# How strong are the tides raised by Io on Jupiter relative to the ones raised by the Moon on Earth?

There seems to be 2 ways of calculating tidal forces that appear contradictory. Either:

1. By calculating the difference of Io's gravitational acceleration on a point on Jupiter's near side and the gravitational acceleration felt by a point on Jupiter's far side, using the equation:

$$\frac{GM_{io}}{(d-r_{jupiter})^{2}}-\frac{GM_{io}}{(d+r_{jupiter})^{2}}$$

For the Io-Jupiter system, this is $$24.0807\times10^{-6}$$ Newtons, while for the Moon-Earth system it is $$2.2009 \times 10^{-6}$$ Newtons.

So if you were to calculate tidal forces that way, the tides raised on Jupiter by Io would be $$10.94$$ times as strong as the tides raised by the Moon on Earth

However there is also another way I found, which is:

1. By using the derivative of Newton's law of universal gravitation, using the equation:

$$\frac{-2G\times M_{jupiter}\times{m_{io}}}{d^{3}}$$

For Io, this is $$3.015 \times 10^{14}$$ Newtons per meter, whereas for the Moon it is $$1.031 \times 10^{12}$$ Newtons per meter.

So if you were to calculate tidal forces that way, the tides raised on Jupiter by Io would be $$292.5$$ times as strong as the tides raised by the Moon on Earth

So, which one is it? Are the tides raised on Jupiter by Io 10.94 or 292.5 times as strong as the ones raised by the Moon on Earth? My intuition leads me to believe the first answer is correct since it takes the primary's radius into account but I just want a concise answer.

• Your first expression gives quantities in m/s², not Newtons. – antispinwards Jun 15 at 17:41
• @antispinwards If we multiply his $\frac{m}{s}$ by the mass of Io, does it match up with the second calculation? – Carl Witthoft Jun 15 at 17:57
• Hi new user: please provide more information. In particular, identify what d is in both formulas, and provide a reference (preferably with link) to sources of both those formulas. – Carl Witthoft Jun 15 at 17:59

Both expressions are incorrect. The first should be $$\frac{GM_{\text{moon}}}{(R_{\text{moon}}-r_{\text{planet}})^2} - \frac{GM_{\text{moon}}}{{R_{\text{moon}}}^2}\tag{1b}$$ or $$\frac{GM_{\text{moon}}}{{R_{\text{moon}}}^2} - \frac{GM_{\text{moon}}}{(R_{\text{moon}}+r_{\text{planet}})^2}\tag{1a}$$ where $$R_{\text{moon}}$$ is the distance between the center of the planet and the center of the moon. Equation (1a) pertains to the point on the surface of the planet closest to the moon while (1b) pertains to the point on the surface of the planet furthest from the moon.

Note that the results have units of acceleration rather than force. The tidal force on a small mass on the surface of the planet is the product of the tidal acceleration and the mass of that small mass.

Since $$R_{\text{moon}} \gg r_\text{planet}$$ in the case of Io and Jupiter, and also in the case of the Moon and the Earth, the two expressions are nearly equal to one another, and both in turn are nearly equal to $$2\frac{GM_{\text{moon}} r_{\text{planet}}}{{R_{\text{moon}}}^3}\tag{2}$$ This is the correct version of your second expression.

On plugging in the numbers one finds that the tidal acceleration exerted by the Moon at the surface of the Earth is 1.068 μm/s^2 while the tidal acceleration exerted by Io at the surface of Jupiter is 9.014 μm/s^2 -- over 8 times the tidal acceleration caused by the Moon at the surface of the Earth. This does not quite show the extent of the tides raised by Io on Jupiter (or by the Moon on the Earth).

It is important to note that equations (1) and (2) pertain only to two special points, the submoon point and its antipode. The tidal acceleration at the point where the moon is on the horizon is half the value of equation (2) and is directed inward. What would be nice would be a potential model whose gradient results equation (2) and also in the inward acceleration at the horizon. This potential function is, to first order, $$V = \frac{GM_{\text{moon}} {r_{\text{planet}}}^2}{{R_{\text{moon}}}^3}\frac{3\cos\phi^2-1}{2}$$ where $$\phi$$ is the angle between the line segment between the center of the planet and the center of the moon and the line segment between the center of the planet and the point in question on the surface of the planet.

If the planet consisted of a fictitious frictionless fluid, the fluid would form a prolate spheroid whose height at some angle $$phi$$ differs from what would be the planet's moonless spherical shape by $$V(\phi) / g_\text{planet}$$. The height of the tidal bulge is the difference between the maximum at $$\phi=0$$ and the minimum at $$\phi=\pi/2$$, or $$h = \frac32 \frac{M_{\text{moon}}}{M_{\text{planet}}} \frac{{r_{\text{planet}}}^4}{{R_{\text{moon}}}^3}$$

On plugging in the numbers one finds that the tidal bulge created by the Moon on a fictitious frictionless fluid Earth would be 0.5327 meters while the tidal bulge created by Io on a fictitious frictionless fluid Jupiter would be 22.52 meters -- over 42 times the size of the tides raised by the Moon on the Earth.

While neither Jupiter nor the Earth is composed of such a fictitious fluid, the tidal bulge height nonetheless does provide a nice picture of how large the tides that result from a moon can be.

• Your derivation doesn’t agree with one I did from first principles (there’s a factor of 3) so I will have great fun going back and seeing where I was wrong! One point though: I usually say frictionless inertialess fluid and I do think you need the inertialessness to make this 0th-order tide height estimate true. – Martin Kochanski Jun 17 at 4:24
• @MartinKochanski - I included inertialess in my use of fictitious. Or perhaps I should have used fictional. Newton's tidal bulge does not exist. Nonetheless, the calculation does yield results that do show how significant the tides can be. – David Hammen Jun 17 at 4:47