5
$\begingroup$

Watched this video: https://fb.watch/1U0vCFBr0L/

Ok, so the distances are huge. And the sizes are much different. But I wonder how then the sun keeps the planets etc. orbiting it. If we swap out all those objects in the video (soccer ball for the sun, grapes, pin heads, etc.) for equivalently sized balls of iron, equally dense objects, they would have a negligible pull on each other.

A ball of iron the size of a football could never keep a grape 4 football fields away orbiting it. Or could it? Is gravity proportionate? At larger star and planet sizes, it pulls harder?

What is the explanation here?

$\endgroup$
  • 3
    $\begingroup$ I think this is a great question, basically "If we kept the average densities of solar system bodies the same but scaled diameters and distances down until the Sun were the size of a soccer ball, would the planets still orbit? If so, how would orbital periods change relative to the spatial scaling?" $\endgroup$ – uhoh Nov 21 at 3:23
  • 1
    $\begingroup$ I think this can have a much better answer, please consider not accepting anything for a few days and waiting to see what other answers are posted. I'll add one within 24 hours myself if it's not properly answered quickly by someone else (which I expect can easily happen!) $\endgroup$ – uhoh Nov 21 at 3:47
  • 1
    $\begingroup$ For a 50kg football-sized piece of iron, you're looking at an orbital period of a little under 30 years for a grape in a circular orbit 400m away, in the Newtonian Physics ideal situation. (Both objects perfectly spherically symmetric, only force in the universe is their mutual gravitation.) $\endgroup$ – notovny Nov 21 at 6:14
  • $\begingroup$ @notovny If we maintain density and scale all linear dimensions, the period doesn't change at all. $\endgroup$ – uhoh Nov 21 at 13:38
3
$\begingroup$

Many questions can be answered using the vis-viva equation:

$$v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right)$$

which gives the velocity of an object in a Keplerian orbit at distance $r$ from a body of mass $M$ and with a semi-major axis $a$. $G$ is the gravitational constant. And for convenience and accuracy, the product $GM$ or standard gravitational parameter for the Sun and for the Earth are 1.327×1020 and 3.986×1014 m3/s2.

For circular orbits set $r=a$ and get

$$v^2 = GM/a.$$

The orbit's circumference $C=2\pi a$ and the time for one orbit (period) is $T=C/v = C=2\pi a / v$ so

$$T^2 = 4 \pi^2 \frac{a^3}{GM}$$

The mass of a sphere is

$$M = \frac{4}{3} \pi R^3 \rho$$

and we're going to keep the density of the Sun fixed at 1408 kg/m3 which is only 41% higher than water! (see At what depth below the Sun's surface does the density reach that of water?) So:

$$T^2 = 4 \pi^2 \frac{a^3}{GM}$$

$$T^2 = \frac{3 \pi}{G \rho} \left(\frac{a}{R}\right)^3$$

or

$$T = \sqrt{\frac{3 \pi}{G \rho}} \ \ \left(\frac{a}{R}\right)^{3/2}$$


punch line: So the period is going to be one year, i.e. about 365 days whether we use the current values for $a$ and $R$ or scale them up or down by any factor!!

In other words, while:

in plain English, in fact a sun-density sized soccer ball will keep a grape-sized object (of the same density as the planet it represents) in the same scale orbit with the same orbital period. This does in fact all scale down.

...is nearly correct. If the Sun were a 22 cm diameter ball with the same 1.4 g/cm^3 average density, and the sesame seed-sized Earth was 47.4 meters away with a diameter of 2 millimeters and the same average density of 5.5 g/cm^2, then it would orbit the soccer ball-sized Sun once a year, unless there were external forces pulling on it from other astronomical object.

Alternatively you could keep the Sun and Earth and all the planets the same sizes and distances but make them a hundred times less dense, and the orbital periods would be $\sqrt{\text{100}} = $ 10 times longer.


This is actually variant of the rule of thumb that the period of a low orbit around a spherical body is just inversely related to the square root of density. So a dust particle orbiting a 1 meter diameter spherical chunk of "average Earth" will orbit in about 90 minutes just like the ISS orbits around the whole Earth in about 90 minutes.

But you can always replace a spherically symmetric mass distribution with a smaller spherically symmetric mass distribution, (even a point).

Not the same, but similar to what's discussed in this answer to Delta-V required for lift-off from a planet/ asteroid

| improve this answer | |
$\endgroup$
  • $\begingroup$ So are you saying that, in plain English, in fact a sun-density sized soccer ball will keep a grape-sized object (of the same density as the planet it represents) in the same scale orbit with the same orbital period? That this does in fact all scale down? $\endgroup$ – richard Nov 23 at 2:13
  • $\begingroup$ @richard yes indeed it does! I've added a section near the end called "punch line". There are no surprises or sudden losses of stability. Gravity works the same at all scales. The only thing you need to worry about is if the soccer ball and sesame seed-sized Earth get too close to a normal size planet, tidal forces would disturb them. $\endgroup$ – uhoh Nov 23 at 3:16
  • 1
    $\begingroup$ @uhoh This is one of the best answers I've ever seen on astronomy stack exchange. Outstanding! Kepler's 3rd law makes a subtle appearance in your last equation, in which you cleverly dispose of mass. Getting rid of mass in that equation demonstrates scale by density, distance, and SMA. Do you think there might actually be some systems like this in the asteroid belt or other parts of our solar system? My intuition says that gravity differentials from massive distant bodies don't vary much inside such smaller orbits, but reality has an ugly tendency to differ from my intuition. $\endgroup$ – Connor Garcia Nov 23 at 5:56
  • $\begingroup$ @ConnorGarcia we can admire gravity together! There are certainly orbiting non-contact binary asteroids known through light curves and a few by Doppler radar (range-rate), and I think I have a few posts about them, but they're much bigger than this. However there must be theoretical estimates about the expected frequency of small systems like this. I think that that would make for an excellent new question! "What's known or predicted..." Go for it! $\endgroup$ – uhoh Nov 23 at 6:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.