According to <a href="https://ssd.jpl.nasa.gov/api/horizons.api?format=text&amp;MAKE_EPHEM=NO&amp;COMMAND=599" rel="noreferrer">JPL Horizons</a>, <span class="math-container">$GM$</span> for Jupiter is <span class="math-container">$126686531.9\,\rm{km^3/s^2}$</span>.</p>
<p>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
<span class="math-container">$$a = GM\left(\frac1{(R+\Delta r)^2} - 
\frac1{R^2}\right)$$</span>
where <span class="math-container">$R$</span> is the orbital radius (<span class="math-container">$128000$</span> km), and <span class="math-container">$\Delta r$</span> is the radius of the Earth (<span class="math-container">$6378.1$</span> km).</p>
<p>Wikipedia gives an approximation equation, which is ok when <span class="math-container">$\Delta r$</span> is much smaller than <span class="math-container">$R$</span>:
<span class="math-container">$$a = 2\Delta r GM / R^3$$</span></p>
<p>We're using units of km, so we need to multiply the results by <span class="math-container">$1000$</span> to convert the acceleration to <span class="math-container">$\rm{m/s^2}$</span>. And we can divide by <span class="math-container">$g = 9.80665\,\rm{m/s^2}$</span> to express the result in terms of standard Earth gravity.</p>
<p>Using that formula, I get
<span class="math-container">$$a \approx 0.0785780\,\rm{m/s^2} = 0.0785780\,g$$</span>
Thus a <span class="math-container">$100$</span> kg person would feel roughly <span class="math-container">$7.86$</span> kg lighter when Jupiter is directly above or below them.</p>
<p>Using the more accurate equation,
<span class="math-container">$$a = 0.83226 \,\rm{m/s^2} = 0.084867\,g$$</span>
when Jupiter is at the zenith, and
<span class="math-container">$$a = 0.71659 \,\rm{m/s^2} = 0.073072\,g$$</span>
when Jupiter is at the nadir.</p>
<hr />
<p>These calculations are only approximate. If Earth were in that orbit, its shape would change due to the tidal force.</p>
<p>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 <a href="https://link.springer.com/article/10.1007/s10686-021-09801-0" rel="noreferrer">The in-situ exploration of Jupiter’s radiation belts</a> for details.</p>

CC BY-SA 4.0
2022-04-25T14:29:12.313Z
I don't normally answer impossible hypothetical questions, but these equations can be used calculate the tidal acceleration on real moons.
PM 2Ring
CC BY-SA 4.0
2022-04-26T03:16:38.790Z
I was questioning whether this question will fall under [list of hypothetical questions](https://astronomy.meta.stackexchange.com/questions/786/what-exactly-count-as-a-hypothetical-question) but you wrote a fantastic answer. +1
Nilay Ghosh
CC BY-SA 4.0
2022-04-28T10:50:29.230Z
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?
user177107
CC BY-SA 4.0
2022-04-25T14:16:33.133Z
2022-04-25T14:22:15.030Z
2022-04-25T14:22:15.030Z
PM 2Ring
<p><a href="https://en.wikipedia.org/wiki/Metis_(moon)" rel="nofollow noreferrer">Metis</a> 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?</p>
<p>There's no need to ponder over how Earth would interact with other Jovian moons, this is purely a hypothetical.</p>
<p>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.</p>

CC BY-SA 4.0
2022-04-25T11:16:04.950Z
You know the Earth's gravitational strength.  Now calculate Jupiter's strength using $G\frac{m_1m_2}{R}$  and add the two.
Carl Witthoft
<gravity><jupiter><tidal-forces>
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?
user177107
# 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?

(Asked by: username_0 on 2022-04-25)

[Metis](https://en.wikipedia.org/wiki/Metis_(moon)) 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.

**Tags:** gravity, jupiter, tidal-forces

> **Comments:**
> **username_1**: You know the Earth's gravitational strength.  Now calculate Jupiter's strength using $G\frac{m_1m_2}{R}$  and add the two.  

---

## 1 answers

### Answer by username_2 on 2022-04-25. Score: 8 | Accepted answer

### TL; DR

Around $8$ kg.

---

You can calculate the [tidal force](https://en.wikipedia.org/wiki/Tidal_force) using [Newton's law of gravitation](https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation),
$$F = G\frac{m\_1m\_2}{r^2}$$
where $G$ is the [universal gravitational constant](https://en.wikipedia.org/wiki/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](https://ssd.jpl.nasa.gov/api/horizons.api?format=text&MAKE_EPHEM=NO&COMMAND=599), $GM$ 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](https://link.springer.com/article/10.1007/s10686-021-09801-0) for details.

> **Comments:**
> **username_2**: I don't normally answer impossible hypothetical questions, but these equations can be used calculate the tidal acceleration on real moons.  
> **username_3**: I was questioning whether this question will fall under [list of hypothetical questions](https://astronomy.meta.stackexchange.com/questions/786/what-exactly-count-as-a-hypothetical-question) but you wrote a fantastic answer. +1  
> **username_0**: 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?  

---

